RS Aggarwal Solutions for Class 12 Chapter 26 Fundamental Concepts of 3-Dimensional Geometry

Question 1. Find the direction cosines of a line segment whose direction ratios are:
(i) 2, - 6, 3
(ii) 2, - 1, - 2
(iii) - 9, 6, - 2
Answer: Direction cosines are unit vectors aligned with a line. If the direction ratios are given, divide each ratio by the square root of the sum of their squares to get the direction cosines.

(i) For direction ratios (2, -6, 3):
First, compute \( \sqrt{2^2 + (-6)^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \)
Then, \( l = \frac{2}{7}, \quad m = \frac{-6}{7}, \quad n = \frac{3}{7} \)
The direction cosines are: \( \left( \frac{2}{7}, \frac{-6}{7}, \frac{3}{7} \right) \)

(ii) For direction ratios (2, -1, -2):
Compute \( \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \)
Then, \( l = \frac{2}{3}, \quad m = \frac{-1}{3}, \quad n = \frac{-2}{3} \)
The direction cosines are: \( \left( \frac{2}{3}, \frac{-1}{3}, \frac{-2}{3} \right) \)

(iii) For direction ratios (-9, 6, -2):
Compute \( \sqrt{(-9)^2 + 6^2 + (-2)^2} = \sqrt{81 + 36 + 4} = \sqrt{121} = 11 \)
Then, \( l = \frac{-9}{11}, \quad m = \frac{6}{11}, \quad n = \frac{-2}{11} \)
The direction cosines are: \( \left( \frac{-9}{11}, \frac{6}{11}, \frac{-2}{11} \right) \)
In simple words: Direction cosines measure how a line points along each axis. To find them from direction ratios, divide each ratio by the square root of the sum of all squared ratios.

Exam Tip: Always verify that your final direction cosines satisfy \( l^2 + m^2 + n^2 = 1 \) - this is the key property that separates direction cosines from plain direction ratios.

Question 2. Find the direction ratios and the direction cosines of the line segment joining the points:
(i) A(1, 0, 0) and B(0, 1, 1)
(ii) A(5, 6, -3) and B(1, -6, 3)
(iii) A(-5, 7, -9) and B(-3, 4, -6)
Answer: To find direction ratios between two points, subtract the coordinates of the first point from the second. Then normalize these to get direction cosines.

(i) For points A(1, 0, 0) and B(0, 1, 1):
Direction ratios: \( \overrightarrow{AB} = (0 - 1, 1 - 0, 1 - 0) = (-1, 1, 1) \)
Magnitude: \( \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3} \)
Direction cosines: \( \left( \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \)

(ii) For points A(5, 6, -3) and B(1, -6, 3):
Direction ratios: \( \overrightarrow{AB} = (1 - 5, -6 - 6, 3 - (-3)) = (-4, -12, 6) \)
Simplify by dividing by -2: (2, 6, -3)
Magnitude: \( \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \)
Direction cosines: \( \left( \frac{2}{7}, \frac{6}{7}, \frac{-3}{7} \right) \)

(iii) For points A(-5, 7, -9) and B(-3, 4, -6):
Direction ratios: \( \overrightarrow{AB} = (-3 - (-5), 4 - 7, -6 - (-9)) = (2, -3, 3) \)
Magnitude: \( \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{4 + 9 + 9} = \sqrt{22} \)
Direction cosines: \( \left( \frac{2}{\sqrt{22}}, \frac{-3}{\sqrt{22}}, \frac{3}{\sqrt{22}} \right) \)
In simple words: Direction ratios are the raw differences between point coordinates. Direction cosines are the same differences divided by the distance between the points, turning them into unit direction values.

Exam Tip: Always simplify direction ratios to their lowest terms before finding the magnitude, and double-check your arithmetic when subtracting negative coordinates.

Question 3. Show that the line joining the points A(1, -1, 2) and B(3, 4, -2) is perpendicular to the line joining the points C(0, 3, 2) and D(3, 5, 6).
Answer: Two lines are perpendicular when their direction vectors have a dot product equal to zero. We compute the direction vectors and verify this condition.

For line AB: \( \overrightarrow{AB} = (3 - 1, 4 - (-1), -2 - 2) = (2, 5, -4) \)
For line CD: \( \overrightarrow{CD} = (3 - 0, 5 - 3, 6 - 2) = (3, 2, 4) \)

Dot product: \( \overrightarrow{AB} \cdot \overrightarrow{CD} = (2)(3) + (5)(2) + (-4)(4) = 6 + 10 - 16 = 0 \)

Since the dot product is zero, the angle between the lines is 90°, confirming they are perpendicular.
In simple words: When you multiply the components of two direction vectors and add them up, if you get zero, the lines meet at right angles.

Exam Tip: Always compute the full dot product carefully - a single arithmetic error will give you a non-zero result and make you fail to recognize perpendicularity.

Question 4. Show that the line segment joining the origin to the point A(2, 1, 1) is perpendicular to the line segment joining the points B(3, 5, -1) and C(4, 3, -1).
Answer: We find the direction vectors for both line segments and check if their dot product equals zero.

For segment OA: \( \overrightarrow{OA} = (2, 1, 1) \)
For segment BC: \( \overrightarrow{BC} = (4 - 3, 3 - 5, -1 - (-1)) = (1, -2, 0) \)

Dot product: \( \overrightarrow{OA} \cdot \overrightarrow{BC} = (2)(1) + (1)(-2) + (1)(0) = 2 - 2 + 0 = 0 \)

Since the dot product is zero, the two line segments are perpendicular to each other.
In simple words: The origin is at (0, 0, 0). We check if the path from origin to A is at right angles to the path from B to C by testing if their dot product is zero - and it is.

Exam Tip: Remember that the origin O is always (0, 0, 0), so \( \overrightarrow{OA} \) is simply the coordinates of A itself without any subtraction.

Question 5. Find the value of p for which the line through the points A(3, 5, -1) and B(5, p, 0) is perpendicular to the line through the points C(2, 1, 1) and D(3, 3, -1).
Answer: For perpendicularity, the dot product of the two direction vectors must be zero. We use this condition to solve for p.

Direction vector AB: \( \overrightarrow{AB} = (5 - 3, p - 5, 0 - (-1)) = (2, p - 5, 1) \)
Direction vector CD: \( \overrightarrow{CD} = (3 - 2, 3 - 1, -1 - 1) = (1, 2, -2) \)

For perpendicularity: \( \overrightarrow{AB} \cdot \overrightarrow{CD} = 0 \)
\( (2)(1) + (p - 5)(2) + (1)(-2) = 0 \)
\( 2 + 2p - 10 - 2 = 0 \)
\( 2p - 10 = 0 \)
\( p = 5 \)
In simple words: We need the dot product to equal zero. Multiplying corresponding components and adding them gives an equation in p, which we solve to get p = 5.

Exam Tip: Always expand the dot product fully before simplifying - rushing through the algebra is a common source of errors here.

Question 6. If O is the origin and P(2, 3, 4) and Q(1, -2, 1) be any two points show that OP ⊥ OQ.
Answer: We demonstrate that segments OP and OQ are perpendicular by confirming their direction vectors have a zero dot product.

Direction vector OP: \( \overrightarrow{OP} = (2, 3, 4) \)
Direction vector OQ: \( \overrightarrow{OQ} = (1, -2, 1) \)

Dot product: \( \overrightarrow{OP} \cdot \overrightarrow{OQ} = (2)(1) + (3)(-2) + (4)(1) = 2 - 6 + 4 = 0 \)

Since the dot product equals zero, we have proven that OP is perpendicular to OQ.
In simple words: When two segments from the origin have a dot product of zero, they form a right angle at the origin.

Exam Tip: Use the notation ⊥ clearly in your proof statement and always conclude by explicitly restating what the zero dot product means geometrically.

Question 7. Show that the line segment joining the points A(1, 2, 3) and B(4, 5, 7) is parallel to the segment joining the points C(-4, 3, -6) and D(2, 9, 2).
Answer: Two line segments are parallel when one direction vector is a scalar multiple of the other. We find both direction vectors and verify this relationship.

Direction vector AB: \( \overrightarrow{AB} = (4 - 1, 5 - 2, 7 - 3) = (3, 3, 4) \)
Direction vector CD: \( \overrightarrow{CD} = (2 - (-4), 9 - 3, 2 - (-6)) = (6, 6, 8) \)

We observe that \( \overrightarrow{CD} = 2 \cdot \overrightarrow{AB} \), since (6, 6, 8) = 2(3, 3, 4).

Since one vector is a scalar multiple of the other, the two segments are parallel.
In simple words: If you can multiply one direction vector by a constant number to get the other direction vector, the lines must be parallel.

Exam Tip: Always check that ALL three components scale by the same constant factor - if only some components match the ratio, the lines are not parallel.

Question 8. If the line segment joining the points A(7, p, 2) and B(q, -2, 5) be parallel to the line segment joining the points C(2, -3, 5) and D(-6, -15, 11), find the values of p and q.
Answer: For parallel lines, one direction vector equals a scalar multiple of the other. We use this to establish equations for p and q.

Direction vector AB: \( \overrightarrow{AB} = (q - 7, -2 - p, 5 - 2) = (q - 7, -2 - p, 3) \)
Direction vector CD: \( \overrightarrow{CD} = (-6 - 2, -15 - (-3), 11 - 5) = (-8, -12, 6) \)

For parallel segments: \( \overrightarrow{AB} = k \cdot \overrightarrow{CD} \) for some scalar k.

From the third component: \( 3 = 6k \), so \( k = \frac{1}{2} \)

From the first component: \( q - 7 = \frac{1}{2} \cdot (-8) = -4 \), so \( q = 3 \)

From the second component: \( -2 - p = \frac{1}{2} \cdot (-12) = -6 \), so \( -2 - p = -6 \), giving \( p = 4 \)
In simple words: We set up equations by matching each component of AB to the same component of CD multiplied by a scale factor k. Solving these gives us the unknowns p and q.

Exam Tip: Find the scalar k first from one component (pick the simplest), then use it to solve for the unknowns from the other components.

Question 9. Show that the points A(2, 3, 4), B(-1, -2, 1) and C(5, 8, 7) are collinear.
Answer: Three points are collinear if they all lie on the same straight line. We use a parametric line equation through two points and check if the third point satisfies it.

Direction vector from A to B: \( \overrightarrow{AB} = (-1 - 2, -2 - 3, 1 - 4) = (-3, -5, -3) \)

The parametric equation of line AB through point A(2, 3, 4) is:
\( R(a) = (2, 3, 4) + a(-3, -5, -3) = (2 - 3a, 3 - 5a, 4 - 3a) \)

For point C(5, 8, 7) to lie on this line, we need:
\( (5, 8, 7) = (2 - 3a, 3 - 5a, 4 - 3a) \)

From the first coordinate: \( 5 = 2 - 3a \Rightarrow a = -1 \)
From the second coordinate: \( 8 = 3 - 5(-1) = 3 + 5 = 8 \) ✓
From the third coordinate: \( 7 = 4 - 3(-1) = 4 + 3 = 7 \) ✓

Since all three coordinates satisfy the line equation with a = -1, point C lies on the line AB. Therefore, the three points are collinear.
In simple words: We write the equation of the line through A and B. Then we check if C satisfies this equation for some value of the parameter a - if it does, all three points are on the same line.

Exam Tip: Always verify your parameter value by substituting back into all three coordinate equations - finding a for one coordinate but having it fail for another means the points are not collinear.

Question 10. Show that the points A(-2, 4, 7), B(3, -6, -8) and C(1, -2, -2) are collinear.
Answer: We verify collinearity by finding the parametric equation of the line through A and B, then checking if C satisfies it.

Direction vector from A to B: \( \overrightarrow{AB} = (3 - (-2), -6 - 4, -8 - 7) = (5, -10, -15) \)

The parametric equation of line AB through A(-2, 4, 7) is:
\( R(a) = (-2, 4, 7) + a(5, -10, -15) = (-2 + 5a, 4 - 10a, 7 - 15a) \)

For point C(1, -2, -2) to lie on the line:
\( (1, -2, -2) = (-2 + 5a, 4 - 10a, 7 - 15a) \)

From the first coordinate: \( 1 = -2 + 5a \Rightarrow a = \frac{3}{5} \)
From the second coordinate: \( -2 = 4 - 10 \cdot \frac{3}{5} = 4 - 6 = -2 \) ✓
From the third coordinate: \( -2 = 7 - 15 \cdot \frac{3}{5} = 7 - 9 = -2 \) ✓

All three coordinates match when \( a = \frac{3}{5} \), so C lies on line AB. The three points are collinear.
In simple words: We find the value of the parameter that places the line at C's first coordinate, then check if the same parameter value also gives us the correct second and third coordinates.

Exam Tip: When working with fractional parameter values, be extra careful with arithmetic - multiply through to clear fractions when solving and double-check your verification steps.

Question 11. Find the value of p for which the points A(-1, 3, 2), B(-4, 2, -2), and C(5, 5, p) are collinear.
Answer: We set up the parametric equation of the line through A and B, then use the requirement that C lies on this line to solve for p.

Direction vector from A to B: \( \overrightarrow{AB} = (-4 - (-1), 2 - 3, -2 - 2) = (-3, -1, -4) \)

The parametric equation of line AB through A(-1, 3, 2) is:
\( R(a) = (-1, 3, 2) + a(-3, -1, -4) = (-1 - 3a, 3 - a, 2 - 4a) \)

For point C(5, 5, p) to satisfy this equation:
\( (5, 5, p) = (-1 - 3a, 3 - a, 2 - 4a) \)

From the first coordinate: \( 5 = -1 - 3a \Rightarrow -3a = 6 \Rightarrow a = -2 \)
From the second coordinate: \( 5 = 3 - (-2) = 3 + 2 = 5 \) ✓

From the third coordinate with \( a = -2 \):
\( p = 2 - 4(-2) = 2 + 8 = 10 \)
In simple words: We find the parameter value from the first coordinate that satisfies that equation, verify it works for the second coordinate, then use it to calculate p from the third coordinate.

Exam Tip: Always verify your parameter value with at least one other coordinate before using it to find the unknown - this catches algebra errors early.

Question 12. Find the angle between the two lines whose direction cosines are: \( \frac{2}{3}, \frac{-1}{3}, \frac{-2}{3} \) and \( \frac{3}{7}, \frac{2}{7}, \frac{6}{7} \)
Answer: The angle between two lines is found using the dot product of their direction cosine vectors. Since direction cosines are already unit vectors, the magnitude is 1.

Direction vector 1: \( \vec{R_1} = \frac{2}{3}\hat{i} - \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k} \)
Direction vector 2: \( \vec{R_2} = \frac{3}{7}\hat{i} + \frac{2}{7}\hat{j} + \frac{6}{7}\hat{k} \)

Dot product: \( \vec{R_1} \cdot \vec{R_2} = \left(\frac{2}{3}\right)\left(\frac{3}{7}\right) + \left(\frac{-1}{3}\right)\left(\frac{2}{7}\right) + \left(\frac{-2}{3}\right)\left(\frac{6}{7}\right) \)
\( = \frac{6}{21} - \frac{2}{21} - \frac{12}{21} = \frac{6 - 2 - 12}{21} = \frac{-8}{21} \)

Since both vectors are unit vectors (|\vec{R_1}| = |\vec{R_2}| = 1):
\( \cos \theta = \frac{-8}{21} \)
\( \theta = \cos^{-1}\left(\frac{-8}{21}\right) \)
In simple words: Direction cosines are special because they already have length 1. So the angle between the lines comes directly from their dot product without needing to divide by magnitudes.

Exam Tip: A negative cosine means the angle is obtuse (greater than 90°). The angle between two lines is conventionally taken as the acute angle, so you may need to use the supplementary angle or take the absolute value - check your problem's instructions.

Question 13. Find the angle between the two lines whose direction ratios are: a, b, c and (b - c), (c - a), (a - b).
Answer: We compute the angle using the direction ratio formula. The dot product of the two direction vectors gives us information about their angle.

Direction vector 1: \( \vec{R_1} = a\hat{i} + b\hat{j} + c\hat{k} \)
Direction vector 2: \( \vec{R_2} = (b - c)\hat{i} + (c - a)\hat{j} + (a - b)\hat{k} \)

Dot product: \( \vec{R_1} \cdot \vec{R_2} = a(b - c) + b(c - a) + c(a - b) \)
\( = ab - ac + bc - ab + ca - cb = 0 \)

Since the dot product equals zero: \( \cos \theta = 0 \)

Therefore: \( \theta = \frac{\pi}{2} \)

The two lines are perpendicular to each other.
In simple words: When you expand the dot product algebraically, all terms cancel out perfectly, leaving zero. This means the angle between these two direction vectors is always 90 degrees, regardless of what a, b, and c are.

Exam Tip: Recognize that this is an algebraic identity - the cancellation is not a coincidence but a fundamental property. Always expand the products fully to see this structure.

Question 14. Find the angle between the lines whose direction ratios are: 2, -3, 4 and 1, 2, 1.
Answer: The angle between two lines is determined by the cosine formula applied to their direction vectors. We compute the dot product and magnitudes.

Direction vector 1: \( \vec{R_1} = 2\hat{i} - 3\hat{j} + 4\hat{k} \)
Direction vector 2: \( \vec{R_2} = \hat{i} + 2\hat{j} + \hat{k} \)

Dot product: \( \vec{R_1} \cdot \vec{R_2} = (2)(1) + (-3)(2) + (4)(1) = 2 - 6 + 4 = 0 \)

Since the dot product is zero: \( \cos \theta = 0 \)

Therefore: \( \theta = \frac{\pi}{2} \)

The two lines are perpendicular.
In simple words: Multiply matching components together, add the products, and if you get zero, the lines meet at right angles. No need to compute magnitudes when the dot product itself is zero.

Exam Tip: If the dot product is already zero, you can immediately conclude the angle is 90° without further calculation - this saves time and reduces arithmetic errors.

Question 15. Find the angle between the lines whose direction ratios are: 1, 1, 2 and (\(\sqrt{3} - 1\)), (-\(\sqrt{3} - 1\)), 4
Answer: We apply the angle formula for lines with given direction ratios, computing the dot product and magnitudes.

Direction vector 1: \( \vec{R_1} = \hat{i} + \hat{j} + 2\hat{k} \)
Direction vector 2: \( \vec{R_2} = (\sqrt{3} - 1)\hat{i} - (\sqrt{3} + 1)\hat{j} + 4\hat{k} \)

Dot product: \( \vec{R_1} \cdot \vec{R_2} = (1)(\sqrt{3} - 1) + (1)(-\sqrt{3} - 1) + (2)(4) \)
\( = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6 \)

Magnitude of \( \vec{R_1} \): \( |\vec{R_1}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6} \)

Magnitude of \( \vec{R_2} \): \( |\vec{R_2}| = \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + 4^2} \)
\( = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} = \sqrt{24} = 2\sqrt{6} \)

\( \cos \theta = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{12} = \frac{1}{2} \)

Therefore: \( \theta = \frac{\pi}{3} \)
In simple words: The radical terms cancel when we compute the dot product. Then we find the magnitudes using the Pythagorean formula, and the angle comes from the ratio of dot product to the product of magnitudes.

Exam Tip: When expanding expressions with \( \sqrt{3} \), watch for terms that cancel - \( \sqrt{3} - \sqrt{3} = 0 \) - this often simplifies the dot product. Also, (\(\sqrt{3} + 1\))² = 3 + 2\(\sqrt{3}\) + 1, not just 3 + 1.

Question 16. Find the angle between the vectors \( \vec{r_1} = (3\hat{i} - 2\hat{j} + \hat{k}) \) and \( \vec{r_2} = (4\hat{i} + 5\hat{j} + 7\hat{k}) \)
Answer: The angle between two vectors is found by computing their dot product and dividing by the product of their magnitudes.

Dot product: \( \vec{r_1} \cdot \vec{r_2} = (3)(4) + (-2)(5) + (1)(7) = 12 - 10 + 7 = 9 \)

Magnitude of \( \vec{r_1} \): \( |\vec{r_1}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \)

Magnitude of \( \vec{r_2} \): \( |\vec{r_2}| = \sqrt{4^2 + 5^2 + 7^2} = \sqrt{16 + 25 + 49} = \sqrt{90} = 3\sqrt{10} \)

\( \cos \theta = \frac{9}{\sqrt{14} \cdot 3\sqrt{10}} = \frac{9}{3\sqrt{140}} = \frac{3}{\sqrt{140}} = \frac{3}{2\sqrt{35}} \)

Therefore: \( \theta = \cos^{-1}\left(\frac{3}{2\sqrt{35}}\right) \)
In simple words: Compute the dot product of the two vectors. Then find how long each vector is using the square root formula. The angle comes from dividing the dot product by these two lengths.

Exam Tip: Always simplify radicals at the end - \( \sqrt{90} = 3\sqrt{10} \) and \( \sqrt{140} = 2\sqrt{35} \) make the final expression cleaner and easier to work with.

Question 17. Find the angles made by the following vectors with the coordinate axes:
(i) \( (\hat{i} - \hat{j} + \hat{k}) \)
(ii) \( (\hat{j} - \hat{k}) \)
(iii) \( (\hat{i} - 4\hat{j} + 8\hat{k}) \)
Answer:
(i) For vector \( \vec{v_1} = \hat{i} - \hat{j} + \hat{k} \):

Magnitude: \( |\vec{v_1}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \)

Angle with x-axis (using direction \( \hat{i} \)):
\( \cos \alpha = \frac{(1)}{1 \cdot \sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow \alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \)

Angle with y-axis (using direction \( \hat{j} \)):
\( \cos \beta = \frac{(-1)}{1 \cdot \sqrt{3}} = \frac{-1}{\sqrt{3}} \Rightarrow \beta = \cos^{-1}\left(\frac{-1}{\sqrt{3}}\right) \)

Angle with z-axis (using direction \( \hat{k} \)):
\( \cos \gamma = \frac{(1)}{1 \cdot \sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow \gamma = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \)

(ii) For vector \( \vec{v_2} = \hat{j} - \hat{k} \):

Magnitude: \( |\vec{v_2}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2} \)

Angle with x-axis:
\( \cos \alpha = \frac{0}{\sqrt{2}} = 0 \Rightarrow \alpha = \frac{\pi}{2} \)

Angle with y-axis:
\( \cos \beta = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow \beta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \)

Angle with z-axis:
\( \cos \gamma = \frac{-1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \Rightarrow \gamma = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4} \)

(iii) For vector \( \vec{v_3} = \hat{i} - 4\hat{j} + 8\hat{k} \):

Magnitude: \( |\vec{v_3}| = \sqrt{1^2 + (-4)^2 + 8^2} = \sqrt{1 + 16 + 64} = \sqrt{81} = 9 \)

Angle with x-axis:
\( \cos \alpha = \frac{1}{9} \Rightarrow \alpha = \cos^{-1}\left(\frac{1}{9}\right) \)

Angle with y-axis:
\( \cos \beta = \frac{-4}{9} \Rightarrow \beta = \cos^{-1}\left(\frac{-4}{9}\right) \)

Angle with z-axis:
\( \cos \gamma = \frac{8}{9} \Rightarrow \gamma = \cos^{-1}\left(\frac{8}{9}\right) \)
In simple words: To find the angle a vector makes with each axis, divide the vector's component along that axis by the vector's total length. Then take the inverse cosine of that ratio.

Exam Tip: For the x-axis angle, use only the i-component; for y-axis, use only the j-component; for z-axis, use only the k-component. A component of zero means the angle with that axis is 90°.

Question 18. 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: The foot of the perpendicular is the point R on line BC such that AR is perpendicular to BC. We use the parametric equation of the line and the perpendicularity condition.

Direction vector of BC: \( \vec{BC} = (2 - 0, -3 - (-1), -1 - 3) = (2, -2, -4) \)

Parametric equation of line BC through B(0, -1, 3):
\( R(a) = (0, -1, 3) + a(2, -2, -4) = (2a, -1 - 2a, 3 - 4a) \)

Vector from A to R:
\( \vec{AR} = (2a - 1, -1 - 2a - 8, 3 - 4a - 4) = (2a - 1, -9 - 2a, -1 - 4a) \)

For perpendicularity: \( \vec{AR} \cdot \vec{BC} = 0 \)
\( (2a - 1)(2) + (-9 - 2a)(-2) + (-1 - 4a)(-4) = 0 \)
\( 4a - 2 + 18 + 4a + 4 + 16a = 0 \)
\( 24a + 20 = 0 \)
\( a = -\frac{5}{6} \)

Substituting \( a = -\frac{5}{6} \) into the parametric equation:
\( R = \left( 2 \cdot \left(-\frac{5}{6}\right), -1 - 2 \cdot \left(-\frac{5}{6}\right), 3 - 4 \cdot \left(-\frac{5}{6}\right) \right) \)
\( = \left( -\frac{5}{3}, -1 + \frac{5}{3}, 3 + \frac{10}{3} \right) \)
\( = \left( -\frac{5}{3}, \frac{2}{3}, \frac{19}{3} \right) \)
In simple words: We write the line BC in parametric form with a variable a. Then we find the value of a where the vector from A to a point on the line is perpendicular to the line's direction. That value of a gives us the foot of the perpendicular.

Exam Tip: Always expand the dot product equation fully before collecting like terms - missing or misplacing a term will lead to an incorrect value of the parameter and thus the wrong final coordinates.