Question

A unit vector in the plane of $\hat{i}+2\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+2\hat{k}$ and perpendicular to $2\hat{i}+\hat{j}+\hat{k}$, is

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

Answer 1

Answer: (D)

$\frac{\hat{j}-\hat{k}}{\sqrt{2}}$

Let $a\hat{i}+b\hat{j}+c\hat{k}$ be the required unit vector. $\therefore a^2+b^2+c^2=1$

Also it is perpendicular to $2\hat{i}+\hat{j}+\hat{k}$

$\therefore 2a+b+c=0$

It lies in the plane of $(\hat{i}+2\hat{j}+\hat{k})$ & $(\hat{i}+\hat{j}+2\hat{k})$

$\therefore$ the vector perpendicular to normal of plane.

normal$=(\hat{i}+2\hat{j}+\hat{k})\times (\hat{i}+\hat{j}+2\hat{k})$

$=3\hat{i}-\hat{j}-\hat{k}$

$\therefore 3a-b-c=0$

onn solving, $a=0,b=-c$

$\therefore b=\frac{1}{\sqrt{2}}$

The vector is $\frac{1}{\sqrt{2}}\hat{j}-\frac{1}{\sqrt{2}}\hat{k}$