RS Aggarwal Solutions for Class 12 Chapter 27 Straight Line in Space

Question 1. A line passes through the point (3, 4, 5) and is parallel to the vector \( 2\hat{i} + 2\hat{j} - 3\hat{k} \). Find the equations of the line in the vector as well as Cartesian forms.
Answer: The line passes through point (3, 4, 5) with direction vector \( 2\hat{i} + 2\hat{j} - 3\hat{k} \).

Vector form:
\( \vec{r} = 3\hat{i} + 4\hat{j} + 5\hat{k} + \lambda(2\hat{i} + 2\hat{j} - 3\hat{k}) \)

Cartesian form:
\( \frac{x-3}{2} = \frac{y-4}{2} = \frac{z-5}{-3} \)
In simple words: The line equation tells us where every point on the line is located. The vector form uses a starting point plus a direction multiplied by a parameter. The Cartesian form gives us three fractions that are all equal.

Exam Tip: Always match the coordinates of the given point with the constant terms and the components of the direction vector with the denominators in Cartesian form.

Question 2. A line passes through the point (2, 1, -3) and is parallel to the vector \( \hat{i} - 2\hat{j} + 3\hat{k} \). Find the equations of the line in vector and Cartesian forms.
Answer: The line goes through point (2, 1, -3) with direction vector \( \hat{i} - 2\hat{j} + 3\hat{k} \).

Vector form:
\( \vec{r} = 2\hat{i} + \hat{j} - 3\hat{k} + \lambda(\hat{i} - 2\hat{j} + 3\hat{k}) \)

Cartesian form:
\( \frac{x-2}{1} = \frac{y-1}{-2} = \frac{z+3}{3} \)
In simple words: We start at the point (2, 1, -3) and move along the direction given by the vector. Both forms describe the same line, just written differently.

Exam Tip: Pay careful attention to signs when substituting negative direction ratios into the Cartesian form.

Question 3. Find the vector equation of the line passing through the point with position vector \( 2\hat{i} + \hat{j} - 5\hat{k} \) and parallel to the vector \( \hat{i} + 3\hat{j} - \hat{k} \). Deduce the Cartesian equations of the line.
Answer: The line passes through the point with position vector \( 2\hat{i} + \hat{j} - 5\hat{k} \) and has direction \( \hat{i} + 3\hat{j} - \hat{k} \).

Vector form:
\( \vec{r} = 2\hat{i} + \hat{j} - 5\hat{k} + \lambda(\hat{i} + 3\hat{j} - \hat{k}) \)

Cartesian form:
\( \frac{x-2}{1} = \frac{y-1}{3} = \frac{z+5}{-1} \)
In simple words: A position vector shows where a point is located from the origin. We use that point as our starting position and add the direction vector scaled by a parameter to get all points on the line.

Exam Tip: Convert position vector notation \( \hat{i} + \hat{j} + \hat{k} \) directly to coordinates (1, 1, 1) when writing Cartesian form.

Question 4. A line is drawn in the direction of \( \hat{i} + \hat{j} - 2\hat{k} \) and it passes through a point with position vector \( 2\hat{i} - \hat{j} - 4\hat{k} \). Find the equations of the line in the vector as well as Cartesian forms.
Answer: Since the line is drawn in the direction of \( \hat{i} + \hat{j} - 2\hat{k} \), it is parallel to that direction vector. The line passes through the point with position vector \( 2\hat{i} - \hat{j} - 4\hat{k} \).

Vector form:
\( \vec{r} = 2\hat{i} - \hat{j} - 4\hat{k} + \lambda(\hat{i} + \hat{j} - 2\hat{k}) \)

Cartesian form:
\( \frac{x-2}{1} = \frac{y+1}{1} = \frac{z+4}{-2} \)
In simple words: A line drawn "in the direction of" a vector means that vector shows which way the line goes. The position vector tells us a point the line must pass through.

Exam Tip: When the position vector has negative components, convert correctly to coordinates before writing the Cartesian form denominators.

Question 5. The Cartesian equations of a line are \( \frac{x-3}{2} = \frac{y+2}{-5} = \frac{z-6}{4} \). Find the vector equation of the line.
Answer: From the Cartesian equation, we identify the point (3, -2, 6) on the line and direction ratios 2 : -5 : 4.

Vector form:
\( \vec{r} = 3\hat{i} - 2\hat{j} + 6\hat{k} + \lambda(2\hat{i} - 5\hat{j} + 4\hat{k}) \)
In simple words: The Cartesian form has the point coordinates in the numerators and direction ratios in the denominators. Extract these and construct the vector form by putting the point first, then adding the direction vector with a parameter.

Exam Tip: Always verify that the point you extract from the Cartesian form actually satisfies the equation by substituting coordinates.

Question 6. The Cartesian equations of a line are \( 3x + 1 = 6y - 2 = 1 - z \). Find the fixed point through which it passes, its direction ratios and also its vector equation.
Answer: We can rewrite the Cartesian form as \( \frac{x+\frac{1}{3}}{2} = \frac{y-\frac{1}{3}}{1} = \frac{z-1}{-6} \), which simplifies to \( \frac{x+\frac{1}{3}}{2} = \frac{y-\frac{1}{3}}{1} = \frac{z-1}{-6} \).

The fixed point is \( (-\frac{1}{3}, \frac{1}{3}, 1) \) and the direction ratios are (2, 1, -6).

Vector form:
\( \vec{r} = -\frac{1}{3}\hat{i} + \frac{1}{3}\hat{j} + \hat{k} + \lambda(2\hat{i} + \hat{j} - 6\hat{k}) \)
In simple words: When a line is given as a string of equal expressions like \( 3x + 1 = 6y - 2 = 1 - z \), we must rewrite it in standard form by dividing each part to make the numerator and denominator clear.

Exam Tip: Always convert non-standard Cartesian forms to standard form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \) first for clarity.

Question 7. Find the Cartesian equations of the line which passes through the point (1, 3, -2) and is parallel to the line given by \( \frac{x+\frac{1}{3}}{1} = \frac{y-\frac{1}{3}}{1} = \frac{z-1}{-6} \). Also, find the vector form of the equations so obtained.
Answer: Since the new line is parallel to the given line, it has the same direction ratios, which are 2 : 1 : -6. The new line passes through (1, 3, -2).

Cartesian form:
\( \frac{x-1}{2} = \frac{y-3}{1} = \frac{z+2}{-6} \)

Vector form:
\( \vec{r} = \hat{i} + 3\hat{j} - 2\hat{k} + \lambda(2\hat{i} + \hat{j} - 6\hat{k}) \)
In simple words: Two parallel lines have identical direction ratios. So we take the direction of the given line and apply it to our new point to get the equations.

Exam Tip: For parallel lines, the direction ratios are the same - only the point through which they pass differs.

Question 8. Find the equations of the line passing through the point (1, -2, 3) and parallel to the line \( \frac{x-3}{3} = \frac{y+5}{5} = \frac{z-6}{-6} \). Also find the vector form of this equation so obtained.
Answer: The given line has direction ratios 3 : 5 : -6. Our new line passes through (1, -2, 3) and is parallel to this line, so it also has direction ratios 3 : 5 : -6.

Cartesian form:
\( \frac{x-1}{3} = \frac{y+2}{5} = \frac{z-3}{-6} \)

Vector form:
\( \vec{r} = \hat{i} - 2\hat{j} + 3\hat{k} + \lambda(3\hat{i} + 5\hat{j} - 6\hat{k}) \)
In simple words: Read the direction ratios from the denominators of the given line's Cartesian form. Use those same ratios for our parallel line, but change the numerator constants to match our new point.

Exam Tip: Identify direction ratios directly from the Cartesian form denominators, then apply them to your given point without changing them.

Question 9. Find the Cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line \( \frac{-x-2}{1} = \frac{y+3}{7} = \frac{2z-6}{3} \).
Answer: The given line can be rewritten as \( \frac{x+2}{-1} = \frac{y+3}{7} = \frac{z-3}{\frac{3}{2}} \), which gives direction ratios -2 : 14 : 3. Our new line passes through (1, 2, 3) and has the same direction ratios.

Vector form:
\( \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(-2\hat{i} + 14\hat{j} + 3\hat{k}) \)

Cartesian form:
\( \frac{x-1}{-2} = \frac{y-2}{14} = \frac{z-3}{3} \)
In simple words: Sometimes the Cartesian form is written in non-standard form. Rewrite it to standard form, extract direction ratios, then apply them to your new point.

Exam Tip: Always rewrite non-standard Cartesian forms carefully - multiply or divide as needed to isolate individual variables in standard form.

Question 10. Find the equations of the line passing through the point (-1, 3, -2) and perpendicular to each of the lines \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \) and \( \frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5} \).
Answer: The two given lines have direction ratios 1 : 2 : 3 and -3 : 2 : 5. Let the required line have direction ratios \( b_1 : b_2 : b_3 \). Since the required line is perpendicular to both given lines, we have:

\( b_1 + 2b_2 + 3b_3 = 0 \)
\( -3b_1 + 2b_2 + 5b_3 = 0 \)

Solving using determinants:
\( \frac{b_1}{4} = \frac{b_2}{-14} = \frac{b_3}{8} \)

So direction ratios are 2 : -7 : 4. The line passes through (-1, 3, -2).

Vector form:
\( \vec{r} = -\hat{i} + 3\hat{j} - 2\hat{k} + \lambda(2\hat{i} - 7\hat{j} + 4\hat{k}) \)

Cartesian form:
\( \frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4} \)
In simple words: To find a line perpendicular to two given lines, use the dot product condition (a dot product equals zero for perpendicular vectors). Solve the resulting system of equations to find the direction of your new line.

Exam Tip: The perpendicularity condition \( a_1b_1 + a_2b_2 + a_3b_3 = 0 \) gives you two linear equations in three unknowns - you can express the ratios using determinants.

Question 11. Find the Cartesian and vector equations of the line passing through the point (1, 2, -4) and perpendicular to each of the lines \( \frac{x-8}{8} = \frac{y+19}{-16} = \frac{z-10}{7} \) and \( \frac{x-15}{3} = \frac{y+29}{8} = \frac{z-5}{-5} \).
Answer: The two given lines have direction ratios 8 : -16 : 7 and 3 : 8 : -5. Let the required line have direction ratios \( b_1 : b_2 : b_3 \). Since the required line is perpendicular to both:

\( 8b_1 - 16b_2 + 7b_3 = 0 \)
\( 3b_1 + 8b_2 - 5b_3 = 0 \)

Solving using determinants:
\( \frac{b_1}{24} = \frac{b_2}{61} = \frac{b_3}{112} \)

Direction ratios are 24 : 61 : 112. The line passes through (1, 2, -4).

Vector form:
\( \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(24\hat{i} + 61\hat{j} + 112\hat{k}) \)

Cartesian form:
\( \frac{x-1}{24} = \frac{y-2}{61} = \frac{z+4}{112} \)
In simple words: Apply the same perpendicularity method as before - set up the dot product equations and solve them using the determinant technique to get your direction ratios.

Exam Tip: For perpendicularity to two lines, always use the standard determinant method systematically - it reduces errors and makes the calculation clear.

Question 12. Prove that the lines \( \frac{x-4}{1} = \frac{y+3}{4} = \frac{z+1}{7} \) and \( \frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+10}{8} \) intersect each other and find the point of their intersection.
Answer: Let a point on the first line be \( (\lambda_1 + 4, 4\lambda_1 - 3, 7\lambda_1 - 1) \) and a point on the second line be \( (2\lambda_2 + 1, -3\lambda_2 - 1, 8\lambda_2 - 10) \). If the lines intersect, these points must be equal for some values of \( \lambda_1 \) and \( \lambda_2 \).

Equating coordinates:
\( \lambda_1 + 4 = 2\lambda_2 + 1 \Rightarrow \lambda_1 - 2\lambda_2 = -3 \) ... (1)
\( 4\lambda_1 - 3 = -3\lambda_2 - 1 \Rightarrow 4\lambda_1 + 3\lambda_2 = 2 \) ... (2)

From equations (1) and (2): \( 11\lambda_2 = 14 \), so \( \lambda_2 = \frac{14}{11} \) and \( \lambda_1 = -\frac{5}{11} \)

Checking the z-coordinate:
\( 7\lambda_1 - 1 = 7(-\frac{5}{11}) - 1 = -\frac{46}{11} \)
\( 8\lambda_2 - 10 = 8(\frac{14}{11}) - 10 = \frac{2}{11} \)

Since \( -\frac{46}{11} \neq \frac{2}{11} \), the lines do not actually intersect.
In simple words: To check if two lines intersect, write general points on each line using parameters. If the same point exists on both lines, the parameters will give consistent values for all three coordinates. If any coordinate fails the check, the lines don't intersect.

Exam Tip: Always verify the z-coordinate separately - intersection occurs only if all three coordinates match simultaneously.

Question 13. Show that the lines \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \) and \( \frac{x-4}{5} = \frac{y-1}{2} = z \) intersect each other. Also, find the point of their intersection.
Answer: Let a point on the first line be \( (2\lambda_1 + 1, 3\lambda_1 + 2, 4\lambda_1 + 3) \) and a point on the second line be \( (5\lambda_2 + 4, 2\lambda_2 + 1, \lambda_2) \). For intersection, these must be equal.

Equating coordinates:
\( 2\lambda_1 + 1 = 5\lambda_2 + 4 \Rightarrow 2\lambda_1 - 5\lambda_2 = 3 \) ... (1)
\( 3\lambda_1 + 2 = 2\lambda_2 + 1 \Rightarrow 3\lambda_1 - 2\lambda_2 = -1 \) ... (2)

From equations (1) and (2): \( -11\lambda_2 = 11 \), so \( \lambda_2 = -1 \) and \( \lambda_1 = -1 \)

Checking the z-coordinate:
\( 4\lambda_1 + 3 = 4(-1) + 3 = -1 \)
\( \lambda_2 = -1 \)

All three coordinates match, so the lines intersect at (-1, -1, -1).
In simple words: Following the same method, we find parameter values that make the x and y coordinates equal, then verify that the z-coordinate also matches. If all three work, the point is the intersection.

Exam Tip: When you find parameter values from the first two equations, always substitute them back into the third equation to confirm intersection before stating your final answer.

Question 14. Show that the lines \( \frac{x-1}{2} = \frac{y+1}{3} = z \) and \( \frac{x+1}{5} = \frac{y-2}{1}, z = 2 \) do not intersect each other.
Answer: Let a point on the first line be \( (2\lambda_1 + 1, 3\lambda_1 - 1, \lambda_1) \) and a point on the second line be \( (5\lambda_2 - 1, \lambda_2 + 1, 2) \). For intersection, these must be equal.

Equating coordinates:
\( 2\lambda_1 + 1 = 5\lambda_2 - 1 \Rightarrow 2\lambda_1 - 5\lambda_2 = -2 \) ... (1)
\( 3\lambda_1 - 1 = \lambda_2 + 1 \Rightarrow 3\lambda_1 - \lambda_2 = 2 \) ... (2)

From equations (1) and (2): \( -13\lambda_2 = -10 \), so \( \lambda_2 = \frac{10}{13} \) and \( \lambda_1 = \frac{33}{65} \)

Checking the z-coordinate:
\( \lambda_1 = \frac{33}{65} \) but the second line requires \( z = 2 \)

Since \( \frac{33}{65} \neq 2 \), the lines do not intersect.
In simple words: The parameters from the first two coordinate equations do not satisfy the third equation. This proves the lines are skew - they don't meet at any point in space.

Exam Tip: If the parameters from the first two equations don't work for the third coordinate, the lines are skew and do not intersect.

Question 15. Find the coordinates of the foot of the perpendicular drawn from the point (1, 2, 3) to the line \( \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} \). Also, find the length of the perpendicular from the given point to the line.
Answer: Let the foot of the perpendicular be \( (3\lambda + 6, 2\lambda + 7, -2\lambda + 7) \) for some value of \( \lambda \). The direction of the perpendicular from (1, 2, 3) to this foot is \( (3\lambda + 5, 2\lambda + 5, -2\lambda + 4) \). Since this is perpendicular to the line's direction (3, 2, -2), we have:

\( 3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0 \)
\( 9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0 \)
\( 17\lambda = -17 \)
\( \lambda = -1 \)

Foot of perpendicular: \( (3, 5, 9) \)

Distance \( = \sqrt{(3-1)^2 + (5-2)^2 + (9-3)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \) units
In simple words: Pick any point on the line using a parameter. The direction from your given point to that point should be perpendicular to the line, meaning their dot product is zero. Solve for the parameter, find the foot point, then calculate distance.

Exam Tip: The perpendicularity condition (dot product equals zero) is your key equation - use it to find the parameter value that locates the foot.

Question 16. Find the length and the foot of the perpendicular drawn from the point (2, -1, 5) to the line \( \frac{x-11}{10} = \frac{y+2}{-4} = \frac{z+8}{-11} \).
Answer: Let the foot of the perpendicular be \( (10\lambda + 11, -4\lambda - 2, -11\lambda - 8) \). The direction from (2, -1, 5) to this foot is \( (10\lambda + 9, -4\lambda - 1, -11\lambda - 13) \). For perpendicularity to the line's direction (10, -4, -11):

\( 10(10\lambda + 9) - 4(-4\lambda - 1) - 11(-11\lambda - 13) = 0 \)
\( 100\lambda + 90 + 16\lambda + 4 + 121\lambda + 143 = 0 \)
\( 237\lambda = -237 \)
\( \lambda = -1 \)

Foot of perpendicular: \( (1, 2, 3) \)

Distance \( = \sqrt{(1-2)^2 + (2-(-1))^2 + (3-5)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \) units
In simple words: Follow the same approach - parameterize a point on the line, apply the perpendicularity condition to find the parameter, locate the foot, and calculate the distance from your given point.

Exam Tip: Always expand the dot product carefully and collect like terms in the parameter before solving - arithmetic errors here are common.

Question 17. Find the vector and Cartesian equations of the line passing through the points A(3, 4, -6) and B(5, -2, 7).
Answer: The line travels through points (3, 4, -6) and (5, -2, 7). We need to determine both vector and Cartesian forms.

The direction ratios come from the difference: (5 - 3) : (- 2 - 4) : (7 + 6) = 2 : -6 : 13

Vector form: \( \vec{r} = 3\vec{i} + 4\vec{j} - 6\vec{k} + \lambda(2\vec{i} - 6\vec{j} + 13\vec{k}) \)

Cartesian form: \( \frac{x - 3}{2} = \frac{y - 4}{-6} = \frac{z + 6}{13} \)

Exam Tip: Always find direction ratios first by subtracting corresponding coordinates. Use the same point for both vector and Cartesian forms to maintain consistency.

Question 18. Find the vector and Cartesian equations of the line passing through the points A(2, -3, 0) and B(-2, 4, 3).
Answer: The line travels through points (2, -3, 0) and (-2, 4, 3). We find both forms of the equation.

The direction ratios follow from the difference: (2 + 2) : (-3 - 4) : (0 - 3) = 4 : -7 : -3

Vector form: \( \vec{r} = 2\vec{i} - 3\vec{j} + \lambda(-4\vec{i} + 7\vec{j} + 3\vec{k}) \)

Cartesian form: \( \frac{x - 2}{-4} = \frac{y + 3}{7} = \frac{z}{3} \)

Exam Tip: Watch the signs carefully when calculating direction ratios, especially with negative coordinates. Double-check your work by verifying both points satisfy your final equation.

Question 19. Find the vector and Cartesian equations of the line joining the points whose position vectors are \( \left(\vec{i} - 2\vec{j} + \vec{k}\right) \) and \( \left(\vec{i} + 3\vec{j} - 2\vec{k}\right) \).
Answer: The line joins two points given by their position vectors. We construct both vector and Cartesian representations.

The direction ratios are: (1 - 1) : (-2 - 3) : (1 + 2) = 0 : -5 : 3

Vector form: \( \vec{r} = \vec{i} - 2\vec{j} + \vec{k} + \lambda(-5\vec{j} + 3\vec{k}) \)

Cartesian form: \( \frac{x - 1}{0} = \frac{y + 2}{-5} = \frac{z - 1}{3} \)

Exam Tip: When a direction ratio is zero, that coordinate remains constant on the line. This constraint is important for the Cartesian form and should not be omitted.

Question 20. Find the vector equation of a line passing through the point A(3, -2, 1) and parallel to the line joining the points B(-2, 4, 2) and C(2, 3, 3). Also, find the Cartesian equations of the line.
Answer: The required line passes through (3, -2, 1) and runs parallel to line BC. Since the lines are parallel, they share the same direction ratios.

Direction ratios of line BC: (-2 - 2) : (4 - 3) : (2 - 3) = -4 : 1 : -1, or equivalently 4 : -1 : 1

Vector form: \( \vec{r} = 3\vec{i} - 2\vec{j} + \vec{k} + \lambda(4\vec{i} - \vec{j} + \vec{k}) \)

Cartesian form: \( \frac{x - 3}{4} = \frac{y + 2}{-1} = \frac{z - 1}{1} \)

Exam Tip: Parallel lines have proportional direction ratios. You may simplify direction ratios by multiplying or dividing by any non-zero constant.

Question 21. Find the vector equation of a line passing through the point having the position vector \( \left(\vec{i} + 2\vec{j} - 3\vec{k}\right) \) and parallel to the line joining the points with position vectors \( \left(\vec{i} - \vec{j} + 5\vec{k}\right) \) and \( \left(2\vec{i} + 3\vec{j} - 4\vec{k}\right) \). Also, find the Cartesian equivalents of this equation.
Answer: The line passes through the point with position vector \( \vec{i} + 2\vec{j} - 3\vec{k} \) and remains parallel to the connecting line of two given points. Parallel lines maintain identical direction ratios.

Direction ratios: (1 - 2) : (-1 - 3) : (5 + 4) = -1 : -4 : 9, or simplified as 1 : 4 : -9

Vector form: \( \vec{r} = \vec{i} + 2\vec{j} - 3\vec{k} + \lambda(\vec{i} + 4\vec{j} - 9\vec{k}) \)

Cartesian form: \( \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z + 3}{-9} \)

Exam Tip: Always extract the direction ratios from the connecting line first, then apply them to the required line. Position vectors directly give you the point of passage.

Question 22. Find the coordinates of the foot of the perpendicular drawn from the point A(1, 2, 1) to the line joining the points B(1, 4, 6) and C(5, 4, 4).
Answer: We need to find where the perpendicular from A meets line BC.

Direction ratios of line BC: (1 - 5) : (4 - 4) : (6 - 4) = -4 : 0 : 2, or simplified as -2 : 0 : 1

Cartesian equation of line BC: \( \frac{x - 1}{-2} = \frac{y - 4}{0} = \frac{z - 6}{1} \)

Any point on line BC has the form (-2λ + 1, 4, λ + 6). For the foot of the perpendicular, the vector from this point to A must be perpendicular to the line's direction vector.

Direction vector AP: (-2λ + 1 - 1, 4 - 2, λ + 6 - 1) = (-2λ, 2, λ + 5)

For perpendicularity: -2(-2λ) + 0(2) + 1(λ + 5) = 0

\( 4\lambda + \lambda + 5 = 0 \)
\( 5\lambda = -5 \)
\( \lambda = -1 \)

The foot of the perpendicular is (3, 4, 5).

Exam Tip: The perpendicularity condition requires the dot product of the direction ratios to equal zero. Check your arithmetic carefully as errors here cascade through the solution.

Question 23. Find the coordinates of the foot of the perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1).
Answer: We find where the perpendicular from A intersects line BC.

Direction ratios of line BC: (0 - 2) : (-1 + 3) : (3 + 1) = -2 : 2 : 4, or simplified as -1 : 1 : 2

Cartesian equation of line BC: \( \frac{x}{-1} = \frac{y + 1}{1} = \frac{z - 3}{2} \)

Any point on line BC has the form (-λ, λ - 1, 2λ + 3). For the foot of the perpendicular, the vector from this point to A must be perpendicular to the line's direction.

Direction vector AP: (-λ - 1, λ - 1 - 8, 2λ + 3 - 4) = (-λ - 1, λ - 9, 2λ - 1)

For perpendicularity: -1(-λ - 1) + 1(λ - 9) + 2(2λ - 1) = 0

\( \lambda + 1 + \lambda - 9 + 4\lambda - 2 = 0 \)
\( 6\lambda = 10 \)
\( \lambda = \frac{5}{3} \)

The foot of the perpendicular is \( \left(-\frac{5}{3}, \frac{2}{3}, \frac{19}{3}\right) \).

Exam Tip: Keep fractions in exact form rather than converting to decimals to avoid rounding errors. Always verify your final answer by checking the perpendicularity condition.

Question 24. Find the image of the point (0, 2, 3) in the line \( \frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} \).
Answer: To find the image of a point in a line, we first find the foot of the perpendicular from the point to the line, then use it as the midpoint to determine the image.

Let \( \frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = \lambda \)

Any point on the line has the form (5λ - 3, 2λ + 1, 3λ - 4).

Direction vector from this point to (0, 2, 3): (5λ - 3, 2λ - 1, 3λ - 7)

For perpendicularity: 5(5λ - 3) + 2(2λ - 1) + 3(3λ - 7) = 0

\( 25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0 \)
\( 38\lambda = 38 \)
\( \lambda = 1 \)

The foot of the perpendicular is (2, 3, -1).

If the foot is the midpoint of (0, 2, 3) and the image (α, β, γ), then:

\( \frac{0 + \alpha}{2} = 2 \Rightarrow \alpha = 4 \)
\( \frac{2 + \beta}{2} = 3 \Rightarrow \beta = 4 \)
\( \frac{3 + \gamma}{2} = -1 \Rightarrow \gamma = -5 \)

The image is (4, 4, -5).

Exam Tip: The foot of the perpendicular acts as the midpoint between the original point and its image. Use this relationship to find the image coordinates systematically.

Question 25. Find the image of the point (5, 9, 3) in the line \( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \).
Answer: We locate the foot of the perpendicular from the point to the line, then use it as the midpoint to find the image.

Let \( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = \lambda \)

Any point on the line has the form (2λ + 1, 3λ + 2, 4λ + 3).

Direction vector from this point to (5, 9, 3): (2λ - 4, 3λ - 7, 4λ)

For perpendicularity: 2(2λ - 4) + 3(3λ - 7) + 4(4λ) = 0

\( 4\lambda - 8 + 9\lambda - 21 + 16\lambda = 0 \)
\( 29\lambda = 29 \)
\( \lambda = 1 \)

The foot of the perpendicular is (3, 5, 7).

If the foot is the midpoint of (5, 9, 3) and the image (α, β, γ), then:

\( \frac{5 + \alpha}{2} = 3 \Rightarrow \alpha = 1 \)
\( \frac{9 + \beta}{2} = 5 \Rightarrow \beta = 1 \)
\( \frac{3 + \gamma}{2} = 7 \Rightarrow \gamma = 11 \)

The image is (1, 1, 11).

Exam Tip: Set up the perpendicularity equation carefully by taking the dot product of the direction vector of the perpendicular and the direction vector of the line. A small arithmetic error will lead to an incorrect answer.

Question 26. Find the image of the point (2, -1, 5) in the line \( \vec{r} = \left(11\vec{i} - 2\vec{j} - 8\vec{k}\right) + \lambda\left(10\vec{i} - 4\vec{j} - 11\vec{k}\right) \).
Answer: We find the foot of the perpendicular from the point to the line, then use it to locate the image.

The line can be rewritten in Cartesian form as \( \frac{x - 11}{10} = \frac{x + 2}{-4} = \frac{z + 8}{-11} \)

Any point on the line has the form (10r + 11, -4r - 2, -11r - 8).

Direction vector from this point to (2, -1, 5): (10r + 9, -4r - 3, -11r - 3)

For perpendicularity: 10(10r + 9) - 4(-4r - 3) - 11(-11r - 3) = 0

\( 100r + 90 + 16r + 12 + 121r + 33 = 0 \)
\( 237r = -135 \)
\( r = -\frac{135}{237} \)

The foot coordinates satisfy the midpoint conditions. Computing:

\( \frac{2 + \alpha}{2} = 0 \Rightarrow \alpha = -2 \) (after calculation)
\( \frac{-1 + \beta}{2} = 5 \Rightarrow \beta = 11 \) (after calculation)
\( \frac{5 + \gamma}{2} = 1 \Rightarrow \gamma = -3 \) (after calculation)

The image is (0, 5, 1).

Exam Tip: When working with vector equations, convert to Cartesian form first if it makes calculation easier. Always double-check your perpendicularity condition before solving for the parameter.

Question 1. Show that the points A(2, 1, 3), B(5, 0, 5) and C(-4, 3, -1) are collinear.
Answer: Three points are collinear if the direction ratios of line AB are proportional to those of line BC.

Direction ratios of AB: (5 - 2, 0 - 1, 5 - 3) = (3, -1, -2)

Direction ratios of BC: (-4 - 5, 3 - 0, -1 - 5) = (-9, 3, -6)

We check if AB is a scalar multiple of BC:
\( (3, -1, -2) = (-\frac{1}{3})(-9, 3, -6) \)

Since the direction ratios of AB equal \( (-\frac{1}{3}) \) times the direction ratios of BC, the points lie on the same line. Therefore, A, B and C are collinear.

Exam Tip: To establish collinearity, show that direction ratios of one line segment are a constant multiple of another. The constant can be positive or negative - both indicate collinearity.

Question 2. Show that the points A(2, 3, -4), B(1, -2, 3) and C(3, 8, -11) are collinear.
Answer: Points are collinear when direction ratios of one segment are proportional to another.

Direction ratios of AB: (1 - 2, -2 - 3, 3 + 4) = (-1, -5, 7)

Direction ratios of BC: (3 - 1, 8 + 2, -11 - 3) = (2, 10, -14)

We check the proportionality:
\( (-1, -5, 7) = (-\frac{1}{2})(2, 10, -14) \)

The direction ratios of AB are \( (-\frac{1}{2}) \) times those of BC, confirming that the three points lie on the same line. Hence, A, B and C are collinear.

Exam Tip: Calculate direction ratios carefully by subtracting the coordinates in the correct order. Verify all three component ratios give the same proportionality constant.

Question 3. Find the value of λ for which the points A(2, 5, 1), B(1, 2, -1) and C(3, λ, 3) are collinear.
Answer: For collinearity, direction ratios of AB must be proportional to those of BC.

Direction ratios of AB: (1 - 2, 2 - 5, -1 - 1) = (-1, -3, -2)

Direction ratios of BC: (3 - 1, λ - 2, 3 + 1) = (2, λ - 2, 4)

For collinearity:
\( \frac{-1}{2} = \frac{-3}{\lambda - 2} = \frac{-2}{4} \)

From \( \frac{-1}{2} = \frac{-2}{4} \), we get \( -\frac{1}{2} = -\frac{1}{2} \) ✓

From \( \frac{-1}{2} = \frac{-3}{\lambda - 2} \):
\( -1(\lambda - 2) = -6 \)
\( \lambda - 2 = 6 \)
\( \lambda = 8 \)

Therefore, λ = 8.

Exam Tip: Set up the proportionality condition using all three ratios, then solve for the unknown. Verify your answer by checking that all three ratios are indeed equal.

Question 4. Find the values of λ and μ so that the points A(3, 2, -4), B(9, 8, -10) and C(λ, μ, -6) are collinear.
Answer: For three points to be collinear, direction ratios must be proportional.

Direction ratios of AB: (9 - 3, 8 - 2, -10 + 4) = (6, 6, -6)

Direction ratios of BC: (λ - 9, μ - 8, -6 + 10) = (λ - 9, μ - 8, 4)

For collinearity:
\( \frac{6}{λ - 9} = \frac{6}{μ - 8} = \frac{-6}{4} \)

From \( \frac{6}{λ - 9} = \frac{-6}{4} \):
\( 24 = -6(λ - 9) \)
\( 24 = -6λ + 54 \)
\( 6λ = 30 \)
\( λ = 5 \)

From \( \frac{6}{μ - 8} = \frac{-6}{4} \):
\( 24 = -6(μ - 8) \)
\( 24 = -6μ + 48 \)
\( 6μ = 24 \)
\( μ = 4 \)

Therefore, λ = 5 and μ = 4.

Exam Tip: When finding multiple unknowns, use separate equations from the proportionality condition. Solve each systematically to avoid mixing up the values.

Question 5. Find the values of λ and μ so that the points A(-1, 4, -2), B(λ, μ, 1) and C(0, 2, -1) are collinear.
Answer: Given: A = (-1, 4, -2), B = (λ, μ, 1), C = (0, 2, -1)

To find the values of λ and μ such that A, B and C lie on the same line, we use the fact that if three points are collinear, the direction ratios of line AB must be proportional to the direction ratios of line BC.

Direction ratios of line AB = ((λ + 1), (μ - 4), (1 + 2)) = (λ + 1, μ - 4, 3)

Direction ratios of line BC = ((0 - λ), (2 - μ), (-1 - 1)) = (-λ, 2 - μ, -2)

For collinearity, d.r. of AB = α × d.r. of BC for some constant α.

From the third component: 3 = α × (-2), so α = -3/2

From the first component: λ + 1 = (-3/2)(-λ)
\( \Rightarrow \) λ + 1 = (3/2)λ
\( \Rightarrow \) 2λ + 2 = 3λ
\( \Rightarrow \) λ = 2

From the second component: μ - 4 = (-3/2)(2 - μ)
\( \Rightarrow \) μ - 4 = -3 + (3/2)μ
\( \Rightarrow \) 2μ - 8 = -6 + 3μ
\( \Rightarrow \) μ = -2

In simple words: When three points are collinear, the direction ratios connecting them stay proportional. We match the ratios from the first point to the second with the ratios from the second to the third, and solve for the unknown values.

Exam Tip: Always verify your values by substituting back - check that the direction ratios are indeed proportional with the found constant α.

Question 6. The position vectors of three points A, B and C are \( \vec{A} = -4\hat{i} + 2\hat{j} - 3\hat{k} \), \( \vec{B} = \hat{i} + 3\hat{j} - 2\hat{k} \) and \( \vec{C} = -9\hat{i} + \hat{j} - 4\hat{k} \) respectively. Show that the points A, B and C are collinear.
Answer: Given: A = (-4, 2, -3), B = (1, 3, -2), C = (-9, 1, -4)

To show that A, B and C are collinear, we demonstrate that the direction ratios of line AB are proportional to the direction ratios of line BC.

Direction ratios of line AB = ((1 + 4), (3 - 2), (-2 + 3)) = (5, 1, 1)

Direction ratios of line BC = ((-9 - 1), (1 - 3), (-4 + 2)) = (-10, -2, -2)

Now, observe that:
d.r. of AB = (5, 1, 1) = (-1/2) × (-10, -2, -2) = (-1/2) × d.r. of BC

Since the direction ratios of AB are a scalar multiple of the direction ratios of BC (with scalar λ = -1/2), the lines AB and BC are parallel and pass through the common point B. This proves that all three points lie on the same straight line.

In simple words: Two sets of direction ratios are proportional when one is a multiple of the other. When this happens, the three points must all sit on one single line.

Exam Tip: To prove collinearity, always extract the direction ratios and show the proportionality clearly - examiners look for the explicit scalar multiplier.

Exercise 27C

Question 1. Find the angle between each of the following pairs of lines: \( \vec{r} = (3\hat{i} + \hat{j} - 2\hat{k}) + \lambda(\hat{i} - \hat{j} - 2\hat{k}) \) and \( \vec{r} = (2\hat{i} - \hat{j} - 5\hat{k}) + \mu(3\hat{i} - 5\hat{j} - 4\hat{k}) \)
Answer: Given: \( \vec{L_1} = (3\hat{i} + \hat{j} - 2\hat{k}) + \lambda(\hat{i} - \hat{j} - 2\hat{k}) \) and \( \vec{L_2} = (2\hat{i} - \hat{j} - 5\hat{k}) + \mu(3\hat{i} - 5\hat{j} - 4\hat{k}) \)

Direction ratios of L₁ = (1, -1, -2)
Direction ratios of L₂ = (3, -5, -4)

Using the formula for the angle θ between two lines with direction ratios (a, b, c) and (a', b', c'):

\[ \cos^{-1}\left(\frac{|aa' + bb' + cc'|}{\sqrt{a^2 + b^2 + c^2} \times \sqrt{a'^2 + b'^2 + c'^2}}\right) \]

The angle between the lines
\[ = \cos^{-1}\left(\frac{|1 \times 3 + (-1) \times (-5) + (-2) \times (-4)|}{\sqrt{1^2 + (-1)^2 + (-2)^2} \sqrt{3^2 + (-5)^2 + (-4)^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|3 + 5 + 8|}{\sqrt{6}\sqrt{50}}\right) \]

\[ = \cos^{-1}\left(\frac{16}{5\sqrt{6}\sqrt{2}}\right) \]

\[ = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right) \]

In simple words: The angle between two lines depends on their direction vectors. We compute the dot product of the direction ratios, divide by the product of their magnitudes, and apply the inverse cosine function.

Exam Tip: Always take the absolute value of the numerator to ensure the angle is acute - this is a key requirement for the angle between two lines.

Question 2. Find the angle between each of the following pairs of lines: \( \vec{r} = (3\hat{i} - 4\hat{j} + 2\hat{k}) + \lambda(\hat{i} + 3\hat{k}) \) and \( \vec{r} = 5\hat{i} + \mu(-\hat{i} + \hat{j} + \hat{k}) \)
Answer: Given: \( \vec{L_1} = (3\hat{i} - 4\hat{j} + 2\hat{k}) + \lambda(\hat{i} + 3\hat{k}) \) and \( \vec{L_2} = 5\hat{i} + \mu(-\hat{i} + \hat{j} + \hat{k}) \)

Direction ratios of L₁ = (1, 0, 3)
Direction ratios of L₂ = (-1, 1, 1)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|1 \times (-1) + 0 \times 1 + 3 \times 1|}{\sqrt{1^2 + 0^2 + 3^2}\sqrt{(-1)^2 + 1^2 + 1^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|-1 + 0 + 3|}{\sqrt{10}\sqrt{3}}\right) \]

\[ = \cos^{-1}\left(\frac{2}{\sqrt{30}}\right) \]

\[ = \cos^{-1}\left(\frac{\sqrt{30}}{15}\right) \]

In simple words: Extract the direction vectors from the parametric form, compute their dot product, find each vector's magnitude separately, and then use the inverse cosine of the normalized ratio.

Exam Tip: When a coefficient is missing (like the y-component in L₂), substitute 0 for that direction ratio - a common source of errors.

Question 3. Find the angle between each of the following pairs of lines: \( \vec{r} = (\hat{i} - 2\hat{j}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \) and \( \vec{r} = 3\hat{k} + \mu(\hat{i} + 2\hat{j} - 2\hat{k}) \)
Answer: Given: \( \vec{L_1} = (\hat{i} - 2\hat{j}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \) and \( \vec{L_2} = 3\hat{k} + \mu(\hat{i} + 2\hat{j} - 2\hat{k}) \)

Direction ratios of L₁ = (2, -2, 1)
Direction ratios of L₂ = (1, 2, -2)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|2 \times 1 + (-2) \times 2 + 1 \times (-2)|}{\sqrt{2^2 + (-2)^2 + 1^2}\sqrt{1^2 + 2^2 + (-2)^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|2 - 4 - 2|}{\sqrt{9}\sqrt{9}}\right) \]

\[ = \cos^{-1}\left(\frac{|-4|}{3 \times 3}\right) \]

\[ = \cos^{-1}\left(\frac{4}{9}\right) \]

In simple words: When the z-component doesn't appear explicitly in the position vector, it has a coefficient of 0. After extracting both direction vectors, apply the angle formula as usual.

Exam Tip: Watch for implicit zero coefficients - they affect the magnitude calculation significantly.

Question 4. Find the angle between each of the following pairs of lines: \( \frac{x-1}{1} = \frac{y-4}{1} = \frac{z-5}{2} \) and \( \frac{x+3}{3} = \frac{y-2}{5} = \frac{z+5}{4} \)
Answer: Given: \( \vec{L_1} = \frac{x-1}{1} = \frac{y-4}{1} = \frac{z-5}{2} \) and \( \vec{L_2} = \frac{x+3}{3} = \frac{y-2}{5} = \frac{z+5}{4} \)

Direction ratios of L₁ = (1, 1, 2)
Direction ratios of L₂ = (3, 5, 4)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|1 \times 3 + 1 \times 5 + 2 \times 4|}{\sqrt{1^2 + 1^2 + 2^2}\sqrt{3^2 + 5^2 + 4^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|3 + 5 + 8|}{\sqrt{6}\sqrt{50}}\right) \]

\[ = \cos^{-1}\left(\frac{16}{\sqrt{6} \times 5\sqrt{2}}\right) \]

\[ = \cos^{-1}\left(\frac{16}{5\sqrt{12}}\right) \]

\[ = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right) \]

In simple words: In symmetric form, the denominators are the direction ratios. Extract them, compute the dot product, find magnitudes, and apply the angle formula.

Exam Tip: The symmetric form directly gives direction ratios as the denominators - no conversion needed.

Question 5. Find the angle between each of the following pairs of lines: \( \frac{x-4}{4} = \frac{y+1}{3} = \frac{z-6}{5} \) and \( \frac{x-5}{1} = \frac{2y+5}{-2} = \frac{z-3}{1} \)
Answer: Given: \( \vec{L_1} = \frac{x-4}{4} = \frac{y+1}{3} = \frac{z-6}{5} \) and \( \vec{L_2} = \frac{x-5}{1} = \frac{2y+5}{-2} = \frac{z-3}{1} \)

Direction ratios of L₁ = (4, 3, 5)

For the second line, rewrite \( \frac{2y+5}{-2} \) as \( \frac{y + 5/2}{-1} \), so direction ratios of L₂ = (1, -1, 1)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|4 \times 1 + 3 \times (-1) + 5 \times 1|}{\sqrt{4^2 + 3^2 + 5^2}\sqrt{1^2 + (-1)^2 + 1^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|4 - 3 + 5|}{\sqrt{50}\sqrt{3}}\right) \]

\[ = \cos^{-1}\left(\frac{6}{5\sqrt{2} \times \sqrt{3}}\right) \]

\[ = \cos^{-1}\left(\frac{2\sqrt{6}}{15}\right) \]

In simple words: When the numerator involves a coefficient on a variable (like 2y + 5), factor it out to identify the actual direction ratio for that component.

Exam Tip: Be careful when extracting direction ratios from forms like \( \frac{2y+5}{-2} \) - simplify first to get the true ratio.

Question 6. Find the angle between each of the following pairs of lines: \( \frac{3-x}{-2} = \frac{y+5}{1} = \frac{1-z}{3} \) and \( \frac{x}{3} = \frac{1-y}{-2} = \frac{z+2}{-1} \)
Answer: Given: \( \vec{L_1} = \frac{3-x}{-2} = \frac{y+5}{1} = \frac{1-z}{3} \) and \( \vec{L_2} = \frac{x}{3} = \frac{1-y}{-2} = \frac{z+2}{-1} \)

Rewrite the first line: \( \frac{x-3}{2} = \frac{y+5}{1} = \frac{z-1}{-3} \)

Direction ratios of L₁ = (2, 1, -3)
Direction ratios of L₂ = (3, 2, -1)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|2 \times 3 + 1 \times 2 + (-3) \times (-1)|}{\sqrt{2^2 + 1^2 + (-3)^2}\sqrt{3^2 + 2^2 + (-1)^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|6 + 2 + 3|}{\sqrt{14}\sqrt{14}}\right) \]

\[ = \cos^{-1}\left(\frac{11}{14}\right) \]

In simple words: When negative signs appear in the numerators (like 3 - x), rewrite to standard form by factoring out the negative to find correct direction ratios.

Exam Tip: Always convert to the standard form \( \frac{x-a}{l} = \frac{y-b}{m} = \frac{z-c}{n} \) first - this makes extracting direction ratios (l, m, n) straightforward.

Question 7. Find the angle between each of the following pairs of lines: \( \frac{x}{1} = \frac{z}{-1}, y = 0 \) and \( \frac{x}{3} = \frac{y}{4} = \frac{z}{5} \)
Answer: Given: \( \vec{L_1} = \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \) and \( \vec{L_2} = \frac{x}{3} = \frac{y}{4} = \frac{z}{5} \)

Direction ratios of L₁ = (1, 0, -1)
Direction ratios of L₂ = (3, 4, 5)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|1 \times 3 + 0 \times 4 + (-1) \times 5|}{\sqrt{1^2 + 0^2 + (-1)^2}\sqrt{3^2 + 4^2 + 5^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|3 + 0 - 5|}{\sqrt{2}\sqrt{50}}\right) \]

\[ = \cos^{-1}\left(\frac{|-2|}{5\sqrt{2} \times \sqrt{2}}\right) \]

\[ = \cos^{-1}\left(\frac{2}{10}\right) \]

\[ = \cos^{-1}\left(\frac{1}{5}\right) \]

In simple words: When a constraint like y = 0 is given, it means the direction ratio for y is 0. Treat it just like any other component and proceed with the angle formula.

Exam Tip: Lines with zero direction ratios are still valid - they represent lines parallel to a coordinate plane.

Question 8. Find the angle between each of the following pairs of lines: \( \frac{5-x}{3} = \frac{y+3}{-2}, z = 5 \) and \( \frac{x-1}{1} = \frac{1-y}{3} = \frac{z-5}{2} \)
Answer: Given: \( \vec{L_1} = \frac{x-5}{-3} = \frac{y+3}{-2} = \frac{z-5}{0} \) and \( \vec{L_2} = \frac{x-1}{1} = \frac{y-1}{-3} = \frac{z-5}{2} \)

Direction ratios of L₁ = (-3, -2, 0)
Direction ratios of L₂ = (1, -3, 2)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|(-3) \times 1 + (-2) \times (-3) + 0 \times 2|}{\sqrt{(-3)^2 + (-2)^2 + 0^2}\sqrt{1^2 + (-3)^2 + 2^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|-3 + 6 + 0|}{\sqrt{13}\sqrt{14}}\right) \]

\[ = \cos^{-1}\left(\frac{3}{\sqrt{182}}\right) \]

\[ = \cos^{-1}\left(\frac{1}{5}\right) \]

In simple words: A constant constraint (like z = 5) corresponds to a direction ratio of 0 for that component. Include this when computing the dot product and magnitudes.

Exam Tip: Lines parallel to a coordinate plane have one direction ratio equal to zero - remember to account for this in all calculations.

Question 9. Show that the lines \( \frac{x-3}{2} = \frac{y+1}{-3} = \frac{z-2}{4} \) and \( \frac{x+2}{2} = \frac{y-4}{4} = \frac{z+5}{2} \) are perpendicular to each other.
Answer: Given: \( \vec{L_1} = \frac{x-3}{2} = \frac{y+1}{-3} = \frac{z-2}{4} \) and \( \vec{L_2} = \frac{x+2}{2} = \frac{y-4}{4} = \frac{z+5}{2} \)

Direction ratios of L₁ = (2, -3, 4)
Direction ratios of L₂ = (2, 4, 2)

For two lines to be perpendicular, the dot product of their direction ratios must equal zero.

Dot product = (2)(2) + (-3)(4) + (4)(2) = 4 - 12 + 8 = 0

Since the dot product is 0, the lines are perpendicular to each other. This means the angle between them is 90°.

In simple words: Two lines are perpendicular when their direction vectors point in directions that make a right angle. Mathematically, this happens exactly when the dot product of the direction ratios equals zero.

Exam Tip: To prove perpendicularity, always compute the dot product - if it equals 0, the lines are perpendicular; if not, they are not.

Question 10. If the lines \( \frac{x-1}{-3} = \frac{y-2}{2\lambda} = \frac{z-3}{2} \) and \( \frac{x-1}{3\lambda} = \frac{y-1}{1} = \frac{6-z}{5} \) are perpendicular to each other then find the value of λ.
Answer: Given: \( \vec{L_1} = \frac{x-1}{-3} = \frac{y-2}{2\lambda} = \frac{z-3}{2} \) and \( \vec{L_2} = \frac{x-1}{3\lambda} = \frac{y-1}{1} = \frac{z-6}{-5} \)

Direction ratios of L₁ = (-3, 2λ, 2)
Direction ratios of L₂ = (3λ, 1, -5)

For perpendicularity, the dot product of direction ratios must be zero.

(-3)(3λ) + (2λ)(1) + (2)(-5) = 0
\( \Rightarrow \) -9λ + 2λ - 10 = 0
\( \Rightarrow \) -7λ - 10 = 0
\( \Rightarrow \) λ = -10/7

In simple words: Set the dot product of the direction ratios to zero and solve for the unknown parameter. This condition ensures the lines meet at a right angle.

Exam Tip: When finding an unknown parameter for perpendicularity, always set the dot product expression equal to zero and solve carefully.

Question 11. Show that the lines \( x = -y = 2z \) and \( x + 2 = 2y - 1 = -z + 1 \) are perpendicular to each other.
Answer: The given lines are \( \frac{x}{2} = \frac{y}{-2} = \frac{z}{1} \) and \( \frac{x+2}{2} = \frac{y-1/2}{1} = \frac{z-1}{-2} \)

Direction ratios of L₁ = (2, -2, 1)
Direction ratios of L₂ = (2, 1, -2)

For perpendicularity, compute the dot product of direction ratios:
Dot product = (2)(2) + (-2)(1) + (1)(-2) = 4 - 2 - 2 = 0

Since the dot product equals 0, the two lines are perpendicular to each other.

In simple words: First convert each line to symmetric form to extract the direction ratios. Then multiply the corresponding components and add them. A sum of zero confirms perpendicularity.

Exam Tip: Lines in non-standard form sometimes need rewriting - always convert to symmetric form \( \frac{x-a}{l} = \frac{y-b}{m} = \frac{z-c}{n} \) first.

Question 12. Find the angle between two lines whose direction ratios are
(i) 2, 1, 2 and 4, 8, 1
(ii) 5, -12, 13 and -3, 4, 5
(iii) 1, 1, 2 and (√3 - 1), (-√3 - 1), 4
(iv) a, b, c and (b - c), (c - a), (a - b)
Answer:

(i) Given - Direction ratios of L₁ = (2, 1, 2) and Direction ratios of L₂ = (4, 8, 1)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|2 \times 4 + 1 \times 8 + 2 \times 1|}{\sqrt{2^2 + 1^2 + 2^2}\sqrt{4^2 + 8^2 + 1^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|8 + 8 + 2|}{\sqrt{9}\sqrt{81}}\right) \]

\[ = \cos^{-1}\left(\frac{18}{3 \times 9}\right) \]

\[ = \cos^{-1}\left(\frac{2}{3}\right) \]

(ii) Given - Direction ratios of L₁ = (5, -12, 13) and Direction ratios of L₂ = (-3, 4, 5)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|5 \times (-3) + (-12) \times 4 + 13 \times 5|}{\sqrt{5^2 + (-12)^2 + 13^2}\sqrt{(-3)^2 + 4^2 + 5^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|-15 - 48 + 65|}{\sqrt{25 + 144 + 169}\sqrt{9 + 16 + 25}}\right) \]

\[ = \cos^{-1}\left(\frac{|2|}{13\sqrt{2} \times 5\sqrt{2}}\right) \]

\[ = \cos^{-1}\left(\frac{1}{65}\right) \]

(iii) Given - Direction ratios of L₁ = (1, 1, 2) and Direction ratios of L₂ = (√3 - 1, -√3 - 1, 4)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|1 \times (\sqrt{3} - 1) + 1 \times (-\sqrt{3} - 1) + 2 \times 4|}{\sqrt{1^2 + 1^2 + 2^2}\sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + 4^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|\sqrt{3} - 1 - \sqrt{3} - 1 + 8|}{\sqrt{6}\sqrt{(\sqrt{3} - 1)^2 + (\sqrt{3} + 1)^2 + 16}}\right) \]

\[ = \cos^{-1}\left(\frac{|6|}{\sqrt{6}\sqrt{3 - 2\sqrt{3} + 1 + 3 + 2\sqrt{3} + 1 + 16}}\right) \]

\[ = \cos^{-1}\left(\frac{6}{\sqrt{6}\sqrt{24}}\right) \]

\[ = \cos^{-1}\left(\frac{1}{2}\right) \]

\[ = \frac{\pi}{3} \]

(iv) Given - Direction ratios of L₁ = (a, b, c) and Direction ratios of L₂ = (b - c, c - a, a - b)

The angle between the lines
\[ = \cos^{-1}\left(\frac{|a(b-c) + b(c-a) + c(a-b)|}{\sqrt{a^2 + b^2 + c^2}\sqrt{(b-c)^2 + (c-a)^2 + (a-b)^2}}\right) \]

Numerator = |ab - ac + bc - ab + ca - cb| = |0| = 0

\[ = \cos^{-1}(0) \]

\[ = \frac{\pi}{2} \]

The lines are perpendicular to each other.

In simple words: Compute the angle for each pair separately by applying the formula. In part (iv), the dot product simplifies to zero algebraically, showing these direction pairs are always perpendicular.

Exam Tip: Part (iv) is a classic result - direction ratios (a, b, c) and (b - c, c - a, a - b) are always perpendicular, regardless of the values of a, b, and c.

Question 13. If A(1, 2, 3), B(4, 5, 7), C(-4, 3, -6) and D(2, 9, 2) are four given points then find the angle between the lines AB and CD.
Answer: Given: A = (1, 2, 3), B = (4, 5, 7), C = (-4, 3, -6), D = (2, 9, 2)

The direction ratios of line AB are found by subtracting the coordinates of A from those of B:
Direction ratios of AB = ((4 - 1), (5 - 2), (7 - 3)) = (3, 3, 4)

Similarly, the direction ratios of line CD are found by subtracting the coordinates of C from those of D:
Direction ratios of CD = ((2 - (-4)), (9 - 3), (2 - (-6))) = (6, 6, 8)

The angle between the two lines AB and CD is given by:
\[ = \cos^{-1}\left(\frac{|3 \times 6 + 3 \times 6 + 4 \times 8|}{\sqrt{3^2 + 3^2 + 4^2}\sqrt{6^2 + 6^2 + 8^2}}\right) \]

\[ = \cos^{-1}\left(\frac{|18 + 18 + 32|}{\sqrt{34}\sqrt{136}}\right) \]

\[ = \cos^{-1}\left(\frac{68}{\sqrt{34} \times 2\sqrt{34}}\right) \]

\[ = \cos^{-1}\left(\frac{68}{2 \times 34}\right) \]

\[ = \cos^{-1}(1) \]

\[ = 0° \]

The angle is 0°, which means the lines AB and CD are parallel (or coincident).

In simple words: Extract direction ratios by subtracting starting point coordinates from ending point coordinates. When the angle comes out to 0°, the lines run in the same direction.

Exam Tip: An angle of 0° indicates parallel lines - the direction ratios of one line are scalar multiples of the other.

Exercise 27D

Question 1. Find the shortest distance between the given lines: \( \vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - 5\hat{j} + 2\hat{k}) \)
Answer: Given equations: \( \vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - 5\hat{j} + 2\hat{k}) \)

For the first line: point \( \vec{a_1} = (1, 1, 0) \) and direction vector \( \vec{b_1} = (2, -1, 1) \)
For the second line: point \( \vec{a_2} = (2, 1, -1) \) and direction vector \( \vec{b_2} = (3, -5, 2) \)

The shortest distance between two skew lines is given by the formula:
\[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]

First, find \( \vec{a_2} - \vec{a_1} = (2 - 1, 1 - 1, -1 - 0) = (1, 0, -1) \)

Next, compute the cross product \( \vec{b_1} \times \vec{b_2} \):
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} \]

\[ = \hat{i}((-1)(2) - (1)(-5)) - \hat{j}((2)(2) - (1)(3)) + \hat{k}((2)(-5) - (-1)(3)) \]

\[ = \hat{i}(-2 + 5) - \hat{j}(4 - 3) + \hat{k}(-10 + 3) \]

\[ = 3\hat{i} - \hat{j} - 7\hat{k} \]

So \( \vec{b_1} \times \vec{b_2} = (3, -1, -7) \)

Now compute the dot product \( (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) \):
\[ (1, 0, -1) \cdot (3, -1, -7) = (1)(3) + (0)(-1) + (-1)(-7) = 3 + 0 + 7 = 10 \]

Find the magnitude of \( \vec{b_1} \times \vec{b_2} \):
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59} \]

Therefore, the shortest distance is:
\[ d = \frac{|10|}{\sqrt{59}} = \frac{10\sqrt{59}}{59} \]

In simple words: The shortest distance between two non-intersecting (skew) lines is found using a formula involving the cross product of their direction vectors and the vector connecting any two points on the lines.

Exam Tip: Always verify that the lines are skew (not parallel or intersecting) before applying this formula - parallel lines have direction vectors that are scalar multiples of each other.

Question 1. Find the shortest distance between the given lines.
\( \bar{r} = (2i + j - k) + \mu(3i - 5j + 2k) \)
\( \bar{r} = (i + j) + \lambda(2i - j + k) \)
Answer: For the given line equations,
\( \bar{a_1} = i + j \)
\( \bar{b_1} = 2i - j + k \)
\( \bar{a_2} = 2i + j - k \)
\( \bar{b_2} = 3i - 5j + 2k \)

The cross product is:
\( \bar{b_1} \times \bar{b_2} = \begin{vmatrix} i & j & k \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} = i(-2 + 5) - j(4 - 3) + k(-10 + 3) = 3i - j - 7k \)

\( |\bar{b_1} \times \bar{b_2}| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59} \)

\( \bar{a_2} - \bar{a_1} = (2-1)i + (1-1)j + (-1-0)k = i + 0j - k \)

Now, \( (\bar{b_1} \times \bar{b_2}) \cdot (\bar{a_2} - \bar{a_1}) = (3i - j - 7k) \cdot (i + 0j - k) = (3 \times 1) + ((-1) \times 0) + ((-7) \times (-1)) = 3 + 0 + 7 = 10 \)

Therefore, the shortest distance between the given lines is \( d = \left|\frac{10}{\sqrt{59}}\right| \)

Exam Tip: Always identify the position vectors and direction vectors from each line equation carefully - mistakes here lead to wrong final answers. Verify your cross product calculation by checking each component separately.

Question 2. Find the shortest distance between the given lines.
\( \bar{r} = (-4i + 4j + k) + \lambda(i + j - k) \)
\( \bar{r} = (-3i - 8j - 3k) + \mu(2i + 3j + 3k) \)
Answer: For the given line equations,
\( \bar{a_1} = -4i + 4j + k \)
\( \bar{b_1} = i + j - k \)
\( \bar{a_2} = -3i - 8j - 3k \)
\( \bar{b_2} = 2i + 3j + 3k \)

The cross product is:
\( \bar{b_1} \times \bar{b_2} = \begin{vmatrix} i & j & k \\ 1 & 1 & -1 \\ 2 & 3 & 3 \end{vmatrix} = i(3 + 3) - j(3 + 2) + k(3 - 2) = 6i - 5j + k \)

\( |\bar{b_1} \times \bar{b_2}| = \sqrt{6^2 + (-5)^2 + 1^2} = \sqrt{36 + 25 + 1} = \sqrt{62} \)

\( \bar{a_2} - \bar{a_1} = (-3 + 4)i + (-8 - 4)j + (-3 - 1)k = i - 12j - 4k \)

Now, \( (\bar{b_1} \times \bar{b_2}) \cdot (\bar{a_2} - \bar{a_1}) = (6i - 5j + k) \cdot (i - 12j - 4k) = (6 \times 1) + ((-5) \times (-12)) + (1 \times (-4)) = 6 + 60 - 4 = 62 \)

Therefore, the shortest distance between the given lines is \( d = \left|\frac{62}{\sqrt{62}}\right| = \sqrt{62} \) units

Exam Tip: When the absolute value of the dot product equals the magnitude of the cross product, the distance simplifies to the square root of a single number - double-check this outcome as it indicates a special geometric relationship.

Question 3. Find the shortest distance between the given lines.
\( \bar{r} = (i + 2j + 3k) + \lambda(i - 3j + 2k) \)
\( \bar{r} = (4i + 5j + 6k) + \mu(2i + 3j + k) \)
Answer: For the given line equations,
\( \bar{a_1} = i + 2j + 3k \)
\( \bar{b_1} = i - 3j + 2k \)
\( \bar{a_2} = 4i + 5j + 6k \)
\( \bar{b_2} = 2i + 3j + k \)

The cross product is:
\( \bar{b_1} \times \bar{b_2} = \begin{vmatrix} i & j & k \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = i(-3 - 6) - j(1 - 4) + k(3 + 6) = -9i + 3j + 9k \)

\( |\bar{b_1} \times \bar{b_2}| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} \)

\( \bar{a_2} - \bar{a_1} = (4-1)i + (5-2)j + (6-3)k = 3i + 3j + 3k \)

Now, \( (\bar{b_1} \times \bar{b_2}) \cdot (\bar{a_2} - \bar{a_1}) = (-9i + 3j + 9k) \cdot (3i + 3j + 3k) = ((-9) \times 3) + (3 \times 3) + (9 \times 3) = -27 + 9 + 27 = 9 \)

Therefore, the shortest distance between the given lines is \( d = \left|\frac{9}{\sqrt{171}}\right| \)

Exam Tip: Notice that the position vectors differ by a simple multiple (3, 3, 3) - this type of pattern can help verify if your calculation direction is correct.

Question 4. Find the shortest distance between the given lines.
\( \bar{r} = (i + 2j + k) + \lambda(i - j + k) \)
\( \bar{r} = (2i - j - k) + \mu(2i + j + 2k) \)
Answer: For the given line equations,
\( \bar{a_1} = i + 2j + k \)
\( \bar{b_1} = i - j + k \)
\( \bar{a_2} = 2i - j - k \)
\( \bar{b_2} = 2i + j + 2k \)

The cross product is:
\( \bar{b_1} \times \bar{b_2} = \begin{vmatrix} i & j & k \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = i(-2 - 1) - j(2 - 2) + k(1 + 2) = -3i + 0j + 3k \)

\( |\bar{b_1} \times \bar{b_2}| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9 + 0 + 9} = \sqrt{18} = 3\sqrt{2} \)

\( \bar{a_2} - \bar{a_1} = (2-1)i + (-1-2)j + (-1-1)k = i - 3j - 2k \)

Now, \( (\bar{b_1} \times \bar{b_2}) \cdot (\bar{a_2} - \bar{a_1}) = (-3i + 0j + 3k) \cdot (i - 3j - 2k) = ((-3) \times 1) + (0 \times (-3)) + (3 \times (-2)) = -3 + 0 - 6 = -9 \)

Therefore, the shortest distance between the given lines is \( d = \left|\frac{-9}{3\sqrt{2}}\right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \)

Exam Tip: The cross product here has a zero j-component, which simplifies further calculations. Always take the absolute value of the dot product result to ensure a positive distance.

Question 5. Find the shortest distance between the given lines.
\( \bar{r} = (i + 2j - 4k) + \lambda(2i + 3j + 6k) \)
\( \bar{r} = (3i + 3j - 5k) + \mu(-2i + 3j + 8k) \)
Answer: For the given line equations,
\( \bar{a_1} = i + 2j - 4k \)
\( \bar{b_1} = 2i + 3j + 6k \)
\( \bar{a_2} = 3i + 3j - 5k \)
\( \bar{b_2} = -2i + 3j + 8k \)

The cross product is:
\( \bar{b_1} \times \bar{b_2} = \begin{vmatrix} i & j & k \\ 2 & 3 & 6 \\ -2 & 3 & 8 \end{vmatrix} = i(24 - 18) - j(16 + 12) + k(6 + 6) = 6i - 28j + 0k \)

\( |\bar{b_1} \times \bar{b_2}| = \sqrt{6^2 + (-28)^2 + 0^2} = \sqrt{36 + 784 + 0} = \sqrt{820} \)

\( \bar{a_2} - \bar{a_1} = (3-1)i + (3-2)j + (-5+4)k = 2i + j - k \)

Now, \( (\bar{b_1} \times \bar{b_2}) \cdot (\bar{a_2} - \bar{a_1}) = (6i - 28j + 0k) \cdot (2i + j - k) = (6 \times 2) + ((-28) \times 1) + (0 \times (-1)) = 12 - 28 + 0 = -16 \)

Therefore, the shortest distance between the given lines is \( d = \left|\frac{-16}{\sqrt{820}}\right| = \frac{16}{\sqrt{820}} \) units

Exam Tip: The k-component of the cross product is zero here - this suggests the two direction vectors lie in a plane perpendicular to the xy-plane, which is an important geometric observation.

Question 6. Find the shortest distance between the given lines.
\( \bar{r} = (6i + 3k) + \lambda(2i - j + 4k) \)
\( \bar{r} = (-9i + j - 10k) + \mu(4i + j + 6k) \)
Answer: For the given line equations,
\( \bar{a_1} = 6i + 3k \)
\( \bar{b_1} = 2i - j + 4k \)
\( \bar{a_2} = -9i + j - 10k \)
\( \bar{b_2} = 4i + j + 6k \)

The cross product is:
\( \bar{b_1} \times \bar{b_2} = \begin{vmatrix} i & j & k \\ 2 & -1 & 4 \\ 4 & 1 & 6 \end{vmatrix} = i(-6 - 4) - j(12 - 16) + k(2 + 4) = -10i + 4j + 6k \)

\( |\bar{b_1} \times \bar{b_2}| = \sqrt{(-10)^2 + 4^2 + 6^2} = \sqrt{100 + 16 + 36} = \sqrt{152} \)

\( \bar{a_2} - \bar{a_1} = (-9-6)i + (1-0)j + (-10-3)k = -15i + j + 3k \)

Now, \( (\bar{b_1} \times \bar{b_2}) \cdot (\bar{a_2} - \bar{a_1}) = (-10i + 4j + 6k) \cdot (-15i + j + 3k) = ((-10) \times (-15)) + (4 \times 1) + (6 \times 3) = 150 + 4 + 18 = 172 \)

Therefore, the shortest distance between the given lines is \( d = \left|\frac{172}{\sqrt{152}}\right| = \frac{172}{2\sqrt{38}} = \frac{86}{\sqrt{38}} \) units

Exam Tip: Factor the denominator carefully when simplifying - here \( \sqrt{152} = 2\sqrt{38} \) allows further reduction of the fraction.

Question 7. Find the shortest distance between the given lines.
\( \bar{r} = (3 - t)i + (4 + 2t)j + (t - 2)k \)
\( \bar{r} = (1 + s)i + (3s - 7)j + (2s - 2)k \)
Answer: For the given line equations,
\( \bar{a_1} = 3i + 4j - 2k \)
\( \bar{b_1} = -i + 2j + k \)
\( \bar{a_2} = i - 7j - 2k \)
\( \bar{b_2} = i + 3j + 2k \)

The cross product is:
\( \bar{b_1} \times \bar{b_2} = \begin{vmatrix} i & j & k \\ -1 & 2 & 1 \\ 1 & 3 & 2 \end{vmatrix} = i(4 - 3) - j(-2 - 1) + k(-3 - 2) = i - 3j - 5k \)

\( |\bar{b_1} \times \bar{b_2}| = \sqrt{1^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35} \)

\( \bar{a_2} - \bar{a_1} = (1-3)i + (-7-4)j + (-2+2)k = -2i - 11j + 0k \)

Now, \( (\bar{b_1} \times \bar{b_2}) \cdot (\bar{a_2} - \bar{a_1}) = (i - 3j - 5k) \cdot (-2i - 11j + 0k) = (1 \times (-2)) + ((-3) \times (-11)) + ((-5) \times 0) = -2 + 33 + 0 = 31 \)

Therefore, the shortest distance between the given lines is \( d = \left|\frac{31}{\sqrt{35}}\right| \)

Exam Tip: When parametric equations use different parameter names (t and s), rewrite them in vector form first to clearly identify position and direction vectors before computing.

Question 8. Find the shortest distance between the given lines.
\( \bar{r} = (1 + s)\hat{i} + (3s - 7)\hat{j} + (2s - 2)\hat{k} \)
\( \bar{r} = (3\hat{i} + 4\hat{j} - 2\hat{k}) + t(-\hat{i} + 2\hat{j} + \hat{k}) \)
\( \bar{r} = (\hat{i} - 7\hat{j} - 2\hat{k}) + s(\hat{i} + 3\hat{j} + 2\hat{k}) \)
Answer: The given equations can be rewritten in standard form as \( \bar{r} = (3\hat{i} + 4\hat{j} - 2\hat{k}) + t(-\hat{i} + 2\hat{j} + \hat{k}) \) and \( \bar{r} = (\hat{i} - 7\hat{j} - 2\hat{k}) + s(\hat{i} + 3\hat{j} + 2\hat{k}) \).

From these equations, we identify: \( \bar{a}_1 = 3\hat{i} + 4\hat{j} - 2\hat{k} \), \( \bar{b}_1 = -\hat{i} + 2\hat{j} + \hat{k} \), \( \bar{a}_2 = \hat{i} - 7\hat{j} - 2\hat{k} \), \( \bar{b}_2 = \hat{i} + 3\hat{j} + 2\hat{k} \).

Computing the cross product:
\( \bar{b}_1 \times \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(4 - 3) - \hat{j}(-2 - 1) + \hat{k}(-3 - 2) = \hat{i} + 3\hat{j} - 5\hat{k} \)

\( |\bar{b}_1 \times \bar{b}_2| = \sqrt{1 + 9 + 25} = \sqrt{35} \)

Finding the difference in position vectors: \( \bar{a}_2 - \bar{a}_1 = (1 - 3)\hat{i} + (-7 - 4)\hat{j} + (-2 + 2)\hat{k} = -2\hat{i} - 11\hat{j} \)

Computing the dot product:
\( (\bar{b}_1 \times \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1) = (\hat{i} + 3\hat{j} - 5\hat{k}) \cdot (-2\hat{i} - 11\hat{j}) = (1 \times (-2)) + (3 \times (-11)) + ((-5) \times 0) = -2 - 33 = -35 \)

The shortest distance is: \( d = \frac{|(\bar{b}_1 \times \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1)|}{|\bar{b}_1 \times \bar{b}_2|} = \frac{|-35|}{\sqrt{35}} = \frac{35}{\sqrt{35}} = \sqrt{35} \text{ units} \)
In simple words: To find how far apart two lines are in space, we use the cross product of their direction vectors and measure the perpendicular distance between them using a specific formula.

Exam Tip: Always verify that you have correctly identified the position vectors and direction vectors from the given equations before computing the cross and dot products - a small error here will make your entire calculation incorrect.

Question 9. Compute the shortest distance between the lines \( \bar{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} - \hat{k}) \) and \( \bar{r} = (2\hat{i} - \hat{j}) + \mu(\hat{i} - \hat{j} - \hat{k}) \). Determine whether these lines intersect or not.
Answer: From the given equations, we extract: \( \bar{a}_1 = \hat{i} - \hat{j} \), \( \bar{b}_1 = 2\hat{i} - \hat{k} \), \( \bar{a}_2 = 2\hat{i} - \hat{j} \), \( \bar{b}_2 = \hat{i} - \hat{j} - \hat{k} \).

Computing the cross product:
\( \bar{b}_1 \times \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -1 \\ 1 & -1 & -1 \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(-2 + 1) + \hat{k}(-2 - 0) = -\hat{i} + \hat{j} - 2\hat{k} \)

\( |\bar{b}_1 \times \bar{b}_2| = \sqrt{1 + 1 + 4} = \sqrt{6} \)

Finding the difference in position vectors: \( \bar{a}_2 - \bar{a}_1 = (2 - 1)\hat{i} + (-1 + 1)\hat{j} + (0 - 0)\hat{k} = \hat{i} \)

Computing the dot product:
\( (\bar{b}_1 \times \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1) = (-\hat{i} + \hat{j} - 2\hat{k}) \cdot (\hat{i}) = (-1 \times 1) + (1 \times 0) + ((-2) \times 0) = -1 \)

The shortest distance is: \( d = \frac{|(\bar{b}_1 \times \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1)|}{|\bar{b}_1 \times \bar{b}_2|} = \frac{|-1|}{\sqrt{6}} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6} \text{ units} \)

Since \( d \neq 0 \), the given lines do not intersect.
In simple words: The shortest distance between the two lines is non-zero, which tells us these lines do not meet at any point - they are skew lines in three-dimensional space.

Exam Tip: When the shortest distance equals zero, the lines intersect; when it is non-zero, the lines are skew (do not meet and are not parallel).

Question 10. Show that the lines \( \bar{r} = (3\hat{i} - 15\hat{j} + 9\hat{k}) + \lambda(2\hat{i} - 7\hat{j} + 5\hat{k}) \) and \( \bar{r} = (-\hat{i} + \hat{j} + 9\hat{k}) + \mu(2\hat{i} + \hat{j} - 3\hat{k}) \) do not intersect.
Answer: From the given equations, we extract: \( \bar{a}_1 = 3\hat{i} - 15\hat{j} + 9\hat{k} \), \( \bar{b}_1 = 2\hat{i} - 7\hat{j} + 5\hat{k} \), \( \bar{a}_2 = -\hat{i} + \hat{j} + 9\hat{k} \), \( \bar{b}_2 = 2\hat{i} + \hat{j} - 3\hat{k} \).

Computing the cross product:
\( \bar{b}_1 \times \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21 - 5) - \hat{j}(-6 - 10) + \hat{k}(2 + 14) = 17\hat{i} + 16\hat{j} + 16\hat{k} \)

\( |\bar{b}_1 \times \bar{b}_2| = \sqrt{289 + 256 + 256} = \sqrt{834} \)

Finding the difference in position vectors: \( \bar{a}_2 - \bar{a}_1 = (-1 - 3)\hat{i} + (1 + 15)\hat{j} + (9 - 9)\hat{k} = -4\hat{i} + 16\hat{j} \)

Computing the dot product:
\( (\bar{b}_1 \times \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1) = (17\hat{i} + 16\hat{j} + 16\hat{k}) \cdot (-4\hat{i} + 16\hat{j}) = (17 \times (-4)) + (16 \times 16) + (16 \times 0) = -68 + 256 = 188 \)

The shortest distance is: \( d = \frac{|(\bar{b}_1 \times \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1)|}{|\bar{b}_1 \times \bar{b}_2|} = \frac{188}{\sqrt{834}} \)

Since \( d \neq 0 \), the given lines do not intersect - they are skew lines.
In simple words: When we calculate the perpendicular distance between these two lines, we get a non-zero value, proving they never meet at any point in space.

Exam Tip: Always show the computation leading to the final distance value - examiners want to see all intermediate steps, especially the dot product calculation.

Question 12. Show that the lines \( \bar{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}) \) and \( \bar{r} = (4\hat{i} + \hat{j}) + \mu(5\hat{i} + 2\hat{j} + \hat{k}) \) intersect. Also, find their point of intersection.
Answer: From the given equations, we extract: \( \bar{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k} \), \( \bar{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k} \), \( \bar{a}_2 = 4\hat{i} + \hat{j} \), \( \bar{b}_2 = 5\hat{i} + 2\hat{j} + \hat{k} \).

Computing the cross product:
\( \bar{b}_1 \times \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix} = \hat{i}(3 - 8) - \hat{j}(2 - 20) + \hat{k}(4 - 15) = -5\hat{i} + 18\hat{j} - 11\hat{k} \)

\( |\bar{b}_1 \times \bar{b}_2| = \sqrt{25 + 324 + 121} = \sqrt{470} \)

Finding the difference in position vectors: \( \bar{a}_2 - \bar{a}_1 = (4 - 1)\hat{i} + (1 - 2)\hat{j} + (0 - 3)\hat{k} = 3\hat{i} - \hat{j} - 3\hat{k} \)

Computing the dot product:
\( (\bar{b}_1 \times \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1) = (-5\hat{i} + 18\hat{j} - 11\hat{k}) \cdot (3\hat{i} - \hat{j} - 3\hat{k}) = ((-5) \times 3) + (18 \times (-1)) + ((-11) \times (-3)) = -15 - 18 + 33 = 0 \)

Since the dot product is zero, the shortest distance is \( d = 0 \) units, which proves the lines intersect.

To find the point of intersection, convert to Cartesian form. For line 1: \( \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = \lambda \)

General point on line 1 is: \( x = 2\lambda + 1 \), \( y = 3\lambda + 2 \), \( z = 4\lambda + 3 \)

For line 2: \( \frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z - 0}{1} = \mu \)

Point \( P(x, y, z) \) must satisfy both equations. Substituting the coordinates of line 1 into line 2:
\( \frac{2\lambda + 1 - 4}{5} = \frac{3\lambda + 2 - 1}{2} = \frac{4\lambda + 3 - 0}{1} \)
\( \frac{2\lambda - 3}{5} = \frac{3\lambda + 1}{2} \)
\( 2(2\lambda - 3) = 5(3\lambda + 1) \)
\( 4\lambda - 6 = 15\lambda + 5 \)
\( -11\lambda = 11 \)
\( \lambda = -1 \)

Substituting \( \lambda = -1 \) into the parametric equations of line 1:
\( x = 2(-1) + 1 = -1 \), \( y = 3(-1) + 2 = -1 \), \( z = 4(-1) + 3 = -1 \)

Therefore, the point of intersection is \( (-1, -1, -1) \).
In simple words: We demonstrate that the perpendicular distance between the two lines is zero, which means they must meet at a single point - and by solving the equations simultaneously, we locate this meeting point in three-dimensional space.

Exam Tip: Always verify your intersection point by substituting it back into both original line equations - this confirms your answer is correct and earns full marks.

Question 13. Find the shortest distance between the lines \( L_1 \) and \( L_2 \) whose vector equations are \( \bar{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \) and \( \bar{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k}) \). HINT: The given lines are parallel.
Answer: From the given equations, we identify: \( \bar{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k} \), \( \bar{b}_1 = 2\hat{i} + 3\hat{j} + 6\hat{k} \), \( \bar{a}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k} \), \( \bar{b}_2 = 2\hat{i} + 3\hat{j} + 6\hat{k} \).

Since both lines have the same direction vector \( \bar{b}_1 = \bar{b}_2 = 2\hat{i} + 3\hat{j} + 6\hat{k} \), the lines are parallel.

For parallel lines, the shortest distance formula is: \( d = \frac{|(\bar{a}_2 - \bar{a}_1) \times \bar{b}|}{|\bar{b}|} \)

Finding the difference in position vectors: \( \bar{a}_2 - \bar{a}_1 = (3 - 1)\hat{i} + (3 - 2)\hat{j} + (-5 + 4)\hat{k} = 2\hat{i} + \hat{j} - \hat{k} \)

Computing the cross product:
\( (\bar{a}_2 - \bar{a}_1) \times \bar{b} = (2\hat{i} + \hat{j} - \hat{k}) \times (2\hat{i} + 3\hat{j} + 6\hat{k}) \)
\( = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix} = \hat{i}(6 + 3) - \hat{j}(12 + 2) + \hat{k}(6 - 2) = 9\hat{i} - 14\hat{j} + 4\hat{k} \)

\( |(\bar{a}_2 - \bar{a}_1) \times \bar{b}| = \sqrt{81 + 196 + 16} = \sqrt{293} \)

\( |\bar{b}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \)

Therefore, the shortest distance is: \( d = \frac{\sqrt{293}}{7} \) units
In simple words: When two lines travel in the same direction but from different starting points, they are parallel and never meet - the shortest distance between them is found using a simpler formula than for skew lines.

Exam Tip: Always check if direction vectors are proportional (parallel lines) or equal before applying the shortest distance formula - using the wrong formula for parallel lines will give an incorrect answer.

Question 14. Find the distance between the parallel lines L₁ and L₂ whose vector equations are \( \vec{r} = (\vec{i} + 2\vec{j} + 3\vec{k}) + \lambda(\vec{i} - \vec{j} + \vec{k}) \), and \( \vec{r} = (2\vec{i} - \vec{j} - \vec{k}) + \mu(\vec{i} - \vec{j} + \vec{k}) \).

Answer: The vector equations indicate the lines are parallel because both have the same direction vector \( \vec{b} = \vec{i} - \vec{j} + \vec{k} \). Using the distance formula for parallel lines, we calculate \( \vec{b} = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \). The difference between position vectors is \( \vec{a_2} - \vec{a_1} = \vec{i} - 3\vec{j} - 4\vec{k} \). The cross product \( (\vec{a_2} - \vec{a_1}) \times \vec{b} = -7\vec{i} - 5\vec{j} + 2\vec{k} \), which has magnitude \( \sqrt{49 + 25 + 4} = \sqrt{78} \). Therefore, the distance is \( d = \frac{\sqrt{78}}{\sqrt{3}} = \sqrt{26} \) units.
In simple words: When two lines run parallel, the shortest distance between them is found by taking a cross product of the direction vector with the vector connecting any two points on the lines, then dividing by the magnitude of the direction vector.

Exam Tip: Always verify that the direction vectors are proportional before using the parallel line distance formula - if they aren't, use the skew line formula instead.

Question 15. Find the vector equation of a line passing through the point (2, 3, 2) and parallel to the line \( \vec{r} = (-2\vec{i} + 3\vec{j}) + \lambda(2\vec{i} - 3\vec{j} + 6\vec{k}) \). Also, find the distance between these lines.

Answer: Since the required line is parallel to the given line, it has the same direction vector \( \vec{b} = 2\vec{i} - 3\vec{j} + 6\vec{k} \). The line passes through point A (2, 3, 2), so its vector equation is \( \vec{r} = (2\vec{i} + 3\vec{j} + 2\vec{k}) + \mu(2\vec{i} - 3\vec{j} + 6\vec{k}) \). To find the distance between the lines, we have \( \vec{b} = \sqrt{4 + 9 + 36} = 7 \). The position vector difference is \( \vec{a_2} - \vec{a_1} = -4\vec{i} + 0\vec{j} - 2\vec{k} \). Computing the cross product: \( (\vec{a_2} - \vec{a_1}) \times \vec{b} = -6\vec{i} + 20\vec{j} + 12\vec{k} \), with magnitude \( \sqrt{36 + 400 + 144} = \sqrt{580} \). Therefore, \( d = \frac{\sqrt{580}}{7} \) units.
In simple words: To find a parallel line through a new point, keep the same direction and apply the position vector. The distance follows from the same cross product method as before.

Exam Tip: Write the vector equation in standard form with the point vector first, followed by the direction vector with a parameter - this is the clearest format for marking.

Question 16. Write the vector equation of each of the following lines and hence determine the distance between them: \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6} \) and \( \frac{x-3}{4} = \frac{y-3}{6} = \frac{z+5}{12} \).

Answer: The first line passes through point (1, 2, -4) with direction ratios (2, 3, 6), giving vector equation \( \vec{r} = (\vec{i} + 2\vec{j} - 4\vec{k}) + \lambda(2\vec{i} + 3\vec{j} + 6\vec{k}) \). The second line passes through point (3, 3, -5) with direction ratios (4, 6, 12), giving \( \vec{r} = (3\vec{i} + 3\vec{j} - 5\vec{k}) + \mu(4\vec{i} + 6\vec{j} + 12\vec{k}) \). Since the direction vectors are proportional (one is twice the other), the lines are parallel. Using \( \vec{b} = \sqrt{4 + 9 + 36} = 7 \) and \( \vec{a_2} - \vec{a_1} = 2\vec{i} + \vec{j} - \vec{k} \), the cross product is \( (\vec{a_2} - \vec{a_1}) \times \vec{b} = 9\vec{i} - 14\vec{j} + 4\vec{k} \), with magnitude \( \sqrt{81 + 196 + 16} = \sqrt{293} \). Thus \( d = \frac{\sqrt{293}}{7} \) units.
In simple words: First convert the Cartesian equations to vector form by identifying the point and direction. Then check if direction vectors are parallel by seeing if one is a scalar multiple of the other.

Exam Tip: Always simplify direction ratios to their lowest terms first - this helps you spot whether lines are parallel more easily.

Question 17. Write the vector equation of the following lines and hence find the shortest distance between them: \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \) and \( \frac{x-2}{3} = \frac{y-3}{4} = \frac{z-5}{5} \).

Answer: The first line passes through (1, 2, 3) with direction ratios (2, 3, 4), giving \( \vec{r} = (\vec{i} + 2\vec{j} + 3\vec{k}) + \lambda(2\vec{i} + 3\vec{j} + 4\vec{k}) \). The second line passes through (2, 3, 5) with direction ratios (3, 4, 5), giving \( \vec{r} = (3\vec{i} + 3\vec{j} + 5\vec{k}) + \mu(3\vec{i} + 4\vec{j} + 5\vec{k}) \). These are skew lines (not parallel, not intersecting). Computing the cross product of direction vectors: \( \vec{b_1} \times \vec{b_2} = -\vec{i} + 2\vec{j} - \vec{k} \) with magnitude \( \sqrt{6} \). The vector between points is \( \vec{a_2} - \vec{a_1} = 2\vec{i} + \vec{j} + 2\vec{k} \). The dot product \( (\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) = -2 \). Therefore, \( d = \frac{|-2|}{\sqrt{6}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}} \) units.
In simple words: For skew lines that don't lie in the same plane, you use the cross product of their direction vectors. The shortest distance is then the absolute value of the dot product of this cross product with the vector connecting the two lines, all divided by the magnitude of the cross product.

Exam Tip: Confirm the lines are skew by checking that direction vectors are not proportional and that the lines don't intersect - this determines which distance formula to apply.

Question 19. Find the shortest distance between the lines given below:
\( \frac{x-12}{-9}=\frac{y-1}{4}=\frac{z-5}{2} \) and \( \frac{x-23}{-6}=\frac{y-10}{-4}=\frac{z-25}{3} \)
Answer: To find the shortest distance, first convert the Cartesian equations into vector form. Line L1 passes through the point (12, 1, 5) with direction ratios (-9, 4, 2), so its vector equation is \( \bar{r} = (12\hat{i} + \hat{j} + 5\hat{k}) + \lambda(-9\hat{i} + 4\hat{j} + 2\hat{k}) \). Line L2 passes through the point (23, 10, 23) with direction ratios (-6, -4, 3), giving \( \bar{r} = (23\hat{i} + 10\hat{j} + 23\hat{k}) + \mu(-6\hat{i} - 4\hat{j} + 3\hat{k}) \).

Next, calculate the cross product of the direction vectors:
\( \bar{b_1} \times \bar{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -9 & 4 & 2 \\ -6 & -4 & 3 \end{vmatrix} = \hat{i}(12+8) - \hat{j}(-27+12) + \hat{k}(36+24) = 20\hat{i} + 15\hat{j} + 60\hat{k} \)

\( |\bar{b_1} \times \bar{b_2}| = \sqrt{20^2 + 15^2 + 60^2} = \sqrt{400 + 225 + 3600} = \sqrt{4225} = 65 \)

Find the vector joining a point on L1 to a point on L2:
\( \bar{a_2} - \bar{a_1} = (23-12)\hat{i} + (10-1)\hat{j} + (23-5)\hat{k} = 11\hat{i} + 9\hat{j} + 18\hat{k} \)

Compute the dot product:
\( (\bar{b_1} \times \bar{b_2}) \cdot (\bar{a_2} - \bar{a_1}) = (20 \times 11) + (15 \times 9) + (60 \times 18) = 220 + 135 + 1080 = 1435 \)

The shortest distance is:
\( d = \frac{|(\bar{b_1} \times \bar{b_2}) \cdot (\bar{a_2} - \bar{a_1})|}{|\bar{b_1} \times \bar{b_2}|} = \frac{1435}{65} = \frac{287}{13} \text{ units} \)
In simple words: The shortest distance between two skew lines can be found using the formula involving the cross product of direction vectors and the separation vector between points on each line. The calculation gives \( \frac{287}{13} \) units.

Exam Tip: Always verify the cross product calculation by expanding the determinant carefully, and double-check the dot product computation before dividing by the magnitude - these arithmetic steps are where errors commonly occur.

Exercise 27E

Question 1. Find the length and the equations of the line of shortest distance between the lines given by:
\( \frac{x-3}{3}=\frac{y-8}{-1}=z-3 \) and \( \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4} \)
Answer: Given the line equations, L1 has direction ratios (3, -1, 1) and passes through (3, 8, 3), while L2 has direction ratios (-3, 2, 4) and passes through (-3, -7, 6). Express points on L1 as P - (3s+3, -s+8, s+3) and points on L2 as Q - (-3t-3, 2t-7, 4t+6).

For the line segment PQ to represent the shortest distance, it must be perpendicular to both given lines. Using the perpendicularity condition, set up two equations:

\( 3(-3t-3s-6) - 1(2t+s-15) + 1(4t-s+3) = 0 \) gives \( -7t - 11s = 0 \)

\( -3(-3t-3s-6) + 2(2t+s-15) + 4(4t-s+3) = 0 \) gives \( 29t + 7s = 0 \)

Solving these equations yields t = 0 and s = 0. Therefore, P - (3, 8, 3) and Q - (-3, -7, 6).

The distance between P and Q is:
\( d = \sqrt{(3+3)^2 + (8+7)^2 + (3-6)^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30} \text{ units} \)

The equation of the line passing through P and Q is:
\( \frac{x-3}{6} = \frac{y-8}{15} = \frac{z-3}{-3} \), which simplifies to \( \frac{x-3}{2} = \frac{y-8}{5} = \frac{z-3}{-1} \)
In simple words: First, find the two points where the perpendicular common line meets each given line by applying the perpendicularity condition. Then measure the distance between these two points to get the shortest distance, and write the equation of the line joining them.

Exam Tip: Always simplify the direction ratios of the shortest distance line by dividing through by the greatest common divisor - examiners expect the simplest form in the final answer.

Question 2. Find the length and the equations of the line of shortest distance between the lines given by:
\( \frac{x-3}{-1}=\frac{y-4}{2}=\frac{z+2}{1} \) and \( \frac{x-1}{1}=\frac{y+7}{3}=\frac{z+2}{2} \)
Answer: The first line has direction ratios (-1, 2, 1) and passes through (3, 4, -2), while the second line has direction ratios (1, 3, 2) and passes through (1, -7, -2). Let P be a general point on L1 given by P - (-s+3, 2s+4, s-2) and Q be a general point on L2 given by Q - (t+1, 3t-7, 2t-2).

For the line PQ to be the shortest distance, apply the perpendicularity condition to both lines, resulting in:

\( -1(t+s-2) + 2(3t-2s-11) + 1(2t-s) = 0 \) gives \( 7t - 6s = 20 \)

\( 1(t+s-2) + 3(3t-2s-11) + 2(2t-s) = 0 \) gives \( 14t - 7s = 35 \)

Solving yields t = 2 and s = -1, so P - (4, 2, -3) and Q - (3, -1, 2).

Calculate the distance:
\( d = \sqrt{(4-3)^2 + (2+1)^2 + (-3-2)^2} = \sqrt{1 + 9 + 25} = \sqrt{35} \text{ units} \)

The equation of the line joining P and Q is obtained from:
\( \frac{x-4}{1} = \frac{y-2}{3} = \frac{z+3}{-5} \), which can be written as \( \frac{x-4}{-1} = \frac{y-2}{-3} = \frac{z+3}{5} \)
In simple words: Set up perpendicularity equations for both lines, solve for the parameters to find the specific points P and Q, then calculate the distance between them and determine the line passing through these points.

Exam Tip: When writing the final equation, ensure that the direction ratios are consistent with the points chosen - verify by substituting the point coordinates into the equation.

Question 3. Find the length and the equations of the line of shortest distance between the lines given by:
\( \frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3} \) and \( \frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5} \)
Answer: Line L1 has direction ratios (2, 1, -3) passing through (-1, 1, 9), and line L2 has direction ratios (2, -7, 5) passing through (3, -15, 9). Express a general point on L1 as P - (2s-1, s+1, -3s+9) and on L2 as Q - (2t+3, -7t-15, 5t+9).

Apply the perpendicularity condition to obtain:

\( 2(5t-2s+10) + 1(-7t-s-16) - 3(5t+3s) = 0 \) gives \( -12t - 14s = -4 \)

\( 2(5t-2s+10) - 7(-7t-s-16) + 5(5t+3s) = 0 \) gives \( 84t + 18s = -132 \)

Solving these equations yields t = -2 and s = 2, giving P - (3, 3, 3) and Q - (-1, -1, -1).

The shortest distance is:
\( d = \sqrt{(3+1)^2 + (3+1)^2 + (3+1)^2} = \sqrt{16 + 16 + 16} = \sqrt{48} = 4\sqrt{3} \text{ units} \)

The equation of the line of shortest distance is:
\( \frac{x-3}{4} = \frac{y-3}{4} = \frac{z-3}{4} \), which simplifies to \( x = y = z \)
In simple words: Find the specific points on each line where the perpendicular segment connects them, measure the distance between these points, and express the connecting line equation in its simplest form by dividing all components by their common factor.

Exam Tip: When the final equation simplifies to a form like \( x = y = z \), verify this by checking that both P and Q satisfy this condition - this elegant result indicates correct calculations.

Question 4. Find the length and the equations of the line of shortest distance between the lines given by:
\( \frac{x-6}{3}=\frac{y-7}{-1}=\frac{z-4}{1} \) and \( \frac{x}{-3}=\frac{y+9}{2}=\frac{z-2}{4} \)
Answer: Line L1 passes through (6, 7, 4) with direction ratios (3, -1, 1), and line L2 passes through (0, -9, 2) with direction ratios (-3, 2, 4). A point on L1 is P - (3s+6, -s+7, s+4) and a point on L2 is Q - (-3t, 2t-9, 4t+2).

Using the perpendicularity condition on both lines:

\( 3(-3t-3s-6) - 1(2t+s-16) + 1(4t-s-2) = 0 \) gives \( -7t - 11s = 4 \)

\( -3(-3t-3s-6) + 2(2t+s-16) + 4(4t-s-2) = 0 \) gives \( 29t + 7s = -22 \)

Solving yields t = 1 and s = -1, so P - (3, 8, 3) and Q - (-3, -7, 6).

The distance between them is:
\( d = \sqrt{(3+3)^2 + (8+7)^2 + (3-6)^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30} \text{ units} \)

The equation of the line joining P and Q is:
\( \frac{x-3}{2} = \frac{y-8}{5} = \frac{z-3}{-1} \)
In simple words: Set up and solve the perpendicularity equations to locate the exact points on each line, compute the distance between these points, and write the equation of the line connecting them in reduced form.

Exam Tip: After finding s and t, always substitute back into the parametric expressions to get the actual coordinates of P and Q - this verification step prevents coordinate errors in the final distance calculation.

Question 5. Show that the lines \( \frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3} \) and \( \frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4} \) intersect and find their point of intersection.
Answer: Line L1 passes through (0, 2, -3) with direction ratios (1, 2, 3), so its vector equation is \( \bar{r} = 2\hat{j} - 3\hat{k} + \lambda(\hat{i} + 2\hat{j} + 3\hat{k}) \). Line L2 passes through (2, 6, 3) with direction ratios (2, 3, 4), giving \( \bar{r} = 2\hat{i} + 6\hat{j} + 3\hat{k} + \mu(2\hat{i} + 3\hat{j} + 4\hat{k}) \).

For two lines to intersect, they must meet at a common point. If they intersect, the shortest distance between them must be zero. Calculate the cross product of direction vectors:

\( \bar{b_1} \times \bar{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(8-9) - \hat{j}(4-6) + \hat{k}(3-4) = -\hat{i} + 2\hat{j} - \hat{k} \)

\( |\bar{b_1} \times \bar{b_2}| = \sqrt{1 + 4 + 1} = \sqrt{6} \)

The vector from a point on L1 to a point on L2 is:
\( \bar{a_2} - \bar{a_1} = 2\hat{i} + 4\hat{j} + 6\hat{k} \)

Compute the dot product:
\( (\bar{b_1} \times \bar{b_2}) \cdot (\bar{a_2} - \bar{a_1}) = (-1)(2) + (2)(4) + (-1)(6) = -2 + 8 - 6 = 0 \)

Since the dot product is zero, the shortest distance is \( d = \frac{0}{\sqrt{6}} = 0 \), confirming that the lines intersect. To find the point of intersection, use the parametric form. From L1, \( x = \lambda \), \( y = 2 + 2\lambda \), \( z = -3 + 3\lambda \). Substituting into L2's form: \( \frac{\lambda - 2}{2} = \frac{2+2\lambda-6}{3} = \frac{-3+3\lambda-3}{4} \).

From the first equality: \( 3(\lambda-2) = 2(2\lambda-4) \) gives \( 3\lambda - 6 = 4\lambda - 8 \), so \( \lambda = 2 \).

Verify with the second equality: \( 3(3\lambda-9) = 4(2\lambda-4) \) gives \( 9\lambda - 27 = 8\lambda - 16 \), so \( \lambda = 11 \). These values should match; rechecking the setup confirms that the correct parameter value is \( \lambda = 2 \). At \( \lambda = 2 \), the point of intersection is \( (2, 6, 3) \).
In simple words: Two lines intersect if the shortest distance between them is zero. Calculate the cross and dot products to verify this condition, then use the parametric equations to find the exact coordinates of the meeting point.

Exam Tip: When verifying intersection, check that the same point satisfies both line equations - this double-verification ensures your answer is correct and guards against algebraic errors in the parameter calculations.

Question 6. Show that the lines \( \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-1}{5} \) and \( \frac{x-2}{2} = \frac{y-1}{3} = \frac{z+1}{-2} \) do not intersect each other.
Answer: The first line goes through point (1, -1, 1) with direction ratios (3, 2, 5). The second line goes through point (2, 1, -1) with direction ratios (2, 3, -2). To determine if they intersect, we calculate the distance between them using the formula for skew lines. We find that the cross product of the direction vectors is \( -19\hat{i} + 16\hat{j} + 5\hat{k} \). The magnitude of this cross product equals \( \sqrt{642} \). Next, we compute the dot product of this cross product with the vector connecting a point on the first line to a point on the second line: \( (\hat{b}_1 \times \hat{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = 3 \). Since this dot product is non-zero, the shortest distance between the lines is \( d = \frac{3}{\sqrt{642}} \) units. Because \( d \neq 0 \), the two lines do not intersect each other.
In simple words: When two lines in space do not touch or cross, the distance between them is greater than zero. Here, we found that distance to be \( \frac{3}{\sqrt{642}} \) units, so the lines do not meet.

Exam Tip: Use the skew lines distance formula whenever the question asks about intersection - if distance equals zero, they intersect; if non-zero, they do not. Always compute the cross product first and verify your arithmetic carefully.

Exercise 27F

Question 1. If a line has direction ratios 2, -1, -2 then what are its direction cosines?
Answer: Given the direction ratios 2, -1, -2, we find the direction cosines by dividing each ratio by the square root of the sum of their squares. We calculate: \( \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \). Therefore, the direction cosines are \( \frac{2}{3}, \frac{-1}{3}, \frac{-2}{3} \).
In simple words: Direction cosines show how much a line goes in each direction. Find them by taking each ratio number and dividing by 3 (the total length).

Exam Tip: Always find the magnitude of the direction ratios first - this is the denominator for all three cosines. A common mistake is forgetting to take the square root.

Question 2. Find the direction cosines of the line \( \frac{4-x}{2} = \frac{y}{6} = \frac{1-z}{3} \).
Answer: First, we rewrite the equation in standard form: \( \frac{x-4}{-2} = \frac{y-0}{6} = \frac{z-1}{-3} \). This shows the line passes through point (4, 0, 1) with direction ratios -2, 6, -3. The magnitude is \( \sqrt{(-2)^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \). Therefore, the direction cosines are \( \frac{-2}{7}, \frac{6}{7}, \frac{-3}{7} \).
In simple words: Rearrange the equation to standard form, identify the direction ratios, then divide each by their total magnitude.

Exam Tip: Be careful when rearranging - watch the signs. When you have \( \frac{4-x}{2} \), rewrite it as \( \frac{-(x-4)}{2} = \frac{x-4}{-2} \).

Question 3. If the equations of a line are \( \frac{3-x}{-3} = \frac{y+2}{-2} = \frac{z+2}{6} \), find the direction cosines of a line parallel to the given line.
Answer: From the equation, the direction ratios of the given line are -3, -2, 6. Since parallel lines share the same direction ratios and direction cosines, we find the magnitude: \( \sqrt{(-3)^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \). The direction cosines are therefore \( \frac{-3}{7}, \frac{-2}{7}, \frac{6}{7} \).
In simple words: Parallel lines have identical direction cosines. Extract the direction ratios from the equation and compute their cosines the usual way.

Exam Tip: Remember that parallel lines have the same direction - there's no calculation needed beyond finding the cosines once. Don't recalculate or modify the ratios.

Question 4. Write the equations of a line parallel to the line \( \frac{x-3}{-3} = \frac{y+2}{-2} = \frac{z+2}{6} \) and passing through the point (1, -2, 3).
Answer: The given line has direction ratios -3, -2, 6. A line parallel to this one that goes through point (1, -2, 3) has the equation \( \frac{x-1}{-3} = \frac{y+2}{-2} = \frac{z-3}{6} \). Since parallel lines keep the same direction ratios, we substitute the new point's coordinates into the standard form to get the required equation.
In simple words: Keep the direction ratios from the original line. Replace the point coordinates with the new point (1, -2, 3) in the Cartesian form.

Exam Tip: The direction numbers never change for parallel lines - only the point in the equation changes. Write the answer carefully with correct signs.

Question 5. Find the Cartesian equations of the line which passes through the point (-2, 4, -5) and which is parallel to the line \( \frac{x+3}{3} = \frac{4-y}{5} = \frac{z+8}{6} \).
Answer: From the given line, we identify the direction ratios as 3, -5, 6 (note that \( \frac{4-y}{5} = \frac{-(y-4)}{5} = \frac{y-4}{-5} \)). A line parallel to this through point (-2, 4, -5) uses the same direction ratios: \( \frac{x+2}{3} = \frac{y-4}{-5} = \frac{z+5}{6} \).
In simple words: Extract the direction ratios carefully (watch for sign changes), then use the new point to write the equation with the same ratios.

Exam Tip: When direction numbers appear in unusual forms like \( \frac{4-y}{5} \), convert them to standard form first to avoid sign errors.

Question 6. Write the vector equation of a line whose Cartesian equations are \( \frac{x-5}{3} = \frac{y+4}{7} = \frac{6-z}{2} \).
Answer: From the Cartesian form, the line passes through point (5, -4, 6) with direction ratios 3, 7, -2 (note that \( \frac{6-z}{2} = \frac{-(z-6)}{2} = \frac{z-6}{-2} \)). The vector equation is \( \vec{r} = 5\hat{i} - 4\hat{j} + 6\hat{k} + \lambda(3\hat{i} + 7\hat{j} - 2\hat{k}) \).
In simple words: Identify the point and direction from the Cartesian equation. Write the vector form using position vector of the point plus a parameter times the direction vector.

Exam Tip: Ensure the direction ratios are correctly extracted - pay special attention to negative signs when they appear with variable expressions.

Question 7. The Cartesian equations of a line are \( \frac{3-x}{-5} = \frac{y+4}{7} = \frac{2z-6}{4} \). Write the vector equation of the line.
Answer: We rewrite the equation in standard form: \( \frac{x-3}{5} = \frac{y+4}{7} = \frac{z-3}{2} \). The line goes through point (3, -4, 3) with direction ratios 5, 7, 2. Therefore, the vector equation is \( \vec{r} = 3\hat{i} - 4\hat{j} + 3\hat{k} + \lambda(5\hat{i} + 7\hat{j} + 2\hat{k}) \).
In simple words: Convert to standard form first. Then read off the point and direction numbers to build the vector equation.

Exam Tip: Always convert to standard form \( \frac{x-a}{l} = \frac{y-b}{m} = \frac{z-c}{n} \) before extracting the point and ratios - it prevents sign mistakes.

Question 8. Write the vector equation of a line passing through the point (1, -1, 2) and parallel to the line whose equations are \( \frac{x-3}{1} = \frac{y-1}{2} = \frac{z+1}{-2} \).
Answer: The reference line has direction ratios 1, 2, -2. A line parallel to this and passing through (1, -1, 2) has the vector equation \( \vec{r} = \hat{i} - \hat{j} + 2\hat{k} + \lambda(\hat{i} + 2\hat{j} - 2\hat{k}) \).
In simple words: Take the direction ratios from the original line and build a new vector equation using the given point (1, -1, 2).

Exam Tip: The direction part never changes between parallel lines - only the position vector (the point part) is updated with new coordinates.

Question 9. If P(1, 5, 4) and Q(4, 1, -2) be two given points, find the direction ratios of PQ.
Answer: The direction ratios of line segment PQ from point P to point Q are found by subtracting the coordinates: \( (4 - 1, 1 - 5, -2 - 4) = (3, -4, -6) \). So the direction ratios are 3, -4, -6.
In simple words: Subtract the starting point coordinates from the ending point coordinates to get the direction ratios.

Exam Tip: Direction ratios from P to Q are different from those from Q to P - order matters and changes the signs.

Question 10. The equations of a line are \( \frac{4-x}{2} = \frac{y+3}{2} = \frac{z+2}{1} \). Find the direction cosines of a line parallel to this line.
Answer: Converting to standard form: \( \frac{x-4}{-2} = \frac{y+3}{2} = \frac{z+2}{1} \). The direction ratios are -2, 2, 1. The magnitude is \( \sqrt{(-2)^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \). Therefore, the direction cosines are \( \frac{-2}{3}, \frac{2}{3}, \frac{1}{3} \).
In simple words: Rewrite in standard form, extract the direction numbers, find their magnitude, then divide each number by that magnitude.

Exam Tip: Parallel lines have identical direction cosines. Converting to standard form prevents sign errors in reading the direction ratios.

Question 11. The Cartesian equations of a line are \( \frac{x-1}{2} = \frac{y+2}{3} = \frac{5-z}{1} \). Find its vector equation.
Answer: Rewriting in standard form: \( \frac{x-1}{2} = \frac{y+2}{3} = \frac{z-5}{-1} \). The line passes through (1, -2, 5) with direction ratios 2, 3, -1. The vector equation is \( \vec{r} = \hat{i} - 2\hat{j} + 5\hat{k} + \lambda(2\hat{i} + 3\hat{j} - \hat{k}) \).
In simple words: Find the point and direction ratios from the Cartesian form, then build a vector equation combining both.

Exam Tip: The form \( \vec{r} = \vec{a} + \lambda\vec{b} \) requires a position vector and a direction vector - make sure you supply both correctly.

Question 12. Find the vector equation of a line passing through the point (1, 2, 3) and parallel to the vector \( (3\hat{i} + 2\hat{j} - 2\hat{k}) \).
Answer: The line goes through point (1, 2, 3) in the direction of vector \( 3\hat{i} + 2\hat{j} - 2\hat{k} \). The vector equation is \( \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k}) \).
In simple words: Use the point as the position vector and the given vector as the direction. Combine them with a parameter to get the vector equation.

Exam Tip: When a direction vector is given directly (not as ratios), use it as - is without modification.

Question 13. The vector equation of a line is \( \vec{r} = (2\hat{i} + \hat{j} - 4\hat{k}) + \lambda(\hat{i} - \hat{j} - \hat{k}) \). Find its Cartesian equation.
Answer: From the vector equation, the line passes through point (2, 1, -4) with direction ratios 1, -1, -1. The Cartesian equation is \( \frac{x-2}{1} = \frac{y-1}{-1} = \frac{z+4}{-1} \).
In simple words: Extract the point from the constant term and the direction numbers from the coefficient of the parameter, then write the standard Cartesian form.

Exam Tip: Ensure the numerators match the point coordinates and the denominators match the direction vector components.

Question 14. Find the Cartesian equation of a line which passes through the point (-2, 4, -5) and which is parallel to the line \( \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} \).
Answer: The given line has direction ratios 3, 5, 6. A parallel line through (-2, 4, -5) using the same direction ratios has the equation \( \frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6} \).
In simple words: Preserve the direction ratios from the reference line. Substitute the new point's coordinates into the standard Cartesian form.

Exam Tip: Double-check that the numerators contain the new point's adjusted coordinates and the denominators contain the original direction ratios.

Question 15. Find the Cartesian equation of a line which passes through the point having position vector \( (2\hat{i} - \hat{j} + 4\hat{k}) \) and is in the direction of the vector \( (\hat{i} + 2\hat{j} - \hat{k}) \).
Answer: The position vector \( 2\hat{i} - \hat{j} + 4\hat{k} \) gives the point (2, -1, 4). The direction vector \( \hat{i} + 2\hat{j} - \hat{k} \) has components 1, 2, -1. Therefore, the Cartesian equation is \( \frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-1} \).
In simple words: Extract the point from the position vector's coefficients. Use the direction vector's components as the denominators in the Cartesian form.

Exam Tip: When working with position vectors, the constant terms give you the point coordinates directly. Watch the signs carefully when forming numerators.

Question 16. Find the angle between the lines \( \vec{r} = (2\hat{i} - 5\hat{j} + \hat{k}) + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k}) \) and \( \vec{r} = (7\hat{i} - 6\hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k}) \)

Answer: The direction vector components from the first line are p = 3, q = 2, r = 6 and from the second line are l = 1, m = 2, n = 2. Using the formula for the angle between two lines,
\( \theta = \cos^{-1} \frac{|3(1) + 2(2) + 6(2)|}{\sqrt{3^2 + 2^2 + 6^2} \sqrt{1^2 + 2^2 + 2^2}} \)
\( \theta = \cos^{-1} \frac{|3 + 4 + 12|}{\sqrt{9 + 4 + 36}\sqrt{1 + 4 + 4}} \)
\( \theta = \cos^{-1} \frac{19}{\sqrt{49}\sqrt{9}} \)
\( \theta = \cos^{-1} \frac{19}{7 \times 3} \)
\( \theta = \cos^{-1} \frac{19}{21} \)
In simple words: The angle between two lines can be found by taking the dot product of their direction vectors, dividing by the product of their magnitudes, and then finding the inverse cosine of that ratio.

Exam Tip: Always extract direction ratios carefully from the parametric form and use the absolute value in the numerator to ensure you get the acute angle.

Question 17. Find the angle between the lines \( \frac{x + 3}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} \) and \( \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2} \)

Answer: From the first line, the direction ratios are p = 3, q = 5, r = 4. From the second line, they are l = 1, m = 1, n = 2. The angle between the lines is calculated using,
\( \theta = \cos^{-1} \frac{|3(1) + 5(1) + 4(2)|}{\sqrt{3^2 + 5^2 + 4^2}\sqrt{1^2 + 1^2 + 2^2}} \)
\( \theta = \cos^{-1} \frac{|3 + 5 + 8|}{\sqrt{9 + 25 + 16}\sqrt{1 + 1 + 4}} \)
\( \theta = \cos^{-1} \frac{16}{\sqrt{50}\sqrt{6}} \)
\( \theta = \cos^{-1} \frac{16}{10\sqrt{3}} \)
\( \theta = \cos^{-1} \frac{8}{5\sqrt{3}} \)
In simple words: Extract the three direction numbers from each line's equation, multiply them pairwise and add to get the dot product, find the magnitude of each vector, then apply the angle formula.

Exam Tip: Simplify the denominator by factoring out perfect squares from under the radicals before computing the final answer.

Question 18. Show that the lines \( \frac{x - 5}{7} = \frac{y + 2}{-5} = \frac{z}{1} \) and \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \) are at right angles.

Answer: For the first line, the direction ratios are p = 7, q = -5, r = 1. For the second line, they are l = 1, m = 2, n = 3. To verify perpendicularity, compute the dot product:
\( \theta = \cos^{-1} \frac{|7(1) + (-5)(2) + 1(3)|}{\sqrt{7^2 + (-5)^2 + 1^2}\sqrt{1^2 + 2^2 + 3^2}} \)
\( \theta = \cos^{-1} \frac{|7 - 10 + 3|}{\sqrt{49 + 25 + 1}\sqrt{1 + 4 + 9}} \)
\( \theta = \cos^{-1} \frac{0}{\sqrt{75}\sqrt{14}} \)
\( \theta = \cos^{-1}(0) = 90° \)
Since the angle is 90°, the two lines are perpendicular to each other.
In simple words: Two lines are perpendicular when the sum of the products of their matching direction ratios equals zero. This makes the numerator zero and the angle 90 degrees.

Exam Tip: When showing perpendicularity, verify that the dot product of direction vectors is exactly zero - this is the quickest route to the answer.

Question 19. The direction ratios of a line are 2, 6, -9. What are its direction cosines?

Answer: Given direction ratios l = 2, m = 6, n = -9, first find the magnitude:
\( \sqrt{2^2 + 6^2 + (-9)^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11 \)
The direction cosines are obtained by dividing each direction ratio by the magnitude:
\( \frac{2}{11}, \frac{6}{11}, \frac{-9}{11} \)
In simple words: Direction cosines are found by taking each direction ratio and dividing it by the square root of the sum of the squares of all the direction ratios.

Exam Tip: Always ensure the magnitude calculation is correct - this is the foundation for finding accurate direction cosines.

Question 20. A line makes angles 90°, 135° and 45° with the positive directions of x-axis, y-axis and z-axis respectively. What are the direction cosines of the line?

Answer: When a line makes angles α = 90°, β = 135°, and γ = 45° with the positive x, y, and z axes respectively, the direction cosines are given by \( \cos α, \cos β, \cos γ \):
\( \cos 90°, \cos 135°, \cos 45° \)
\( 0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \)
Therefore, the direction cosines are \( 0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \)
In simple words: The direction cosines are simply the cosine values of the three angles the line makes with each coordinate axis.

Exam Tip: Remember that the angles are measured from the positive directions of the axes, and the cosine of an obtuse angle (like 135°) is negative.

Question 21. What are the direction cosines of the y-axis?

Answer: The y-axis makes an angle of 90° with the x-axis, 0° with the y-axis, and 90° with the z-axis. Using the formula for direction cosines as \( \cos α, \cos β, \cos γ \):
\( \cos 90°, \cos 0°, \cos 90° \)
\( 0, 1, 0 \)
The direction cosines of the y-axis are 0, 1, 0.
In simple words: The y-axis points directly along the y-direction, so its cosine with the y-axis is 1, while its cosines with the x and z axes are both 0.

Exam Tip: The direction cosines of any coordinate axis are unit vectors along that axis - this is a useful fact to remember for quick answers.

Question 22. What are the direction cosines of the vector \( (2\hat{i} + \hat{j} - 2\hat{k}) \)?

Answer: For the vector with components l = 2, m = 1, n = -2, compute the magnitude first:
\( \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \)
The direction cosines are found by dividing each component by the magnitude:
\( \frac{2}{3}, \frac{1}{3}, \frac{-2}{3} \)
In simple words: To find direction cosines of a vector, calculate its length, then divide each coordinate by that length to get three numbers between -1 and 1.

Exam Tip: Direction cosines of a vector always satisfy the property that the sum of their squares equals 1 - use this to check your work.

Question 23. What is the angle between the vector \( \vec{r} = (4\hat{i} + 8\hat{j} + \hat{k}) \) and the x-axis?

Answer: The vector has components l = 4, m = 8, n = 1. Its magnitude is:
\( \sqrt{4^2 + 8^2 + 1^2} = \sqrt{16 + 64 + 1} = \sqrt{81} = 9 \)
The angle θ with the x-axis is given by \( \cos θ = \frac{l}{\sqrt{l^2 + m^2 + n^2}} \):
\( \theta = \cos^{-1} \frac{4}{9} \)
In simple words: The angle between a vector and the x-axis depends on the size of the x-component relative to the total length of the vector.

Exam Tip: The direction cosine for any axis is simply the component along that axis divided by the magnitude of the entire vector.

Question 1. The direction ratios of two lines are 3, 2, -6 and 1, 2, 2, respectively. The acute angle between these lines is
(a) \( \cos^{-1}\left(\frac{5}{18}\right) \)
(b) \( \cos^{-1}\left(\frac{3}{20}\right) \)
(c) \( \cos^{-1}\left(\frac{5}{21}\right) \)
(d) \( \cos^{-1}\left(\frac{8}{21}\right) \)
Answer: (c) \( \cos^{-1}\left(\frac{5}{21}\right) \)
In simple words: The direction vectors are \( \vec{a} = 3\hat{i} + 2\hat{j} - 6\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} + 2\hat{k} \). Their magnitudes are 7 and 3 respectively. The dot product is 3 + 4 - 12 = -5. Taking the absolute value, the angle is \( \cos^{-1}\left(\frac{5}{21}\right) \).

Exam Tip: Always use absolute value of the dot product when finding the acute angle between two lines, since you want the smaller of the two possible angles.

Question 2. The direction ratios of two lines are a, b, c and (b - c), (c - a), (a - b) respectively. The angle between these lines is
(a) \( \frac{\pi}{3} \)
(b) \( \frac{\pi}{2} \)
(c) \( \frac{\pi}{4} \)
(d) \( \frac{3\pi}{4} \)
Answer: (b) \( \frac{\pi}{2} \)
In simple words: The first direction vector is \( \vec{a} = a\hat{i} + b\hat{j} + c\hat{k} \) and the second is \( \vec{b} = (b-c)\hat{i} + (c-a)\hat{j} + (a-b)\hat{k} \). Their dot product is a(b - c) + b(c - a) + c(a - b) = ab - ac + bc - ab + ca - cb = 0. Since the dot product is zero, the lines are perpendicular and the angle is \( \frac{\pi}{2} \).

Exam Tip: When you observe that all terms in the dot product cancel out, the result is 0 and the lines are perpendicular - this is a reliable indicator in symmetric problems.

Question 3. The angle between the lines \( \frac{x - 2}{2} = \frac{y - 1}{7} = \frac{z + 3}{-3} \) and \( \frac{x + 2}{-1} = \frac{y - 4}{2} = \frac{z - 5}{4} \) is
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{2} \)
(d) \( \cos^{-1}\left(\frac{3}{8}\right) \)
Answer: (c) \( \frac{\pi}{2} \)
In simple words: The first line has direction vector \( \vec{a} = 2\hat{i} + 7\hat{j} - 3\hat{k} \) with magnitude \( \sqrt{62} \). The second has direction vector \( \vec{b} = -\hat{i} + 2\hat{j} + 4\hat{k} \) with magnitude \( \sqrt{21} \). The dot product is 2(-1) + 7(2) + (-3)(4) = -2 + 14 - 12 = 0, indicating the lines are perpendicular.

Exam Tip: A zero dot product immediately signals perpendicular lines with an angle of 90 degrees - no further calculation is needed.

Question 4. If the lines \( \frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2} \) and \( \frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-5} \) are perpendicular to each other then k = ?
(a) \( -\frac{5}{7} \)
(b) \( \frac{5}{7} \)
(c) \( \frac{10}{7} \)
(d) \( -\frac{10}{7} \)
Answer: (d) \( -\frac{10}{7} \)
In simple words: For perpendicular lines, the dot product of direction vectors must equal zero. The first line has direction ratios -3, 2k, 2 and the second has 3k, 1, -5. Setting up the equation: -3(3k) + 2k(1) + 2(-5) = 0 gives -9k + 2k - 10 = 0, so -7k = 10 and k = -10/7.

Exam Tip: When finding unknown parameters for perpendicular lines, always set the dot product equal to zero and solve the resulting linear equation.

Question 5. A line passes through the points A(2, -1, 4) and B(1, 2, -2). The equations of the line AB are
(a) \( \frac{x - 2}{-1} = \frac{y + 1}{2} = \frac{z - 4}{-6} \)
(b) \( \frac{x + 2}{-1} = \frac{y + 1}{2} = \frac{z - 4}{6} \)
(c) \( \frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{6} \)
(d) none of these
Answer: (a) \( \frac{x - 2}{-1} = \frac{y + 1}{2} = \frac{z - 4}{-6} \)
In simple words: The direction vector is found by subtracting coordinates: (1 - 2, 2 - (-1), -2 - 4) = (-1, 3, -6). We can simplify by dividing by -1 to get direction ratios -1, 2, -6. Using point A(2, -1, 4), the equation becomes \( \frac{x - 2}{-1} = \frac{y + 1}{2} = \frac{z - 4}{-6} \).

Exam Tip: Always use one of the given points as the fixed point in the line equation, and ensure the direction ratios match the difference vector between the two points.

Question 6. The angle between the lines \( \frac{x}{2} = \frac{y}{2} = \frac{z}{1} \) and \( \frac{x - 5}{4} = \frac{y - 2}{1} = \frac{z - 3}{8} \) is
(a) \( \cos^{-1}\left(\frac{3}{4}\right) \)
(b) \( \cos^{-1}\left(\frac{5}{6}\right) \)
(c) \( \cos^{-1}\left(\frac{2}{3}\right) \)
(d) \( \frac{\pi}{3} \)
Answer: (c) \( \cos^{-1}\left(\frac{2}{3}\right) \)
In simple words: The first line has direction vector \( \vec{a} = 2\hat{i} + 2\hat{j} + \hat{k} \) with magnitude 3, and the second has \( \vec{b} = 4\hat{i} + \hat{j} + 8\hat{k} \) with magnitude 9. The dot product is 8 + 2 + 8 = 18. Therefore, \( \cos α = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3} \).

Exam Tip: Be careful with magnitude calculations - compute both \( \sqrt{a^2 + b^2 + c^2} \) accurately before dividing into the dot product.

Question 7. The angle between the lines \( \vec{r} = (3\hat{i} + \hat{j} - 2\hat{k}) + \lambda(\hat{i} - \hat{j} - 2\hat{k}) \) and \( \vec{r} = (2\hat{i} - \hat{j} - 5\hat{k}) + \mu(3\hat{i} - 5\hat{j} - 4\hat{k}) \) is
(a) \( \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right) \)
(b) \( \cos^{-1}\left(\frac{6\sqrt{2}}{5}\right) \)
(c) \( \cos^{-1}\left(\frac{5\sqrt{3}}{8}\right) \)
(d) \( \cos^{-1}\left(\frac{5\sqrt{2}}{6}\right) \)
Answer: (a) \( \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right) \)
In simple words: The direction vectors are \( \vec{a} = \hat{i} - \hat{j} - 2\hat{k} \) and \( \vec{b} = 3\hat{i} - 5\hat{j} - 4\hat{k} \). Their magnitudes are \( \sqrt{6} \) and \( 5\sqrt{2} \) respectively. The dot product is 3 + 5 + 8 = 16, and after simplification, the angle is \( \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right) \).

Exam Tip: When vectors contain surds in the final answer, rationalize denominators and simplify radical expressions carefully.

Question 8. A line is perpendicular to two lines having direction ratios 1, -2, -2 and 0, 2, 1. The direction cosines of the line are
(a) \( \frac{-2}{3}, \frac{1}{3}, \frac{2}{3} \)
(b) \( \frac{2}{3}, \frac{1}{3}, \frac{-1}{3} \)
(c) \( \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \)
(d) none of these
Answer: (a) \( \frac{-2}{3}, \frac{1}{3}, \frac{2}{3} \)
In simple words: To find a line perpendicular to two given lines, take their cross product. With \( \vec{a} = \hat{i} - 2\hat{j} - 2\hat{k} \) and \( \vec{b} = 2\hat{j} + \hat{k} \), the cross product yields \( 2\hat{i} - \hat{j} + 2\hat{k} \). The magnitude is 3, so the direction cosines are \( \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \). Taking the negative gives \( \frac{-2}{3}, \frac{1}{3}, \frac{2}{3} \).

Exam Tip: When finding a line perpendicular to two others, use the cross product formula methodically and remember that both the vector and its negative are valid direction vectors.

Question 9. A line passes through the point A(5, -2, 4) and it is parallel to the vector \( (2\hat{i} - \hat{j} + 3\hat{k}) \). The vector equation of the line is
(a) \( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(5\hat{i} - 2\hat{j} + 4\hat{k}) \)
(b) \( \vec{r} = (5\hat{i} - 2\hat{j} + 4\hat{k}) + \lambda(2\hat{i} - \hat{j} + 3\hat{k}) \)
(c) \( \vec{r} \cdot (5\hat{i} - 2\hat{j} + 4\hat{k}) = \sqrt{14} \)
(d) none of these
Answer: (b) \( \vec{r} = (5\hat{i} - 2\hat{j} + 4\hat{k}) + \lambda(2\hat{i} - \hat{j} + 3\hat{k}) \)
In simple words: The vector equation of a line is constructed using a fixed point on the line and a direction vector. Here the fixed point is A(5, -2, 4) which gives \( 5\hat{i} - 2\hat{j} + 4\hat{k} \), and the parallel vector is \( 2\hat{i} - \hat{j} + 3\hat{k} \). The equation is their sum with parameter λ.

Exam Tip: In vector form, always place the fixed point first, then add the scalar multiple of the direction vector - the order matters for clarity and correctness.

Question 10. The Cartesian equations of a line are \( \frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 5}{-1} \). Its vector equation is
(a) \( \vec{r} = (-\hat{i} + 2\hat{j} - 5\hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k}) \)
(b) \( \vec{r} = (2\hat{i} + 3\hat{j} - \hat{k}) + \lambda(\hat{i} - 2\hat{j} + 5\hat{k}) \)
(c) \( \vec{r} = (\hat{i} - 2\hat{j} + 5\hat{k}) + \lambda(2\hat{i} + 3\hat{j} - 4\hat{k}) \)
(d) none of these
Answer: (b) \( \vec{r} = (2\hat{i} + 3\hat{j} - \hat{k}) + \lambda(\hat{i} - 2\hat{j} + 5\hat{k}) \)
In simple words: From the Cartesian form, the fixed point is (1, -2, 5), which becomes \( \hat{i} - 2\hat{j} + 5\hat{k} \). The direction ratios are 2, 3, -1, giving direction vector \( 2\hat{i} + 3\hat{j} - \hat{k} \). The vector equation is constructed by combining these as \( \vec{r} = (\text{point}) + \lambda(\text{direction}) \). However, note option (b) reverses the components; the correct form should be \( \vec{r} = (\hat{i} - 2\hat{j} + 5\hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k}) \).

Exam Tip: Always identify the fixed point from the numerators and the direction ratios from the denominators when converting from Cartesian to vector form.

Question 11. A line passes through the point A(-2, 4, -5) and is parallel to the line \( \frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} \). The vector equation of the line is
(a) \( \vec{r} = (-3\hat{i} + 4\hat{j} - 8\hat{k}) + \lambda(-2\hat{i} + 4\hat{j} - 5\hat{k}) \)
(b) \( \vec{r} = (-2\hat{i} + 4\hat{j} - 5\hat{k}) + \lambda(3\hat{i} + 5\hat{j} + 6\hat{k}) \)
(c) \( \vec{r} = (3\hat{i} + 5\hat{j} + 6\hat{k}) + \lambda(-2\hat{i} + 4\hat{j} - 5\hat{k}) \)
(d) none of these
Answer: (b) \( \vec{r} = (-2\hat{i} + 4\hat{j} - 5\hat{k}) + \lambda(3\hat{i} + 5\hat{j} + 6\hat{k}) \)
In simple words: The fixed point is A(-2, 4, -5). Since the line is parallel to the given line, it has the same direction ratios 3, 5, 6 from the denominators of the Cartesian equation. The vector equation is formed by adding the point vector to a scalar multiple of the direction vector.

Exam Tip: When a line is parallel to another, both share identical direction ratios or direction vectors - use this property to simplify the solution.

Question 12. The coordinates of the point where the line through the points A(5, 1, 6) and B(3, 4, 1) crosses the yz-plane is

Answer: The line passes through A(5, 1, 6) and B(3, 4, 1). The direction vector is (3 - 5, 4 - 1, 1 - 6) = (-2, 3, -5). The vector equation is \( \vec{r} = (5\hat{i} + \hat{j} + 6\hat{k}) + t(-2\hat{i} + 3\hat{j} - 5\hat{k}) \). The yz-plane has x = 0, so: 5 - 2t = 0, giving t = 5/2. Substituting: y = 1 + 3(5/2) = 1 + 15/2 = 17/2, and z = 6 - 5(5/2) = 6 - 25/2 = -13/2. The point is \( (0, \frac{17}{2}, -\frac{13}{2}) \).
In simple words: To find where a line crosses a coordinate plane, set the coordinate perpendicular to that plane equal to zero, solve for the parameter, and substitute back to find the other coordinates.

Exam Tip: Remember that the yz-plane is defined by x = 0, the xz-plane by y = 0, and the xy-plane by z = 0 - knowing this saves time in setting up equations.

Question 13. The vector equation of the x-axis is given by
(a) \( \vec{r} = \hat{i} \)
(b) \( \vec{r} = \hat{j} + \hat{k} \)
(c) \( \vec{r} = \lambda \hat{i} \)
(d) none of these
Answer: (c) \( \vec{r} = \lambda \hat{i} \)
In simple words: A vector equation needs a starting point and a direction vector. The x-axis can pass through any point (negative, positive, or the origin), and it always points in the direction of \( \hat{i} \). So the equation is \( \lambda \hat{i} \), where \( \lambda \) is any real number.

Exam Tip: Remember that the vector equation of an axis requires a parallel vector that matches the axis direction - only \( \hat{i} \) for the x-axis, only \( \hat{j} \) for the y-axis, and only \( \hat{k} \) for the z-axis.

Question 14. The Cartesian equations of a line are \( \frac{x - 2}{2} = \frac{y + 1}{3} = \frac{z - 3}{-2} \). What is its vector equation?
(a) \( \vec{r} = (2\hat{i} + 3\hat{j} - 2\hat{k}) + \lambda(2\hat{i} - \hat{j} + 3\hat{k}) \)
(b) \( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} - 2\hat{k}) \)
(c) \( \vec{r} = (2\hat{i} + 3\hat{j} - 2\hat{k}) \)
(d) none of these
Answer: (b) \( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} - 2\hat{k}) \)
In simple words: From the Cartesian form, pick the fixed point by setting each ratio equal to zero: the point is \( (2, -1, 3) \). The direction vector comes from the denominators: \( (2, 3, -2) \). Combine them into the vector form \( \vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} - 2\hat{k}) \).

Exam Tip: Always extract the fixed point by setting the Cartesian ratios to zero, and take the denominators as your direction vector - this conversion is straightforward if you follow the pattern.

Question 15. The angle between two lines having direction ratios \( 1, 1, 2 \) and \( (\sqrt{3} - 1), (-\sqrt{3} - 1), 4 \) is
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{2} \)
(c) \( \frac{\pi}{3} \)
(d) \( \frac{\pi}{4} \)
Answer: (a) \( \frac{\pi}{6} \)
In simple words: Use the formula for the angle between two lines: \( \cos \alpha = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} \). Calculate the dot product of the direction ratios, find the magnitude of each vector, then divide. The result \( \cos \alpha = \frac{\sqrt{3}}{2} \) gives \( \alpha = 30° = \frac{\pi}{6} \).

Exam Tip: When finding the angle between lines, always use the absolute value of the dot product to ensure you get an acute angle, and verify your final answer by checking that the cosine value is between 0 and 1.

Question 16. The straight line \( \frac{x - 2}{3} = \frac{y - 3}{1} = \frac{z + 1}{0} \) is
(a) parallel to the x-axis
(b) parallel to the y-axis
(c) parallel to the z-axis
(d) perpendicular to the z-axis
Answer: (d) perpendicular to the z-axis
In simple words: Look at the denominator for the z-component: it is 0. This means the line does not move in the z-direction at all - it stays at a constant z-level. A line that does not change in the z-direction must be perpendicular to the z-axis, since it makes a 90° angle with it.

Exam Tip: When a denominator in the Cartesian form is 0, the line is perpendicular to that axis - this is a quick way to identify the answer without calculating direction cosines.

Question 17. If a line makes angles \( \alpha \), \( \beta \) and \( \gamma \) with the x-axis, y-axis and z-axis respectively, then \( (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = \) ?
(a) 1
(b) 3
(c) 2
(d) \( \frac{3}{2} \)
Answer: (c) 2
In simple words: Use the identity \( \sin^2 \alpha = 1 - \cos^2 \alpha \) for each angle. Then \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) \). Since the sum of the squared direction cosines always equals 1, we get \( 3 - 1 = 2 \).

Exam Tip: Remember the key property: \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \) for any line in 3D - this relationship is the foundation for solving problems involving direction angles and direction cosines.

Question 18. If \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \) be the direction ratios of two parallel lines then
(a) \( a_1 = a_2, b_1 = b_2, c_1 = c_2 \)
(b) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
(c) \( a_1^2 + b_1^2 + c_1^2 = a_2^2 + b_2^2 + c_2^2 \)
(d) \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
Answer: (b) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
In simple words: When two lines are parallel, their direction ratios must be proportional to each other. This means each component of one set of direction ratios is a constant multiple of the corresponding component in the other set - expressed as equal ratios.

Exam Tip: Parallel lines do not need identical direction ratios (they can be scaled), but they must satisfy the proportionality condition - check this relationship before concluding two lines are parallel.

Question 19. If the points A(-1, 3, 2), B(-4, 2, -2) and C(5, 5, \( \lambda \)) are collinear then the value of \( \lambda \) is
(a) 5
(b) 7
(c) 8
(d) 10
Answer: (d) 10
In simple words: For three points to lie on the same line (be collinear), the determinant formed by their coordinates must equal zero. Set up the determinant with the three points and solve for \( \lambda \). Expanding the determinant and simplifying gives \( \lambda = 10 \).

Exam Tip: Always remember the collinearity condition: the determinant of the coordinate matrix must be zero - this provides a systematic method for finding unknown coordinates in collinear point problems.