Exercise 10.2 (Revised) - Chapter 10 - Vector Algebra - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths: Chapter 10 - Vector Algebra Solutions

Ex 10.2 Question 1.

Compute the magnitude of the following vectors:
$
\vec{a}=\hat{i}+\hat{j}+\hat{k}, \quad \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k}, \quad \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}
$

Answer.

Given: $\vec{a}=\hat{i}+\hat{j}+\hat{k}$
$
\therefore|\vec{a}|=\sqrt{x^2+y^2+z^2}=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{1+1+1}=\sqrt{3}
$

And $\vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k}$
$
\therefore|\vec{b}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(-7)^2+(-3)^2}=\sqrt{4+49+9}=\sqrt{62}
$

Also $\vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$
$
\therefore|\vec{c}|=\sqrt{x^2+y^2+z^2}=\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}}=1
$

Ex 10.2 Question 2.

Write two different vectors having same magnitude.

Answer.

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$
Clearly, $\vec{a} \neq \vec{b},[\because$ Coefficients of $\hat{i}$ and $\hat{j}$ are same in vectors $\vec{a}$ and $\vec{b}$ coefficients of $\hat{k}$ in $\vec{a}$ and $\vec{b}$ are unequal as $1 \neq-1$. ]

But $|\vec{a}|=\sqrt{x^2+y^2+z^2}=\sqrt{1+1+1}=\sqrt{3}$
And $|\vec{b}|=\sqrt{x^2+y^2+z^2}=\sqrt{1+1+1}=\sqrt{3}$

 Such possible answers are infinite. 

Ex 10.2 Question 3.

Write two different vectors having same direction.

Answer.

Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b}=2(\hat{i}+2 \hat{j}+3 \hat{k})=2 \vec{a}$
$\therefore \vec{b}=m \vec{a}$ where $m=2>0$
$\therefore$ Vectors $\vec{a}$ and $\vec{b}$ have the same direction.
But $\vec{b} \neq \vec{a}[\because \vec{b}=2 \vec{a} \Rightarrow|\vec{b}|=|2||\vec{a}|=|\vec{a}|]$
Such possible vectors are infinite.

Ex 10.2 Question 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.

Answer.

Given: $2 \hat{i}+3 \hat{j}=x \hat{i}+y \hat{j}$
If vectors are equal, then their respective components are equal.
Comparing coefficients of $\hat{i}$ and $\hat{j}$ on both sides, we have, $x=2$ and $y=3$.

Ex 10.2 Question 5.

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

Answer.

Let $\overrightarrow{\mathrm{AB}}$ be the vector with initial point $\mathrm{A}(2,1)$ and terminal point $\mathrm{B}(-5,7)$.
$\Rightarrow$ Position vector of point $\mathrm{A}$ is $(2,1)=2 \hat{i}+\hat{j}$ and position vector of point $\mathrm{B}$ is $(-5,7)=$ $-5 \hat{i}+7 \hat{j}$
$\therefore \overrightarrow{\mathrm{AB}}=$ Position vector of point $\mathrm{B}-$ Position vector of point $\mathrm{A}$

$
\begin{aligned}
& =(-5 \hat{i}+7 \hat{j})-(2 \hat{i}+\hat{j}) \\
& =-5 \hat{i}+7 \hat{j}-2 \hat{i}-\hat{j} \\
& \Rightarrow \overrightarrow{\mathrm{AB}}=-7 \hat{i}+6 \hat{j}
\end{aligned}
$
$\therefore$ Scalar components of the vectors $\overrightarrow{\mathrm{AB}}$ are coefficients of $\hat{i}$ and $\hat{j}$ in $\overrightarrow{\mathrm{AB}}$ i.e., -7 and 6 and vector components of the vector $\overrightarrow{\mathrm{AB}}$ are $-7 \hat{i}$ and $6 \hat{j}$.

Ex 10.2 Question 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 $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$

Answer.

Given: $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \quad \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$
Adding, $\vec{a}+\vec{b}+\vec{c}=\hat{i}-2 \hat{j}+\hat{k}-2 \hat{i}+4 \hat{j}+5 \hat{k}+\hat{i}-6 \hat{j}-7 \hat{k}$
$
=0 \hat{i}-4 \hat{j}-\hat{k}=-4 \hat{j}-\hat{k}
$

Ex 10.2 Question 7.

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

Answer.

We know that a unit vector in the direction of the vector $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ is
$
\begin{aligned}
& \widehat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{(1)^2+(1)^2+(2)^2}}=\frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{1+1+4}}=\frac{\hat{i+}+\hat{j}+2 \hat{k}}{\sqrt{6}} \\
& \Rightarrow \widehat{a}=\frac{1}{\sqrt{6}} \hat{i}+\frac{1}{\sqrt{6}} \hat{j}+\frac{2}{\sqrt{6}} \hat{k} \\
&
\end{aligned}
$

Ex 10.2 Question 8.

Find the unit vector in the direction of the vector $\overrightarrow{\mathrm{PQ}}$ where $P$ and $Q$ are the points $(1,2,3)$ and $(4,5,6)$ respectively.

Answer.

Given: Points P $(1,2,3)$ and Q $(4,5,6)$
$\therefore$ Position vector of point $\mathrm{P}=\overrightarrow{\mathrm{OP}}=\hat{i}+2 \hat{j}+3 \hat{k}$ and position vector of $\mathrm{Q}=\overrightarrow{\mathrm{OQ}}=$ $4 \hat{i}+5 \hat{j}+6 \hat{k}$, where $\mathrm{O}$ is the origin.

$
\begin{aligned}
& \therefore \overrightarrow{\mathrm{PQ}}=\text { Position vector of } \mathrm{Q}-\text { Position vector of } \mathrm{P}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}} \\
& =4 \hat{i}+5 \hat{j}+6 \hat{k}-\hat{i}-2 \hat{j}-3 \hat{k}=3 \hat{i}+3 \hat{j}+3 \hat{k}
\end{aligned}
$

Therefore, the unit vector in the direction of vector $\overrightarrow{\mathrm{PQ}}=\frac{\overrightarrow{\mathrm{PQ}}}{|\overrightarrow{\mathrm{PQ}}|}$
$
\begin{aligned}
& =\frac{3 \hat{i}+3 \hat{j}+3 \hat{k}}{\sqrt{(3)^2+(3)^2+(3)^2}}=\frac{3 \hat{i}+3 \hat{j}+3 \hat{k}}{\sqrt{9+9+9}}=\frac{3 \hat{i}+3 \hat{j}+3 \hat{k}}{\sqrt{27}}=\frac{3 \hat{i}+3 \hat{j}+3 \hat{k}}{3 \sqrt{3}} \\
& =\frac{3(\hat{i}+\hat{j}+\hat{k})}{3 \sqrt{3}}=\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}
\end{aligned}
$

Ex 10.2 Question 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 $\vec{a}+\vec{b}$.

Answer.

Given: Vectors $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$
$
\therefore \vec{a}+\vec{b}=2 \hat{i}-\hat{j}+2 \hat{k}-\hat{i}+\hat{j}-\hat{k}=\hat{i}+0 \hat{j}+\hat{k}
$

Therefore, $|\vec{a}+\vec{b}|=\frac{\hat{i}+0 \hat{j}+\hat{k}}{\sqrt{(1)^2+(0)^2+(1)^2}}$
$
\Rightarrow|\vec{a}+\vec{b}|=\frac{\hat{i}+0 \hat{j}+\hat{k}}{\sqrt{1+0+1}}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}
$

Ex 10.2 Question 10.

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

Answer.

Let $\vec{a}=5 \hat{i}-\hat{j}+2 \hat{k}$
A vector in the direction of vector $\vec{a}$ which has magnitude 8 units $=8 \hat{a}$
$8 \cdot \frac{\vec{a}}{|\vec{a}|}=\frac{8(5 \hat{i}-\hat{j}+2 \hat{k})}{\sqrt{(5)^2+(-1)^2+(2)^2}}$

$\begin{aligned}
& \Rightarrow 8 \frac{\vec{a}}{|\vec{a}|}=\frac{8(5 \hat{i}-\hat{j}+2 \hat{k})}{\sqrt{25+1+4}}=\frac{8}{\sqrt{30}}(5 \hat{i}-\hat{j}+2 \hat{k}) \\
& =\frac{40}{\sqrt{30}} \hat{i}-\frac{8}{\sqrt{30}} \hat{j}+\frac{16}{\sqrt{30}} \hat{k}
\end{aligned}$

Ex 10.2 Question 11.

Show that the vectors $2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $-4 \hat{i}+6 \hat{j}-8 \hat{k}$ are collinear.

Answer.

Let $\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $\vec{b}=-4 \hat{i}+6 \hat{j}-8 \hat{k}=-2(2 \hat{i}-3 \hat{j}+4 \hat{k})=-2 \vec{a}$ $\Rightarrow \vec{b}=-2 \vec{a}=m \vec{a}$ where $m=-2<0$

Therefore, vectors $2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $-4 \hat{i}+6 \hat{j}-8 \hat{k}$ are collinear.

Ex 10.2 Question 12.

Find the direction cosines of the vector $\hat{i}+2 \hat{j}+3 \hat{k}$

Answer.

The given vector is $(\vec{a})=\hat{i}+2 \hat{j}+3 \hat{k}$
$
\therefore \hat{a}=\frac{\hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{(1)^2+(2)^2+(3)^2}}=\frac{\hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{1+4+9}}=\frac{\hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{14}}=\frac{1}{\sqrt{14}} \hat{i}+\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}
$

We know that the direction cosines of a vector $\vec{a}$ are coefficients of $\hat{i}, \hat{j}, \hat{k}$ in $\hat{a}$ i.e.,
$
\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}
$
Ex 10.2 Question 13.

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

Answer.

Given: Points $\mathrm{A}(1,2,-3)$ and $\mathrm{B}(-1,-2,1)$
$\therefore$ Position vector of point $\mathrm{A}=\overrightarrow{\mathrm{OA}}=\hat{i}+2 \hat{j}-3 \hat{k}$

And Position vector of point $\mathrm{B}=\overrightarrow{\mathrm{OB}}=-\hat{i}-2 \hat{j}+\hat{k}$
$\therefore$ Vector $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=-\hat{i}-2 \hat{j}+\hat{k}-\hat{i}-2 \hat{j}+3 \hat{k}=-2 \hat{i}-4 \hat{j}+4 \hat{k}$
Now $|\overrightarrow{\mathrm{AB}}|=\sqrt{(-2)^2+(-4)^2+(4)^2}=\sqrt{4+16+16}=\sqrt{36}=6$
$\therefore$ A unit vector along $\overrightarrow{\mathrm{AB}}=\frac{\overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AB}}|}=\frac{-2 \hat{i}-4 \hat{j}+4 \hat{k}}{6}$
$=\frac{-2}{6} \hat{i}+\frac{-4}{6} \hat{j}+\frac{4}{6} \hat{k}=\frac{-1}{3} \hat{i}+\frac{-2}{3}+\frac{2}{3} \hat{k}$
Therefore, the direction cosines of vector $\overrightarrow{\mathrm{AB}}=\frac{-1}{3}, \frac{-2}{3}, \frac{2}{3}$

Ex 10.2 Question 14.

Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes $O \mathrm{X}$, oY and $\mathrm{Oz}$.

Answer.

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, then $|\vec{a}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
Let us find angle $\theta_1$ (say) between vector $\vec{a}$ and OX $(\Rightarrow \hat{i})$
$
\begin{aligned}
& \Rightarrow \cos \theta_1=\frac{\vec{a} \hat{i}}{|\vec{a}| \cdot|\cdot|}=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}+0 \hat{j}+0 \hat{k})}{|\hat{i}+\hat{j}+\hat{k}||\hat{i}+0 \hat{j}+0 \hat{k}|}=\frac{1(1)+1(0)+1(0)}{\sqrt{1+1+1} \sqrt{1+0+0}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \theta_1=\cos ^{-1} \frac{1}{\sqrt{3}}
\end{aligned}
$

Similarly angle $\theta_2$ (say) between vector $\vec{a}$ and OY $(\Rightarrow \hat{j})$ is $\cos ^{-1} \frac{1}{\sqrt{3}}$
And angle $\theta_3$ (say) between vector $\vec{a}$ and $\mathrm{OZ}(\Rightarrow \hat{k})$ is $\cos ^{-1} \frac{1}{\sqrt{3}}$

$
\therefore \theta_1=\theta_2=\theta_3
$
$\therefore$ Vector $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ is equally inclined to $\mathrm{OX}$, OY and $\mathrm{OZ}$.

Ex 10.2 Question 15.

Find the position vector of a point $R$ which divides the line joining two points $P$ and $\mathbf{Q}$ whose position vectors are $\hat{i}+2 \hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively, in the ratio $2: 1$ (i) internally (ii) externally.

Answer.

Position vector of $\mathrm{P}$ is $\vec{a}=\hat{i}+2 \hat{j}-\hat{k}$ and Position vector of $\mathrm{Q}$ is $\vec{b}=-\hat{i}+\hat{j}+\hat{k}$
(i) Position vector of point $\mathrm{R}$ dividing $\mathrm{PQ}$ internally (i.e., $\mathrm{R}$ lies within the segment $\mathrm{PQ}$ ) in the
$
\begin{aligned}
& \text { ratio } 2: 1=m: n=\mathrm{PR}: \mathrm{QR} \text { is } \frac{m \vec{b}+n \vec{a}}{m+n} \\
& =\frac{2(-\hat{i}+\hat{j}+\hat{k})+\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}
\end{aligned}
$
(ii) Position vector of point $\mathrm{R}$ dividing $\mathrm{PQ}$ externally (i.e., $\mathrm{R}$ lies outside the segment $\mathrm{PQ}$ and to the right of point $\mathrm{Q}$ because ratio $2: 1>1$ i.e., $\mathrm{PR}: \mathrm{QR}=2: 1$ ) is $\frac{m \vec{b}-n \vec{a}}{m-n}$
$
\begin{aligned}
& =\frac{2(-\hat{i}+\hat{j}+\hat{k})-(\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}
\end{aligned}
$

Ex 10.2 Question 16.

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

Answer.

Given: Point $P(2,3,4)$ and $Q(4,1,-2)$

$\therefore$ Position vector of point $\mathrm{P}$ is $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$
And Position vector of point $\mathrm{Q}$ is $\vec{b}=4 \hat{i}+\hat{j}-2 \hat{k}$
And Position vector of mid-point $\mathrm{R}$ of $\mathrm{PQ}$ is $\frac{\vec{a}+\vec{b}}{2}=\frac{2 \hat{i}+3 \hat{j}+4 \hat{k}+4 \hat{i}+\hat{j}-2 \hat{k}}{2}$
$
=\frac{6 \hat{i}+4 \hat{j}+2 \hat{k}}{2}=3 \hat{i}+2 \hat{j}+\hat{k}
$

Ex 10.2 Question 17.

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

Answer.

Given: Position vector of point $\mathrm{A}$ is $\vec{a}=\overrightarrow{\mathrm{OA}}=3 \hat{i}-4 \hat{j}-4 \hat{k}$
Position vector of point $\mathrm{B}$ is $\vec{b}=\overrightarrow{\mathrm{OB}}=2 \hat{i}-\hat{j}+\hat{k}$ and
Position vector of point $\mathrm{C}$ is $\vec{c}=\overrightarrow{\mathrm{OC}}=\hat{i}-3 \hat{j}-5 \hat{k}$ where $\mathrm{O}$ is the origin.
Now $\overrightarrow{\mathrm{AB}}=$ Position vector of point $\mathrm{B}-$ Position vector of point $\mathrm{A}$
$
\begin{aligned}
& =2 \hat{i}-\hat{j}+\hat{k}-(3 \hat{i}-4 \hat{j}-4 \hat{k})=2 \hat{i}-\hat{j}+\hat{k}-3 \hat{i}+4 \hat{j}+4 \hat{k} \\
& =-\hat{i}+3 \hat{j}+5 \hat{k} \ldots \ldots \ldots \text { (i) }
\end{aligned}
$
$\overrightarrow{\mathrm{BC}}=$ Position vector of point $\mathrm{C}-$ Position vector of point $\mathrm{B}$
$=\hat{i}-3 \hat{j}-5 \hat{k}-(2 \hat{i}-\hat{j}+\hat{k})=\hat{i}-3 \hat{j}-5 \hat{k}-2 \hat{i}+\hat{j}-\hat{k}$
$=-\hat{i}-2 \hat{j}-6 \hat{k}$
$\overrightarrow{\mathrm{AC}}=$ Position vector of point $\mathrm{C}-$ Position vector of point $\mathrm{A}$
$=\hat{i}-3 \hat{j}-5 \hat{k}-(3 \hat{i}-4 \hat{j}-4 \hat{k})=\hat{i}-3 \hat{j}-5 \hat{k}-3 \hat{i}+4 \hat{j}+4 \hat{k}$

$
=-2 \hat{i}+\hat{j}-\hat{k}
$

Adding eq. (i) and (ii),
$
\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=-\hat{i}+3 \hat{j}+5 \hat{k}-\hat{i}-2 \hat{j}-6 \hat{k}=-2 \hat{i}+\hat{j}-\hat{k}=\overrightarrow{\mathrm{AC}} \text { [By eq. (iii)] }
$

Now From eq. (i), $\mathrm{AB}=|\overrightarrow{\mathrm{AB}}|=\sqrt{1+9+25}=\sqrt{35}$
From eq. (ii), $\mathrm{BC}=|\overrightarrow{\mathrm{BC}}|=\sqrt{1+4+36}=\sqrt{41}$
From eq. (iii), $\mathrm{AC}=|\overrightarrow{\mathrm{AC}}|=\sqrt{4+1+1}=\sqrt{6}$
Here, we can observe that (Longest side $\mathrm{BC})^2=(\sqrt{41})^2=41=35+6=\mathrm{AB}^2+\mathrm{AC}^2$
Also, $\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CA}}=(+1)(2)+(-3)(-1)+(-5)(1)=2+3-5=0$
Thus, $\overrightarrow{\mathrm{AB}} \perp \overrightarrow{\mathrm{CA}} \Rightarrow \angle \mathrm{A}=90^{\circ}$
Therefore, A B, C are the vertices of a right angled triangle.

Ex 10.2 Question 18.

In triangle ABC (Fig. below), which of the following is not true: 

(A) $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$
(B) $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}$
(C) $\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}$

Answer.

We know by Triangle law of Addition of vectors that $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}} \Rightarrow$

$
\begin{aligned}
& \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=-\overrightarrow{\mathrm{CA}} \\
& \Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}
\end{aligned}
$

Therefore option (C) is not true because in option (C),
$
\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AC}}=2 \overrightarrow{\mathrm{AC}} \neq 0
$
Ex 10.2 Question 19.

If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect:
(A) $\vec{b}=\lambda \vec{a}$, for some scalar $\lambda$.
(B) $\vec{a}= \pm \vec{b}$
(C) The respective components of $\vec{a}$ and $\vec{b}$ are proportional.
(D) Both the vectors $\vec{a}$ and $\vec{b}$ have same direction, but different magnitudes.

Answer.

Option (D) is not true because two collinear vectors can have different directions and also different magnitudes.

The option (A) and option (C) are true by definition of collinear vectors.

Option (B) is a particular case of option (A) taking $\lambda= \pm 1$.