Question

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

$\therefore \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{a}$

So we get $\overrightarrow{b}\cdot \overrightarrow{c}=0$....$\left(1\right)$

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

Putting values in Eq-$1$

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

$b_1-b_2-b_3=0$

Now given that $\overrightarrow{a}\cdot \overrightarrow{b}=3$
$=b_2-b_3=3$
$b_1=b_2+b_3=3+2b_3$
$\overrightarrow{b}=(3+2b_3)\hat{i}+(3+b_3)\hat{j}+b_3\hat{k}$

Comparing the above equation, we get,

$b_1=-1,$

$b_2=1,$

$b_3=-2$

Hence, option ${}^{\prime }{D}^{\prime }$ is correct.