Question

Let $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{c}=\hat{j}-\hat{k}$ and a vector $\overrightarrow{b}$ be such that $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$ and $\overrightarrow{a}\cdot \overrightarrow{b}=3$. Then $|\overrightarrow{b}|$ equals?

• A. $\sqrt{\frac{11}{3}}$
• B. $\frac{\sqrt{11}}{3}$
• C. $\frac{11}{\sqrt{3}}$
• D. $\frac{11}{3}$

Answer 1

The correct option is A $\sqrt{\frac{11}{3}}$

Given $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{c}=\hat{j}-\hat{k}$

$|\overrightarrow{a}|=\sqrt{1+1+1}=\sqrt{3}$

$|\overrightarrow{c}|=\sqrt{1+1}=\sqrt{2}$

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

$⟹\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|sin\theta =\left|\overrightarrow{c}\right|$

$⟹\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|sin\theta =\sqrt{2}$ ...[1]

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

$⟹\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta =3$ ...[2]

Dividing [1] by [2], we get

$tan\theta =\frac{\sqrt{2}}{3}$

$⟹sin\theta =\frac{\sqrt{2}}{\sqrt{1}1}$

Substituting value of $sin\theta$ in [1] ,we get

$⟹\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|sin\theta =\sqrt{2}$

$⟹\sqrt{3}\left|\overrightarrow{b}\right|\frac{\sqrt{2}}{\sqrt{1}1}=\sqrt{2}$

$\left|\overrightarrow{b}\right|=\frac{\sqrt{11}}{\sqrt{3}}$

Hence, answer is option (A).