NCERT Solutions Class 12 Maths Chapter 10 Ex 10.2 | LearnCBSEHub

Vector Operations

NCERT EXERCISE 10.2 • FULL SOLUTIONS Q1-Q19

💡 Formulas at a Glance

  • Magnitude: If $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, then $|\vec{r}| = \sqrt{x^2+y^2+z^2}$.
  • Unit Vector: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
  • Vector Joining Points: $\vec{PQ} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$.
  • Section Formula: $\vec{r} = \frac{m\vec{b} \pm n\vec{a}}{m \pm n}$.
Question 01
Compute the magnitude of the vectors:
$\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} – 7\hat{j} – 3\hat{k}$, $\vec{c} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} – \frac{1}{\sqrt{3}}\hat{k}$.
$|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

$|\vec{b}| = \sqrt{2^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \sqrt{62}$

$|\vec{c}| = \sqrt{(\frac{1}{\sqrt{3}})^2 + (\frac{1}{\sqrt{3}})^2 + (-\frac{1}{\sqrt{3}})^2} = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{1} = 1$
Question 02
Write two different vectors having same magnitude.

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$.

$|\vec{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$
$|\vec{b}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}$
Here $\vec{a} \neq \vec{b}$ but $|\vec{a}| = |\vec{b}|$.
Question 03
Write two different vectors having same direction.

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 3(\hat{i} + \hat{j} + \hat{k}) = 3\hat{i} + 3\hat{j} + 3\hat{k}$.

The direction cosines of both vectors are $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$. Thus, they have the same direction but different magnitudes.

Question 04
Find values of $x$ and $y$ so that $2\hat{i} + 3\hat{j}$ and $x\hat{i} + y\hat{j}$ are equal.

Two vectors are equal if their corresponding components are equal.

x = 2, y = 3
Question 05
Find scalar and vector components of vector with initial point (2, 1) and terminal point (-5, 7).
Let $A=(2, 1)$ and $B=(-5, 7)$.
$\vec{AB} = (-5 – 2)\hat{i} + (7 – 1)\hat{j} = -7\hat{i} + 6\hat{j}$.

Scalar Components: -7 and 6

Vector Components: $-7\hat{i}$ and $6\hat{j}$

Question 06
Find the sum of vectors $\vec{a} = \hat{i} – 2\hat{j} + \hat{k}$, $\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{c} = \hat{i} – 6\hat{j} – 7\hat{k}$.
Sum $= (1 – 2 + 1)\hat{i} + (-2 + 4 – 6)\hat{j} + (1 + 5 – 7)\hat{k}$
$= 0\hat{i} – 4\hat{j} – 1\hat{k} = -4\hat{j} – \hat{k}$.
Question 07
Find unit vector in direction of $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.
$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$.
$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$.
Question 08
Find unit vector in direction of $\vec{PQ}$ where P(1, 2, 3) and Q(4, 5, 6).
$\vec{PQ} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.
$|\vec{PQ}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$.
Unit Vector = $\frac{3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$.
Question 09
Find unit vector in direction of $\vec{a} + \vec{b}$ for $\vec{a} = 2\hat{i} – \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} – \hat{k}$.
$\vec{c} = \vec{a} + \vec{b} = (2-1)\hat{i} + (-1+1)\hat{j} + (2-1)\hat{k} = \hat{i} + \hat{k}$.
$|\vec{c}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$.
$\hat{c} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$.
Question 10
Find a vector in direction of $5\hat{i} – \hat{j} + 2\hat{k}$ with magnitude 8 units.
Let $\vec{a} = 5\hat{i} – \hat{j} + 2\hat{k}$.
$|\vec{a}| = \sqrt{25 + 1 + 4} = \sqrt{30}$.
$\hat{a} = \frac{1}{\sqrt{30}}(5\hat{i} – \hat{j} + 2\hat{k})$.
Required Vector = $8\hat{a} = \frac{40}{\sqrt{30}}\hat{i} – \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}$.
Question 11
Show vectors $2\hat{i} – 3\hat{j} + 4\hat{k}$ and $-4\hat{i} + 6\hat{j} – 8\hat{k}$ are collinear.
Let $\vec{a} = 2\hat{i} – 3\hat{j} + 4\hat{k}$.
Let $\vec{b} = -4\hat{i} + 6\hat{j} – 8\hat{k} = -2(2\hat{i} – 3\hat{j} + 4\hat{k}) = -2\vec{a}$.
Since $\vec{b} = \lambda \vec{a}$ (where $\lambda = -2$), the vectors are collinear.
Question 12
Find direction cosines of vector $\hat{i} + 2\hat{j} + 3\hat{k}$.
$\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k}$.
$|\vec{r}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$.
Direction Cosines: $(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}})$.
Question 13
Find direction cosines of vector joining A(1, 2, -3) and B(-1, -2, 1) directed from A to B.
$\vec{AB} = (-1-1)\hat{i} + (-2-2)\hat{j} + (1-(-3))\hat{k} = -2\hat{i} – 4\hat{j} + 4\hat{k}$.
$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
DCs: $(\frac{-2}{6}, \frac{-4}{6}, \frac{4}{6}) = (-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})$.
Question 14
Show vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to axes OX, OY, OZ.
Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$. Magnitude $|\vec{a}| = \sqrt{3}$.
Direction Cosines are $l = \frac{1}{\sqrt{3}}, m = \frac{1}{\sqrt{3}}, n = \frac{1}{\sqrt{3}}$.
Since $l = m = n$, the angles $\alpha, \beta, \gamma$ are equal. $\cos\alpha = \cos\beta = \cos\gamma = 1/\sqrt{3}$.
Hence, the vector is equally inclined to the axes.
Question 15
Find position vector of point R dividing P ($\hat{i} + 2\hat{j} – \hat{k}$) and Q ($-\hat{i} + \hat{j} + \hat{k}$) in ratio 2:1.
(i) Internally
$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} – \hat{k})}{2+1}$
$= \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} + \hat{i} + 2\hat{j} – \hat{k}}{3} = \frac{-\hat{i} + 4\hat{j} + \hat{k}}{3} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} + \frac{1}{3}\hat{k}$.
(ii) Externally
$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) – 1(\hat{i} + 2\hat{j} – \hat{k})}{2-1}$
$= \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} – \hat{i} – 2\hat{j} + \hat{k}}{1} = -3\hat{i} + 3\hat{k}$.
Question 16
Find position vector of mid point of vector joining P(2, 3, 4) and Q(4, 1, -2).
Midpoint Formula: $\frac{\vec{a} + \vec{b}}{2}$.
$\vec{m} = \frac{(2\hat{i} + 3\hat{j} + 4\hat{k}) + (4\hat{i} + 1\hat{j} – 2\hat{k})}{2}$
$= \frac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2} = 3\hat{i} + 2\hat{j} + \hat{k}$.
Question 17
Show points A, B, C form a right angled triangle.
$\vec{a} = 3\hat{i} – 4\hat{j} – 4\hat{k}, \quad \vec{b} = 2\hat{i} – \hat{j} + \hat{k}, \quad \vec{c} = \hat{i} – 3\hat{j} – 5\hat{k}$.
Find side vectors:
$\vec{AB} = \vec{b} – \vec{a} = -\hat{i} + 3\hat{j} + 5\hat{k} \implies |\vec{AB}|^2 = 1 + 9 + 25 = 35$.
$\vec{BC} = \vec{c} – \vec{b} = -\hat{i} – 2\hat{j} – 6\hat{k} \implies |\vec{BC}|^2 = 1 + 4 + 36 = 41$.
$\vec{CA} = \vec{a} – \vec{c} = 2\hat{i} – \hat{j} + \hat{k} \implies |\vec{CA}|^2 = 4 + 1 + 1 = 6$.
Since $35 + 6 = 41$ ($|\vec{AB}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2$), it forms a right angled triangle.
Question 18
In triangle ABC, which is not true?
[Image of triangle law of vector addition]

Triangle Law: $\vec{AB} + \vec{BC} = \vec{AC} \implies \vec{AB} + \vec{BC} – \vec{AC} = \vec{0}$.

Also $\vec{AC} = -\vec{CA}$, so $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$.

(C) states $\vec{AB} + \vec{BC} – \vec{CA} = \vec{0}$, which implies $\vec{AB} + \vec{BC} = \vec{CA}$ (Incorrect magnitude direction combination).

Correct Answer: (C)
Question 19
If $\vec{a}, \vec{b}$ collinear, which is incorrect?

(A) $\vec{b} = \lambda\vec{a}$ (True definition).

(B) $\vec{a} = \pm \vec{b}$ (True for specific cases).

(C) respective components are proportional (True).

(D) both vectors have same direction, but different magnitudes (False: can have opposite direction).

Correct Answer: (D)
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