RS Aggarwal Solutions for Class 12 Chapter 22 Vectors and Their Properties

Question 1. Write down the magnitude of each of the following vectors:
(A) \( \vec{a} = \vec{i} + 2\vec{j} + 5\vec{k} \)
(B) \( \vec{b} = 5\vec{i} - 4\vec{j} - 3\vec{k} \)
(C) \( \vec{c} = \left( \frac{1}{\sqrt{3}}\vec{i} - \frac{1}{\sqrt{3}}\vec{j} + \frac{1}{\sqrt{3}}\vec{k} \right) \)
(D) \( \vec{d} = (\sqrt{2}\vec{i} + \sqrt{3}\vec{j} - \sqrt{5}\vec{k}) \)
Answer: For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the magnitude is given by \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

(A) \( \vec{a} = \vec{i} + 2\vec{j} + 5\vec{k} \)
\( \therefore |\vec{a}| = \sqrt{1^2 + 2^2 + 5^2} = \sqrt{30} \text{ units} \)

(B) \( \vec{a} = 5\vec{i} - 4\vec{j} - 3\vec{k} \)
\( \therefore |\vec{a}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{50} = 5\sqrt{2} \text{ units} \)

(C) \( \vec{a} = \frac{1}{\sqrt{3}}\vec{i} - \frac{1}{\sqrt{3}}\vec{j} + \frac{1}{\sqrt{3}}\vec{k} \)
\( \therefore |\vec{a}| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2} = 1 \text{ unit} \)

(D) \( \vec{a} = \sqrt{2}\vec{i} + \sqrt{3}\vec{j} - \sqrt{5}\vec{k} \)
\( \therefore |\vec{a}| = \sqrt{(\sqrt{2})^2 + (\sqrt{3})^2 + (\sqrt{5})^2} = \sqrt{10} \text{ units} \)
In simple words: The magnitude shows how long a vector is. Square each component number, add them all together, then take the square root of that total.

Exam Tip: Always apply the magnitude formula correctly - square each coefficient, sum them, and extract the square root. Watch for negative signs within components.

Question 2. Find a unit vector in the direction of the vector:
(A) \( (3\vec{i} + 4\vec{j} - 5\vec{k}) \)
(B) \( (3\vec{i} - 2\vec{j} + 6\vec{k}) \)
(C) \( (\vec{i} + \vec{k}) \)
(D) \( (2\vec{i} + \vec{j} + 2\vec{k}) \)
Answer: For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the unit vector is given by \( \hat{a} = \frac{a_x\vec{i} + a_y\vec{j} + a_z\vec{k}}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \).

(A) \( \vec{a} = 3\vec{i} + 4\vec{j} - 5\vec{k} \)
\( \therefore \hat{a} = \frac{3\vec{i} + 4\vec{j} - 5\vec{k}}{\sqrt{3^2 + 4^2 + 5^2}} = \frac{3}{5\sqrt{2}}\vec{i} + \frac{4}{5\sqrt{2}}\vec{j} - \frac{5}{5\sqrt{2}}\vec{k} \)

(B) \( \vec{a} = 3\vec{i} - 2\vec{j} + 6\vec{k} \)
\( \therefore \hat{a} = \frac{3\vec{i} - 2\vec{j} + 6\vec{k}}{\sqrt{3^2 + 2^2 + 6^2}} = \frac{3}{7}\vec{i} - \frac{2}{7}\vec{j} + \frac{6}{7}\vec{k} \)

(C) \( \vec{a} = \vec{i} + \vec{k} \)
\( \therefore \hat{a} = \frac{\vec{i} + \vec{k}}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}}\vec{k} \)

(D) \( \vec{a} = 2\vec{i} + \vec{j} + 2\vec{k} \)
\( \therefore \hat{a} = \frac{2\vec{i} + \vec{j} + 2\vec{k}}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{2}{3}\vec{i} + \frac{1}{3}\vec{j} + \frac{2}{3}\vec{k} \)
In simple words: A unit vector has a length of exactly 1. Take your original vector and split each component by the vector's total length (its magnitude).

Exam Tip: Unit vectors always have magnitude 1. Divide every component by the vector's magnitude to normalize it.

Question 3. If \( \vec{a} = (2\vec{i} - 4\vec{j} + 5\vec{k}) \) then find the value of \( \lambda \) so that \( \lambda\vec{a} \) may be a unit vector.
Answer: Given \( \vec{a} = 2\vec{i} - 4\vec{j} + 5\vec{k} \), we construct the scaled vector \( \lambda\vec{a} = 2\lambda\vec{i} - 4\lambda\vec{j} + 5\lambda\vec{k} \).

For a unit vector, its magnitude must equal 1. For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the magnitude is given by \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

\( \therefore |\lambda\vec{a}| = \sqrt{(2\lambda)^2 + (-4\lambda)^2 + (5\lambda)^2} = 1 \)

\( \Rightarrow 45\lambda^2 = 1 \)

\( \Rightarrow \lambda^2 = \frac{1}{45} = \frac{1}{(3\sqrt{5})^2} \)

\( \Rightarrow \lambda = \pm \frac{1}{3\sqrt{5}} \)
In simple words: When you multiply a vector by a scaling factor, the magnitude gets multiplied by that same factor. Set this scaled magnitude equal to 1 and solve for the scaling value.

Exam Tip: When finding a scalar multiplier that produces a unit vector, square the magnitude formula and isolate the squared scalar before taking its square root.

Question 4. If \( \vec{a} = (-\vec{i} + \vec{j} - \vec{k}) \) and \( \vec{b} = (2\vec{i} - \vec{j} + 2\vec{k}) \) then find the unit vector in the direction of \( (\vec{a} + \vec{b}) \).
Answer: We are given \( \vec{a} = -\vec{i} + \vec{j} - \vec{k} \) and \( \vec{b} = 2\vec{i} - \vec{j} + 2\vec{k} \).

First, we find their sum:
\( \vec{a} + \vec{b} = (-\vec{i} + \vec{j} - \vec{k}) + (2\vec{i} - \vec{j} + 2\vec{k}) = \vec{i} + \vec{k} \)

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the unit vector is given by \( \hat{a} = \frac{a_x\vec{i} + a_y\vec{j} + a_z\vec{k}}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \).

\( \therefore \widehat{(\vec{a} + \vec{b})} = \frac{\vec{i} + \vec{k}}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}(\vec{i} + \vec{k}) \)
In simple words: Add the two vectors first, then convert the result into a unit vector by dividing by its magnitude.

Exam Tip: Always add or subtract vectors component-wise first, then apply the unit vector formula to the resulting vector.

Question 5. If \( \vec{a} = (3\vec{i} + \vec{j} - 5\vec{k}) \) and \( \vec{b} = (\vec{i} + 2\vec{j} - \vec{k}) \) then find a unit vector in the direction of \( (\vec{a} - \vec{b}) \).
Answer: Given \( \vec{a} = 3\vec{i} + \vec{j} - 5\vec{k} \) and \( \vec{b} = \vec{i} + 2\vec{j} - \vec{k} \).

First, we find their difference:
\( \vec{a} - \vec{b} = (3\vec{i} + \vec{j} - 5\vec{k}) - (\vec{i} + 2\vec{j} - \vec{k}) = 2\vec{i} - \vec{j} - 4\vec{k} \)

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the unit vector is given by \( \hat{a} = \frac{a_x\vec{i} + a_y\vec{j} + a_z\vec{k}}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \).

\( \therefore \widehat{(\vec{a} - \vec{b})} = \frac{2\vec{i} - \vec{j} - 4\vec{k}}{\sqrt{2^2 + 1^2 + 4^2}} = \frac{1}{\sqrt{21}}(2\vec{i} - \vec{j} - 4\vec{k}) \)
In simple words: Subtract one vector from the other component-wise, then divide by the result's magnitude to get a unit vector.

Exam Tip: When subtracting vectors, carefully distribute the negative sign across all components of the second vector.

Question 6. If \( \vec{a} = (\vec{i} + 2\vec{j} - 3\vec{k}) \) and \( \vec{b} = (2\vec{i} + 4\vec{j} + 9\vec{k}) \) then find a unit vector parallel to \( (\vec{a} + \vec{b}) \).
Answer: Given \( \vec{a} = \vec{i} + 2\vec{j} - 3\vec{k} \) and \( \vec{b} = 2\vec{i} + 4\vec{j} + 9\vec{k} \).

First, we calculate their sum:
\( \vec{a} + \vec{b} = (\vec{i} + 2\vec{j} - 3\vec{k}) + (2\vec{i} + 4\vec{j} + 9\vec{k}) = 3\vec{i} + 6\vec{j} + 6\vec{k} \)

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the unit vector is represented as \( \hat{a} = \frac{a_x\vec{i} + a_y\vec{j} + a_z\vec{k}}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \).

\( \therefore \widehat{(\vec{a} + \vec{b})} = \frac{3\vec{i} + 6\vec{j} + 6\vec{k}}{\sqrt{3^2 + 6^2 + 6^2}} = \frac{1}{9}(3\vec{i} + 6\vec{j} + 6\vec{k}) = \frac{1}{3}(\vec{i} + 2\vec{j} + 2\vec{k}) \)
In simple words: Two vectors are parallel when one is a scaled version of the other. Add the vectors, then normalize by dividing by the magnitude.

Exam Tip: Parallel vectors point in the same or opposite directions. Their unit vectors are also parallel and point in the same direction.

Question 7. Find a vector of magnitude 9 units in the direction of the vector \( (-2\vec{i} + \vec{j} + 2\vec{k}) \).
Answer: Let \( \lambda \) be a scaling constant. The required vector takes the form \( -2\lambda\vec{i} + \lambda\vec{j} + 2\lambda\vec{k} \).

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the magnitude is given by \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

\( \therefore \sqrt{(-2\lambda)^2 + (\lambda)^2 + (2\lambda)^2} = 9 \)

\( \Rightarrow 3\lambda = 9 \)

\( \Rightarrow \lambda = 3 \)

The required vector is \( -6\vec{i} + 3\vec{j} + 6\vec{k} \)
In simple words: Scale the direction vector by a constant so that the resulting vector has the desired length.

Exam Tip: To find a vector with a specific magnitude in a given direction, find the scaling factor by setting the scaled magnitude equal to the desired length.

Question 8. Find a vector of magnitude 8 units in the direction of the vector \( (5\vec{i} - \vec{j} + 2\vec{k}) \).
Answer: Let \( \lambda \) be a scaling constant. The required vector takes the form \( 5\lambda\vec{i} - \lambda\vec{j} + 2\lambda\vec{k} \).

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the magnitude is given by \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

\( \therefore \sqrt{(5\lambda)^2 + (-\lambda)^2 + (2\lambda)^2} = 8 \)

\( \Rightarrow \sqrt{30}\lambda = 8 \)

\( \Rightarrow \lambda = \frac{8}{\sqrt{30}} \)

The required vector is \( \frac{8}{\sqrt{30}}(5\vec{i} - \vec{j} + 2\vec{k}) \)
In simple words: Multiply the given direction vector by a scaling factor such that the final vector's length equals 8 units.

Exam Tip: When the magnitude does not simplify to a whole number, leave the scaling factor in simplified radical form.

Question 9. Find a vector of magnitude 21 units in the direction of the vector \( (2\vec{i} - 3\vec{j} + 6\vec{k}) \).
Answer: Let \( \lambda \) be a scaling constant. The required vector takes the form \( 2\lambda\vec{i} - 3\lambda\vec{j} + 6\lambda\vec{k} \).

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the magnitude is given by \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

\( \therefore \sqrt{(2\lambda)^2 + (-3\lambda)^2 + (6\lambda)^2} = 21 \)

\( \Rightarrow 7\lambda = 21 \)

\( \Rightarrow \lambda = 3 \)

The required vector is \( (6\vec{i} - 9\vec{j} + 18\vec{k}) \)
In simple words: Find the scaling factor by equating the magnitude formula to 21, then multiply each component by this factor.

Exam Tip: Always verify your answer by computing the magnitude of the final vector to confirm it matches the required value.

Question 10. If \( \vec{a} = (\vec{i} - 2\vec{j}) \), \( \vec{b} = (2\vec{i} - 3\vec{j}) \) and \( \vec{c} = (2\vec{i} + 3\vec{k}) \), find \( (\vec{a} + \vec{b} + \vec{c}) \).
Answer: We are given \( \vec{a} = \vec{i} - 2\vec{j} \), \( \vec{b} = 2\vec{i} - 3\vec{j} \), and \( \vec{c} = 2\vec{i} + 3\vec{k} \).

Adding all three vectors:
\( \vec{a} + \vec{b} + \vec{c} = (\vec{i} - 2\vec{j}) + (2\vec{i} - 3\vec{j}) + (2\vec{i} + 3\vec{k}) = 5\vec{i} - 5\vec{j} + 3\vec{k} \)
In simple words: Combine like terms - add all the \( \vec{i} \) components together, all the \( \vec{j} \) components together, and all the \( \vec{k} \) components together.

Exam Tip: Vector addition works component-by-component. Keep track of positive and negative signs carefully when combining.

Question 11. If A(-2, 1, 2) and B(2, -1, 6) are two given points, find a unit vector in the direction of \( \overrightarrow{AB} \).
Answer: We have A = (-2, 1, 2) and B = (2, -1, 6).

First, we determine the vector from A to B:
\( \overrightarrow{AB} = \{2 - (-2)\}\vec{i} + \{(-1) - 1\}\vec{j} + \{6 - 2\}\vec{k} = 4\vec{i} - 2\vec{j} + 4\vec{k} \)

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the unit vector is represented as \( \hat{a} = \frac{a_x\vec{i} + a_y\vec{j} + a_z\vec{k}}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \).

\( \therefore \widehat{AB} = \frac{4\vec{i} - 2\vec{j} + 4\vec{k}}{\sqrt{4^2 + 2^2 + 4^2}} = \frac{4\vec{i} - 2\vec{j} + 4\vec{k}}{6} = \frac{2}{3}\vec{i} - \frac{1}{3}\vec{j} + \frac{2}{3}\vec{k} \)
In simple words: Subtract the coordinates of the starting point from the ending point to get the vector. Then normalize by dividing by its magnitude.

Exam Tip: Always ensure the direction is correct - use endpoint minus starting point, not the reverse.

Question 12. Find the direction ratios and direction cosines of the vector \( \vec{a} = (5\vec{i} - 3\vec{j} + 4\vec{k}) \).
Answer: We are given \( \vec{a} = 5\vec{i} - 3\vec{j} + 4\vec{k} \).

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the direction ratios are represented as (a_x, a_y, a_z). The direction cosines are obtained by dividing each component by the vector's magnitude: \( \frac{a_x}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \), \( \frac{a_y}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \), \( \frac{a_z}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \).

The direction ratios are (5, -3, 4).

The direction cosines are \( \frac{5}{\sqrt{5^2 + (-3)^2 + 4^2}} \), \( \frac{-3}{\sqrt{5^2 + (-3)^2 + 4^2}} \), \( \frac{4}{\sqrt{5^2 + (-3)^2 + 4^2}} \)

\( = \frac{5}{\sqrt{50}} \), \( \frac{-3}{\sqrt{50}} \), \( \frac{4}{\sqrt{50}} \)

\( = \frac{5}{5\sqrt{2}} \), \( \frac{-3}{5\sqrt{2}} \), \( \frac{4}{5\sqrt{2}} \)

\( = \frac{1}{\sqrt{2}} \), \( \frac{-3}{5\sqrt{2}} \), \( \frac{4}{5\sqrt{2}} \)
In simple words: Direction ratios are simply the coefficients of \( \vec{i} \), \( \vec{j} \), and \( \vec{k} \). Direction cosines are these ratios divided by the vector's length.

Exam Tip: Direction cosines always satisfy the property that the sum of their squares equals 1. Use this as a check for your work.

Question 13. Find the direction ratios and the direction cosines of the vector joining the points A(2, 1, -2) and B(3, 5, -4).
Answer: We are given A = (2, 1, -2) and B = (3, 5, -4).

First, we find the vector joining these points:
\( \overrightarrow{AB} = \{3 - 2\}\vec{i} + \{5 - 1\}\vec{j} + \{(-4) - (-2)\}\vec{k} = \vec{i} + 4\vec{j} - 2\vec{k} \)

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the direction ratios are represented as (a_x, a_y, a_z). The direction cosines are obtained by dividing each component by the vector's magnitude: \( \frac{a_x}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \), \( \frac{a_y}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \), \( \frac{a_z}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \).

The direction ratios are (1, 4, -2).

The direction cosines are \( \frac{1}{\sqrt{1^2 + 4^2 + (-2)^2}} \), \( \frac{4}{\sqrt{1^2 + 4^2 + (-2)^2}} \), \( \frac{-2}{\sqrt{1^2 + 4^2 + (-2)^2}} \)

\( = \frac{1}{\sqrt{21}} \), \( \frac{4}{\sqrt{21}} \), \( \frac{-2}{\sqrt{21}} \)
In simple words: Build the vector by subtracting starting coordinates from ending coordinates. The direction ratios are these coefficients. To get direction cosines, divide each ratio by the vector's magnitude.

Exam Tip: Direction ratios are proportional numbers, while direction cosines are the actual cosines of the angles the vector makes with the axes.

Question 14. Show that the points A, B and C having position vectors \( (\vec{i} + 2\vec{j} + 7\vec{k}) \), \( (2\vec{i} + 6\vec{j} + 2\vec{k}) \) and \( (3\vec{i} + 10\vec{j} - 3\vec{k}) \) respectively, are collinear.
Answer: We have \( \vec{A} = \vec{i} + 2\vec{j} + 7\vec{k} \), \( \vec{B} = 2\vec{i} + 6\vec{j} + 2\vec{k} \), and \( \vec{C} = 3\vec{i} + 10\vec{j} - 3\vec{k} \).

Now we find the vectors connecting these points:
\( \overrightarrow{AB} = (2\vec{i} + 6\vec{j} + 2\vec{k}) - (\vec{i} + 2\vec{j} + 7\vec{k}) = \vec{i} + 4\vec{j} - 5\vec{k} \)

\( \overrightarrow{BC} = (3\vec{i} + 10\vec{j} - 3\vec{k}) - (2\vec{i} + 6\vec{j} + 2\vec{k}) = \vec{i} + 4\vec{j} - 5\vec{k} \)

Since \( \overrightarrow{AB} = \overrightarrow{BC} \), the vectors are identical. This indicates that B lies on the line segment AC, and the three points share the same direction. Therefore, the points A, B, and C are collinear.
In simple words: Three points are collinear if the vectors connecting them are parallel (one is a scalar multiple of another) or if moving from one point to another in two steps gives the same displacement as going directly.

Exam Tip: To prove collinearity, show that one displacement vector is a scalar multiple of another, or that the vectors lie on the same line.

Question 15. The position vectors of the points A, B and C are \( (2\vec{i} + \vec{j} - \vec{k}) \), \( (3\vec{i} - 2\vec{j} + \vec{k}) \) and \( (\vec{i} + 4\vec{j} - 3\vec{k}) \) respectively. Show that the points A, B and C are collinear.
Answer: We are given \( \vec{A} = 2\vec{i} + \vec{j} - \vec{k} \), \( \vec{B} = 3\vec{i} - 2\vec{j} + \vec{k} \), and \( \vec{C} = \vec{i} + 4\vec{j} - 3\vec{k} \).

Now we calculate the displacement vectors:
\( \overrightarrow{AB} = (3\vec{i} - 2\vec{j} + \vec{k}) - (2\vec{i} + \vec{j} - \vec{k}) = \vec{i} - 3\vec{j} + 2\vec{k} \)

\( \overrightarrow{BC} = (\vec{i} + 4\vec{j} - 3\vec{k}) - (3\vec{i} - 2\vec{j} + \vec{k}) = -2\vec{i} + 6\vec{j} - 4\vec{k} \)

We observe that \( \overrightarrow{BC} = -2 \overrightarrow{AB} \), which means \( \overrightarrow{BC} \) is a scalar multiple of \( \overrightarrow{AB} \). Since one vector is a scalar multiple of the other, the vectors are parallel. This shows that B lies between A and C on the same line. Therefore, the points A, B, and C are collinear.
In simple words: When one displacement vector is exactly a scaled version of another, the three points must lie on the same straight line.

Exam Tip: Check if \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) (or \( \overrightarrow{AC} \)) are scalar multiples; if so, the points are collinear.

Question 16. If the position vectors of the vertices A, B and C of a triangle ABC be \( (\vec{i} + 2\vec{j} + 3\vec{k}) \), \( (2\vec{i} + 3\vec{j} + \vec{k}) \) and \( (3\vec{i} + \vec{j} + 2\vec{k}) \) respectively, prove that triangle ABC is equilateral.
Answer: We have \( \vec{A} = \vec{i} + 2\vec{j} + 3\vec{k} \), \( \vec{B} = 2\vec{i} + 3\vec{j} + \vec{k} \), and \( \vec{C} = 3\vec{i} + \vec{j} + 2\vec{k} \).

Now we find the side vectors:
\( \overrightarrow{AB} = (2\vec{i} + 3\vec{j} + \vec{k}) - (\vec{i} + 2\vec{j} + 3\vec{k}) = \vec{i} + \vec{j} - 2\vec{k} \)

\( \overrightarrow{BC} = (3\vec{i} + \vec{j} + 2\vec{k}) - (2\vec{i} + 3\vec{j} + \vec{k}) = \vec{i} - 2\vec{j} + \vec{k} \)

\( \overrightarrow{CA} = (\vec{i} + 2\vec{j} + 3\vec{k}) - (3\vec{i} + \vec{j} + 2\vec{k}) = -2\vec{i} + \vec{j} + \vec{k} \)

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the magnitude is given by \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

\( \therefore |\overrightarrow{AB}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6} \)

\( \therefore |\overrightarrow{BC}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6} \)

\( \therefore |\overrightarrow{CA}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{6} \)

Since \( |\overrightarrow{AB}| = |\overrightarrow{BC}| = |\overrightarrow{CA}| \), all three sides have the same length. Therefore, triangle ABC is equilateral.
In simple words: An equilateral triangle has all three sides of equal length. Calculate each side's length and verify they are equal.

Exam Tip: Always compute magnitudes carefully - verify your arithmetic by checking that the sum of squared components matches before taking the square root.

Question 17. Show that the points A, B and C having position vectors \( (3\vec{i} - 4\vec{j} - 4\vec{k}) \), \( (2\vec{i} - \vec{j} + \vec{k}) \) and \( (\vec{i} - 3\vec{j} - 5\vec{k}) \) respectively, form the vertices of a right-angled triangle.
Answer: We have \( \vec{A} = 3\vec{i} - 4\vec{j} - 4\vec{k} \), \( \vec{B} = 2\vec{i} - \vec{j} + \vec{k} \), and \( \vec{C} = \vec{i} - 3\vec{j} - 5\vec{k} \).

First, we find the side vectors:
\( \overrightarrow{AB} = (2\vec{i} - \vec{j} + \vec{k}) - (3\vec{i} - 4\vec{j} - 4\vec{k}) = -\vec{i} + 3\vec{j} + 5\vec{k} \)

\( \overrightarrow{BC} = (\vec{i} - 3\vec{j} - 5\vec{k}) - (2\vec{i} - \vec{j} + \vec{k}) = -\vec{i} - 2\vec{j} - 6\vec{k} \)

\( \overrightarrow{CA} = (3\vec{i} - 4\vec{j} - 4\vec{k}) - (\vec{i} - 3\vec{j} - 5\vec{k}) = 2\vec{i} - \vec{j} + \vec{k} \)

For any two perpendicular vectors \( \vec{a} \) and \( \vec{b} \), their dot product equals zero: \( \vec{a} \cdot \vec{b} = 0 \).

\( \therefore \overrightarrow{AB} \cdot \overrightarrow{CA} = (-\vec{i} + 3\vec{j} + 5\vec{k}) \cdot (2\vec{i} - \vec{j} + \vec{k}) \)
\( = (-1)(2) + (3)(-1) + (5)(1) = -2 - 3 + 5 = 0 \)

Since the dot product of \( \overrightarrow{AB} \) and \( \overrightarrow{CA} \) equals zero, these two vectors are perpendicular. This means the angle at vertex A is a right angle. Therefore, the triangle is right-angled.
In simple words: Two vectors are perpendicular (at a 90-degree angle) if their dot product equals zero. Check if any two sides of the triangle satisfy this condition.

Exam Tip: When checking for perpendicularity, compute the dot product carefully by multiplying corresponding components and summing the results.

Question 18. Using vector method, show that the points A(1, -1, 0), B(4, -3, 1) and C(2, -4, 5) are the vertices of a right-angled triangle.
Answer: We are given A = (1, -1, 0), B = (4, -3, 1), and C = (2, -4, 5).

First, we determine the side vectors:
\( \overrightarrow{AB} = (4 - 1)\vec{i} + (-3 + 1)\vec{j} + (1 - 0)\vec{k} = 3\vec{i} - 2\vec{j} + \vec{k} \)

\( \overrightarrow{BC} = (2 - 4)\vec{i} + (-4 + 3)\vec{j} + (5 - 1)\vec{k} = -2\vec{i} - \vec{j} + 4\vec{k} \)

\( \overrightarrow{CA} = (1 - 2)\vec{i} + (-1 + 4)\vec{j} + (0 - 5)\vec{k} = -\vec{i} + 3\vec{j} - 5\vec{k} \)

For any two perpendicular vectors \( \vec{a} \) and \( \vec{b} \), their dot product equals zero: \( \vec{a} \cdot \vec{b} = 0 \).

\( \therefore \overrightarrow{AB} \cdot \overrightarrow{BC} = (3\vec{i} - 2\vec{j} + \vec{k}) \cdot (-2\vec{i} - \vec{j} + 4\vec{k}) \)
\( = (3)(-2) + (-2)(-1) + (1)(4) = -6 + 2 + 4 = 0 \)

Since the dot product of \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) equals zero, these two vectors are perpendicular. The angle at vertex B is a right angle. Therefore, the triangle is right-angled.
In simple words: Calculate the displacement vectors for each pair of consecutive vertices. If any two of these are perpendicular (dot product zero), the triangle has a right angle.

Exam Tip: Always convert coordinate points to vectors first, then compute displacements before testing for perpendicularity.

Question 19. Find the position vector of the point which divides the join of the points \( (2\vec{a} - 3\vec{b}) \) and \( (3\vec{a} - 2\vec{b}) \) (i) internally and (ii) externally in the ratio 2 : 3.
Answer: We have \( \vec{A} = 2\vec{a} - 3\vec{b} \) and \( \vec{B} = 3\vec{a} - 2\vec{b} \).

The formula for a point dividing a line joining points \( \vec{a} \) and \( \vec{b} \) in ratio m:n internally or externally is given by \( \frac{m\vec{b} + n\vec{a}}{m + n} \) or \( \frac{m\vec{b} - n\vec{a}}{m - n} \) respectively.

(i) The position vector of the point dividing the line internally in the ratio 2 : 3:
\( = \frac{2(3\vec{a} - 2\vec{b}) + 3(2\vec{a} - 3\vec{b})}{2 + 3} \)
\( = \frac{6\vec{a} - 4\vec{b} + 6\vec{a} - 9\vec{b}}{5} \)
\( = \frac{12\vec{a} - 13\vec{b}}{5} \)

(ii) The position vector of the point dividing the line externally in the ratio 2 : 3:
\( = \frac{2(3\vec{a} - 2\vec{b}) - 3(2\vec{a} - 3\vec{b})}{2 - 3} \)
\( = \frac{6\vec{a} - 4\vec{b} - 6\vec{a} + 9\vec{b}}{-1} \)
\( = -5\vec{b} \)
In simple words: For internal division, blend the two position vectors using weights equal to the opposite ratio parts. For external division, use a subtraction formula instead.

Exam Tip: Remember the formulas: internal division uses \( \frac{m\vec{b} + n\vec{a}}{m + n} \); external division uses \( \frac{m\vec{b} - n\vec{a}}{m - n} \).

Question 20. The position vectors of two points A and B are \( (2\vec{a} + \vec{b}) \) and \( (\vec{a} - 3\vec{b}) \) respectively. Find the position vector of a point C which divides AB externally in the ratio 1 : 2. Also, show that A is the mid-point of the line segment CB.
Answer: We are given \( \vec{A} = 2\vec{a} + \vec{b} \) and \( \vec{B} = \vec{a} - 3\vec{b} \).

The formula for a point dividing a line joining points \( \vec{a} \) and \( \vec{b} \) in ratio m:n externally is given by \( \frac{m\vec{b} - n\vec{a}}{m - n} \).

The position vector of the point C dividing the line externally in the ratio 1 : 2:
\( = \frac{1(\vec{a} - 3\vec{b}) - 2(2\vec{a} + \vec{b})}{1 - 2} \)
\( = \frac{\vec{a} - 3\vec{b} - 4\vec{a} - 2\vec{b}}{-1} \)
\( = \frac{-3\vec{a} - 5\vec{b}}{-1} \)
\( = 3\vec{a} + 5\vec{b} \)

Now we verify that A is the mid-point of B and C:
The mid-point of B and C is given by \( \frac{\vec{B} + \vec{C}}{2} = \frac{(\vec{a} - 3\vec{b}) + (3\vec{a} + 5\vec{b})}{2} = \frac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b} \)

This equals \( \vec{A} \), confirming that A is indeed the mid-point of B and C.
In simple words: External division places the dividing point outside the line segment. Check that A's position matches the average of B and C to confirm the mid-point relationship.

Exam Tip: For external division verification, always compute the mid-point of the other two points to confirm the relationship.

Question 21. Find the position vector of a point R which divides the line joining A(-2, 1, 3) and B(3, 5, -2) in the ratio 2 : 1 (i) internally (ii) externally.
Answer: We are given A = (-2, 1, 3) and B = (3, 5, -2).

First, we express these as position vectors:
\( \overrightarrow{OA} = -2\vec{i} + \vec{j} + 3\vec{k} \)
\( \overrightarrow{OB} = 3\vec{i} + 5\vec{j} - 2\vec{k} \)

The formula for a point dividing a line joining points \( \vec{a} \) and \( \vec{b} \) in ratio m:n internally or externally is given by \( \frac{m\vec{b} + n\vec{a}}{m + n} \) or \( \frac{m\vec{b} - n\vec{a}}{m - n} \) respectively.

(i) The position vector of the point dividing the line internally in the ratio 2 : 1:
\( = \frac{2(3\vec{i} + 5\vec{j} - 2\vec{k}) + 1(-2\vec{i} + \vec{j} + 3\vec{k})}{2 + 1} \)
\( = \frac{6\vec{i} + 10\vec{j} - 4\vec{k} - 2\vec{i} + \vec{j} + 3\vec{k}}{3} \)
\( = \frac{4\vec{i} + 11\vec{j} - \vec{k}}{3} \)
\( = \frac{4}{3}\vec{i} + \frac{11}{3}\vec{j} - \frac{1}{3}\vec{k} \)

(ii) The position vector of the point dividing the line externally in the ratio 2 : 1:
\( = \frac{2(3\vec{i} + 5\vec{j} - 2\vec{k}) - 1(-2\vec{i} + \vec{j} + 3\vec{k})}{2 - 1} \)
\( = \frac{6\vec{i} + 10\vec{j} - 4\vec{k} + 2\vec{i} - \vec{j} - 3\vec{k}}{1} \)
\( = 8\vec{i} + 9\vec{j} - 7\vec{k} \)
In simple words: For internal division, use weighted average favoring the farther point. For external division, use a subtraction-based formula that places the point outside the segment.

Exam Tip: Always double-check by verifying that the internal division point lies between the two given points, while the external division point lies outside.

Question 22. Find the position vector of the mid-point of the vector joining the points \( A(3\vec{i} + 2\vec{j} + 6\vec{k}) \) and \( B(\vec{i} + 4\vec{j} - 2\vec{k}) \).
Answer: We are given \( \overrightarrow{OA} = 3\vec{i} + 2\vec{j} + 6\vec{k} \) and \( \overrightarrow{OB} = \vec{i} + 4\vec{j} - 2\vec{k} \).

The formula for the mid-point of a line joining points \( \vec{a} \) and \( \vec{b} \) is given by \( \frac{\vec{a} + \vec{b}}{2} \).

The position vector of the mid-point:
\( = \frac{(3\vec{i} + 2\vec{j} + 6\vec{k}) + (\vec{i} + 4\vec{j} - 2\vec{k})}{2} \)
\( = \frac{4\vec{i} + 6\vec{j} + 4\vec{k}}{2} \)
\( = 2\vec{i} + 3\vec{j} + 2\vec{k} \)
In simple words: The mid-point is found by adding the two position vectors and dividing by 2, which is equivalent to averaging the coordinates.

Exam Tip: The mid-point formula is the simplest division case (ratio 1:1). Always add first, then divide each component by 2.

Question 23. If \( \overrightarrow{AB} = (2\vec{i} + \vec{j} - 3\vec{k}) \) and A(1, 2, -1) is the given point, find the coordinates of B.
Answer: We are given A = (1, 2, -1) and \( \overrightarrow{AB} = 2\vec{i} + \vec{j} - 3\vec{k} \).

Let the coordinates of point B be (b_1, b_2, b_3).

\( \overrightarrow{AB} = 2\vec{i} + \vec{j} - 3\vec{k} \)

\( \Rightarrow [(b_1 - 1)\vec{i} + (b_2 - 2)\vec{j} + (b_3 + 1)\vec{k}] = 2\vec{i} + \vec{j} - 3\vec{k} \)

Comparing the respective coefficients,

b_1 - 1 = 2, therefore b_1 = 3

b_2 - 2 = 1, therefore b_2 = 3

b_3 + 1 = -3, therefore b_3 = -4

The required coordinates of B are (3, 3, -4).
In simple words: A displacement vector shows how far you move from one point to another. Add the displacement components to point A's coordinates to get point B.

Exam Tip: When equating vectors, always match up coefficients of \( \vec{i} \), \( \vec{j} \), and \( \vec{k} \) separately and solve for each unknown coordinate.

Question 24. Write a unit vector in the direction of \( \overrightarrow{PQ} \), where P and Q are the points (1, 3, 0) and (4, 5, 6) respectively.
Answer: We are given P = (1, 3, 0) and Q = (4, 5, 6).

First, we find the displacement vector:
\( \overrightarrow{PQ} = (4 - 1)\vec{i} + (5 - 3)\vec{j} + (6 - 0)\vec{k} = 3\vec{i} + 2\vec{j} + 6\vec{k} \)

For any vector \( \vec{a} = a_x\vec{i} + a_y\vec{j} + a_z\vec{k} \), the unit vector is represented as \( \hat{a} = \frac{a_x\vec{i} + a_y\vec{j} + a_z\vec{k}}{\sqrt{a_x^2 + a_y^2 + a_z^2}} \).

\( \therefore \widehat{PQ} = \frac{3\vec{i} + 2\vec{j} + 6\vec{k}}{\sqrt{3^2 + 2^2 + 6^2}} = \frac{3\vec{i} + 2\vec{j} + 6\vec{k}}{7} = \frac{1}{7}(3\vec{i} + 2\vec{j} + 6\vec{k}) \)
In simple words: Find the vector by subtracting the starting point's coordinates from the ending point's coordinates. Then normalize by dividing by the vector's length.

Exam Tip: Always verify that your final unit vector has magnitude 1 by checking that the sum of the squared components equals 1.