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 $\overrightarrow{a}.\overrightarrow{b}=3$. Then $|\overrightarrow{b}|$ equals :

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

Answer 1

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

Let, $\overrightarrow{b}=x\hat{i}+y\hat{j}+z\hat{k}$

Now,

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

$\Rightarrow (z-y)\hat{i}-(z-x)\hat{j}+(y-x)\hat{k}=0\hat{i}+\hat{j}-\hat{k}$

$\Rightarrow z=y$ ….(i)

$-z+x=1$ ….(ii)

$y-x=-1$ ……(iii)

and, $\overrightarrow{a}\cdot \overrightarrow{b}=3$

$\Rightarrow x+y+z=3$ ….(iv)

Solving equations (i), (ii), (iii) and (iv) we get;

$x=\frac{5}{3};y=z=\frac{2}{3}$

$\therefore |\overrightarrow{b}|=\sqrt{(5/3{)}^2+(2/3{)}^2+(2/3{)}^2}=\sqrt{11/3}$.