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

Answer 1

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

$\overrightarrow{a}=\hat{ı}+\hat{ȷ}+\hat{k}\overrightarrow{c}=\hat{ȷ}-\hat{k}$

Given that $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$ and $\overrightarrow{a}\cdot \overrightarrow{b}=3$

Let $\overrightarrow{b}=x\hat{ı}+y\hat{ȷ}+z\hat{k}$

$\overrightarrow{a}\cdot \overrightarrow{b}=(\hat{i}+\hat{ȷ}+\hat{k})\cdot (x\hat{ı}+y\hat{ȷ}+z\hat{k})$

$3=x+y+z-\left(1\right)$

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{ı}& \hat{ȷ}& \hat{k}\\ 1& 1& 1\\ x& y& z\end{array}\right|=\overrightarrow{c}$

$\Rightarrow \hat{i}(z-y)-\hat{ȷ}(z-x)+\hat{k}(y-x)=\hat{ȷ}-\hat{k}$

Comparing both sides

$z-y=0\Rightarrow z=y-\left(i\right)$

$x-z=1\Rightarrow x=z+1-\left(ii\right)$

$y-x=-1\Rightarrow y=x-1-(iv)$

solving equations (iv) and (i) and (ii)

$x+y+z=3$

$y+1+y+y=3\Rightarrow 3y=2$

$\Rightarrow y=\frac{2}{3}$

$\therefore z=y=\frac{2}{3}$ and $x=y+1=\frac{5}{3}$

$\begin{array}{cc}|\overrightarrow{b}|& =\sqrt{x^2+y^2+z^2}\\ & =\sqrt{{\left(\frac{5}{3}\right)}^2+{\left(\frac{2}{3}\right)}^2+{\left(\frac{2}{3}\right)}^2}\\ & =\sqrt{\frac{33}{9}}=\sqrt{\frac{11}{3}}\\ \text{Answer}& \text{option}\left(C\right)\end{array}$