Get detailed solutions for NCERT Class 12 Maths Exercise 10.2 (Vector Algebra). Master computing vector magnitude, finding scalar and vector components, calculating unit vectors ($\hat{a}$), and determining direction cosines. Includes solving problems on collinearity, vector addition, section formula (internal and external division), and geometric applications (midpoint, right-angled triangles). Essential for securing marks in Chapter 10.

Ex. 10.1

Ex. 10.2

Ex. 10.3

Ex. 10.4

Ex. 10 Misc

This exercise covers the calculation of vector magnitude, vector components, unit vectors, and collinearity.

Rbse Solutions Class 12 Maths Exercise 10.2 | Vector Algebra

Part 1: Vector Computations (Q1-Q10)

1. Compute the magnitude of the following vectors.

The magnitude of a vector $\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}$ is $|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$.

$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$$|\vec{a}| = \sqrt{(1)^2 + (1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \mathbf{\sqrt{3}}$$
$\vec{b} = 2\hat{i} – 7\hat{j} – 3\hat{k}$$$|\vec{b}| = \sqrt{(2)^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \mathbf{\sqrt{62}}$$
$\vec{c} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} – \frac{1}{\sqrt{3}}\hat{k}$$$|\vec{c}| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(-\frac{1}{\sqrt{3}}\right)^2} = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{1} = \mathbf{1}$$

2. Write two different vectors having same magnitude.

Choose any two vectors with different components that result in the same sum of squares.

$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $|\vec{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$
$\vec{b} = 3\hat{i} + 2\hat{j} + \hat{k}$, $|\vec{b}| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14}$These vectors are different but have the same magnitude $\mathbf{\sqrt{14}}$.

3. Write two different vectors having same direction.

Two vectors $\vec{a}$ and $\vec{b}$ have the same direction if they are collinear and have the same sense (i.e., $\vec{b} = \lambda \vec{a}$ where $\lambda > 0$).

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$
Choose $\lambda = 2$: $\vec{b} = 2\vec{a} = 2\hat{i} + 2\hat{j} + 2\hat{k}$These vectors are different (different magnitudes) but have the same direction.

4. Find the values of $x$ and $y$ so that the vectors $2\hat{i} + 3\hat{j}$ and $x\hat{i} + y\hat{j}$ are equal.

Two vectors $\vec{a} = a_1 \hat{i} + a_2 \hat{j}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j}$ are equal if and only if their corresponding components are equal: $a_1 = b_1$ and $a_2 = b_2$.

$$2\hat{i} + 3\hat{j} = x\hat{i} + y\hat{j}$$

By comparing coefficients:

$$x = \mathbf{2}$$

$$y = \mathbf{3}$$

5. Find the scalar and vector components of the vector with initial point $P(2, 1)$ and terminal point $Q(-5, 7)$.

The vector $\vec{PQ}$ is found by subtracting the coordinates of the initial point from the coordinates of the terminal point:

$$\vec{PQ} = (x_2 – x_1)\hat{i} + (y_2 – y_1)\hat{j}$$

$$\vec{PQ} = (-5 – 2)\hat{i} + (7 – 1)\hat{j} = -7\hat{i} + 6\hat{j}$$

Scalar components: The coefficients of $\hat{i}$ and $\hat{j}$, which are $\mathbf{-7}$ and $\mathbf{6}$.
Vector components: The individual vector terms, which are $\mathbf{-7\hat{i}}$ and $\mathbf{6\hat{j}}$.

6. Find the sum of the vectors $\vec{a} = \hat{i} – 2\hat{j} + \hat{k}$, $\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{c} = \hat{i} – 6\hat{j} – 7\hat{k}$.

The sum $\vec{r} = \vec{a} + \vec{b} + \vec{c}$ is found by adding the corresponding components:

$$\vec{r} = (1 + (-2) + 1)\hat{i} + (-2 + 4 + (-6))\hat{j} + (1 + 5 + (-7))\hat{k}$$

$$\vec{r} = (0)\hat{i} + (-4)\hat{j} + (-1)\hat{k}$$

$$\vec{r} = \mathbf{-4\hat{j} – \hat{k}}$$

7. Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.

The unit vector in the direction of $\vec{a}$ is given by $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.

Magnitude of $\vec{a}$:$$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$
Unit Vector $\hat{a}$:$$\hat{a} = \frac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}} = \mathbf{\frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}}$$

8. Find the unit vector in the direction of vector $\vec{PQ}$, where $P(1, 2, 3)$ and $Q(4, 5, 6)$.

Vector $\vec{PQ}$:$$\vec{PQ} = (4 – 1)\hat{i} + (5 – 2)\hat{j} + (6 – 3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$$
Magnitude of $\vec{PQ}$:$$|\vec{PQ}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$$
Unit Vector $\hat{PQ}$:$$\hat{PQ} = \frac{3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}} = \frac{3(\hat{i} + \hat{j} + \hat{k})}{3\sqrt{3}} = \mathbf{\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}}$$

9. For given vectors, $\vec{a} = 2\hat{i} – \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$, find the unit vector in the direction of the vector $\vec{a} + \vec{b}$.

Vector Sum $\vec{r} = \vec{a} + \vec{b}$:$$\vec{r} = (2 + (-1))\hat{i} + (-1 + 1)\hat{j} + (2 + 1)\hat{k} = 1\hat{i} + 0\hat{j} + 3\hat{k} = \hat{i} + 3\hat{k}$$
Magnitude of $\vec{r}$:$$|\vec{r}| = \sqrt{1^2 + 0^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$$
Unit Vector $\hat{r}$:$$\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \mathbf{\frac{1}{\sqrt{10}}\hat{i} + \frac{3}{\sqrt{10}}\hat{k}}$$

10. Find a vector in the direction of vector $5\hat{i} – \hat{j} + 2\hat{k}$ which has magnitude 8 units.

Let $\vec{a} = 5\hat{i} – \hat{j} + 2\hat{k}$. The required vector $\vec{b}$ is $8 \hat{a}$.

Magnitude of $\vec{a}$:$$|\vec{a}| = \sqrt{5^2 + (-1)^2 + 2^2} = \sqrt{25 + 1 + 4} = \sqrt{30}$$
Unit Vector $\hat{a}$:$$\hat{a} = \frac{1}{\sqrt{30}} (5\hat{i} – \hat{j} + 2\hat{k})$$
Required Vector $\vec{b}$:$$\vec{b} = 8 \hat{a} = \mathbf{\frac{8}{\sqrt{30}} (5\hat{i} – \hat{j} + 2\hat{k})}$$

Part 2: Collinearity and Direction Cosines (Q11-Q14)

11. Show that the vectors $\vec{a} = 2\hat{i} – 3\hat{j} + 4\hat{k}$ and $\vec{b} = -4\hat{i} + 6\hat{j} – 8\hat{k}$ are collinear.

Two vectors $\vec{a}$ and $\vec{b}$ are collinear if and only if $\vec{b} = \lambda \vec{a}$ for some scalar $\lambda \neq 0$.

$$\vec{b} = -4\hat{i} + 6\hat{j} – 8\hat{k}$$

Factor out a scalar from $\vec{b}$:

$$\vec{b} = -2(2\hat{i} – 3\hat{j} + 4\hat{k})$$

Since $\vec{b} = -2\vec{a}$, we have $\lambda = -2$.

Thus, $\vec{a}$ and $\vec{b}$ are collinear.

12. Find the direction cosines of the vector $\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k}$.

For a vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, the direction cosines are $l, m, n$:

$$l = \frac{x}{|\vec{r}|}, \quad m = \frac{y}{|\vec{r}|}, \quad n = \frac{z}{|\vec{r}|}$$

Magnitude of $\vec{r}$:$$|\vec{r}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$
Direction Cosines:$$l = \frac{1}{\sqrt{14}}, \quad m = \frac{2}{\sqrt{14}}, \quad n = \frac{3}{\sqrt{14}}$$

The direction cosines are $\mathbf{\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}}$.

13. Find the direction cosines of the vector joining the points $A(1, 2, -3)$ and $B(-1, -2, 1)$, directed from $A$ to $B$.

Vector $\vec{AB}$:$$\vec{AB} = (-1 – 1)\hat{i} + (-2 – 2)\hat{j} + (1 – (-3))\hat{k} = -2\hat{i} – 4\hat{j} + 4\hat{k}$$
Magnitude of $\vec{AB}$:$$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$
Direction Cosines:$$l = \frac{-2}{6}, \quad m = \frac{-4}{6}, \quad n = \frac{4}{6}$$$$l = -\frac{1}{3}, \quad m = -\frac{2}{3}, \quad n = \frac{2}{3}$$

The direction cosines are $\mathbf{-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}}$.

14. Show that the vector $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes $OX, OY$ and $OZ$.

A vector is equally inclined to the axes if its direction cosines are equal, i.e., $l = m = n$.

Magnitude of $\vec{r}$:$$|\vec{r}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$$
Direction Cosines:$$l = \frac{1}{\sqrt{3}}, \quad m = \frac{1}{\sqrt{3}}, \quad n = \frac{1}{\sqrt{3}}$$Since $l = m = n$, the vector is equally inclined to the axes.

Part 3: Position Vectors and Geometry (Q15-Q17)

15. Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $\vec{p} = \hat{i} + 2\hat{j} – \hat{k}$ and $\vec{q} = -\hat{i} + \hat{j} + \hat{k}$ in the ratio $2:1$.

The Section Formula for position vectors dividing a line segment in the ratio $m:n$ is $\vec{r} = \frac{m\vec{q} \pm n\vec{p}}{m \pm n}$. Here $m=2$ and $n=1$.

(i) Internally

$$\vec{r}_{\text{int}} = \frac{m\vec{q} + n\vec{p}}{m + n} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} – \hat{k})}{2 + 1}$$

$$\vec{r}_{\text{int}} = \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} – \hat{k})}{3} = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3}$$

$$\vec{r}_{\text{int}} = \mathbf{-\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}}$$

(ii) Externally

$$\vec{r}_{\text{ext}} = \frac{m\vec{q} – n\vec{p}}{m – n} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) – 1(\hat{i} + 2\hat{j} – \hat{k})}{2 – 1}$$

$$\vec{r}_{\text{ext}} = \frac{(-2\hat{i} + 2\hat{j} + 2\hat{k}) – \hat{i} – 2\hat{j} + \hat{k}}{1} = -3\hat{i} + 0\hat{j} + 3\hat{k}$$

$$\vec{r}_{\text{ext}} = \mathbf{-3\hat{i} + 3\hat{k}}$$

16. Find the position vector of the mid point of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.

The position vectors are $\vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{q} = 4\hat{i} + \hat{j} – 2\hat{k}$.

The mid-point is the internal division in the ratio $1:1$: $\vec{r}_{\text{mid}} = \frac{\vec{p} + \vec{q}}{2}$.

$$\vec{r}_{\text{mid}} = \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + \hat{j} – 2\hat{k})}{2} = \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2}$$

$$\vec{r}_{\text{mid}} = \mathbf{3\hat{i} + 2\hat{j} + \hat{k}}$$

17. Show that the points $A, B$ and $C$ with position vectors $\vec{a} = 3\hat{i} – 4\hat{j} – 4\hat{k}$, $\vec{b} = 2\hat{i} – \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} – 3\hat{j} – 5\hat{k}$ form the vertices of a right angled triangle.

A triangle is right-angled if the square of the magnitude of one side vector equals the sum of the squares of the magnitudes of the other two side vectors (Pythagoras theorem).

Side Vectors:$$\vec{AB} = \vec{b} – \vec{a} = (2-3)\hat{i} + (-1-(-4))\hat{j} + (1-(-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$$$$\vec{BC} = \vec{c} – \vec{b} = (1-2)\hat{i} + (-3-(-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} – 2\hat{j} – 6\hat{k}$$$$\vec{AC} = \vec{c} – \vec{a} = (1-3)\hat{i} + (-3-(-4))\hat{j} + (-5-(-4))\hat{k} = -2\hat{i} + \hat{j} – \hat{k}$$
Square of Magnitudes:$$|\vec{AB}|^2 = (-1)^2 + 3^2 + 5^2 = 1 + 9 + 25 = \mathbf{35}$$$$|\vec{BC}|^2 = (-1)^2 + (-2)^2 + (-6)^2 = 1 + 4 + 36 = \mathbf{41}$$$$|\vec{AC}|^2 = (-2)^2 + 1^2 + (-1)^2 = 4 + 1 + 1 = \mathbf{6}$$
Check Pythagoras:$$|\vec{AC}|^2 + |\vec{AB}|^2 = 6 + 35 = 41$$Since $|\vec{AC}|^2 + |\vec{AB}|^2 = |\vec{BC}|^2$, the triangle $\text{ABC}$ is a right-angled triangle with the right angle at $\text{A}$.

Part 4: Multiple Choice Questions (Q18-Q19)

18. In triangle ABC, which of the following is not true:

In $\triangle \text{ABC}$, the triangle law of vector addition states: $\vec{AB} + \vec{BC} = \vec{AC}$.

(A) $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$: True, because $\vec{AB} + \vec{BC} = \vec{AC}$, and $\vec{AC} + \vec{CA} = \vec{0}$.
(B) $\vec{AB} + \vec{BC} – \vec{AC} = \vec{0}$: True, because $\vec{AB} + \vec{BC} = \vec{AC}$.
(C) $\vec{AB} + \vec{BC} – \vec{CA} = \vec{0}$: False, because $\vec{AB} + \vec{BC} = \vec{AC}$. This means $\vec{AC} – \vec{CA} = \vec{0}$, which simplifies to $\vec{AC} + \vec{AC} = 2\vec{AC} = \vec{0}$. This is only true if $\vec{AC} = \vec{0}$.
(D) $\vec{AB} – \vec{CB} + \vec{CA} = \vec{0}$: True, because $\vec{AB} – \vec{CB} = \vec{AB} + \vec{BC} = \vec{AC}$. This means $\vec{AC} + \vec{CA} = \vec{0}$.

The statement that is not true is (C).

19. If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect:

Collinear vectors mean $\vec{b} = \lambda \vec{a}$.

(A) $\vec{b} = \lambda \vec{a}$, for some scalar $\lambda$: Correct. This is the definition of collinearity.
(B) $\vec{a} = \pm \vec{b}$: Incorrect. This is only true if they are unit vectors or equal in magnitude. In general, $\vec{a}$ and $\vec{b}$ can have any magnitudes.
(C) the respective components of $\vec{a}$ and $\vec{b}$ are not proportional: Incorrect. Since $\vec{b} = \lambda \vec{a}$, their components are proportional ($b_i = \lambda a_i$).
(D) both the vectors have same direction, but different magnitudes: Incorrect. They can have opposite directions ($\lambda < 0$) or the same magnitude ($\lambda = \pm 1$).

The statements that are incorrect are (B), (C), and (D).