Question

Let $\overrightarrow{a}=\hat{j}-\hat{k};\overrightarrow{c}=\hat{i}-\hat{j}-\hat{k}$. Then the vector $\overrightarrow{b}$ satisfying $\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{c}=0$ and $\overrightarrow{a}.\overrightarrow{b}=3$ is

• A. $2\hat{i}-\hat{j}+2\hat{k}$
• B. $\hat{i}-\hat{j}-2\hat{k}$
• C. $\hat{i}-+\hat{j}-2\hat{k}$
• D. $-\hat{i}+\hat{j}-2\hat{k}$

Answer 1

The correct option is D $-\hat{i}+\hat{j}-2\hat{k}$

Given $\overrightarrow{a}=\hat{j}-\hat{k}$

$\overrightarrow{c}=\hat{i}-\hat{j}-\hat{k}$

let $\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$

$\overrightarrow{a}.\overrightarrow{b}=3$

$\Rightarrow (\hat{j}-\hat{k}).(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})=3$

$\Rightarrow b_2-b_3=3$ __ (1)

$\overrightarrow{a}\times \overrightarrow{b}=$ $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& -1\\ b_1& b_2& b_3\end{array}\right|$

$=\hat{i}(b_3+b_2)-\hat{j}\left(b_1\right)+\hat{k}(-b_1)$

$\overrightarrow{a}\times \overrightarrow{b}=(b_3+b_2)\hat{i}-b_1\hat{j}-b_1\hat{k}$

$\overrightarrow{c}+\overrightarrow{a}\times \overrightarrow{b}=0$

$\Rightarrow \overrightarrow{c}+(b_3+b_2)\hat{i}-b_1\hat{j}-b_1\hat{k}=0$

$\Rightarrow \hat{i}-\hat{j}-\hat{k}=-(b_3+b_2)\hat{i}+b_1\hat{j}+b_1\hat{k}$

Equating

$b_1=-1,$

$b_3+b_2=-1$ __ (ii)

$b_2-b_3=3$ __ (i)

________________

$2b_2=2\Rightarrow b_2=1$

$\therefore b_3=-2$

Hence, $\overrightarrow{b}=-\hat{i}+\hat{j}-2\hat{k}$