Question

A unit vector coplanar with $\hat{i}+\hat{j}+2\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k},$ and perpendicular to $\hat{i}+\hat{j}+\hat{k},$ is:

• A. $\left(\frac{\hat{j}-\hat{k}}{\sqrt{2}}\right)$
• B. $\left(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\right)$
• C. $\left(\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}\right)$
• D. $\left(\frac{\hat{i}+2\hat{j}+\hat{k}}{\sqrt{6}}\right)$
• E. $\left(\frac{-\hat{j}+2\hat{k}}{\sqrt{5}}\right)$

Answer 1

Answer: (A)

Let unit vector is $a\hat{i}+b\hat{j}+c\hat{k}$
$\because$ $a\hat{i}+b\hat{j}+c\hat{k}$ is $\perp$ to $\hat{i}+\hat{j}+\hat{k},$
then $a+b+c=0$ ...(i) and $a\hat{i}+b\hat{j}+c\hat{k},(\hat{i}+2\hat{j}+\hat{k})$ and $(\hat{i}+\hat{j}+2\hat{k})$ are coplanar $\therefore$ $\left|\begin{array}{ccc}a& b& c\\ 1& 1& 2\\ 1& 2& 1\end{array}\right|=0$
$\Rightarrow$ $-3a+b+c=0$ ...(ii)
From Eqs. (i) and (ii), we get
$a=0$ and $c=-b$
$\because$ $a\hat{i}+b\hat{j}+c\hat{k}$ is a unit vector, then
$a^2+b^2+c^2=1$
$\Rightarrow$ $0+b^2+b^2=1$ $\Rightarrow$ $b=\frac{1}{\sqrt{2}}$
$\therefore$ $a\hat{i}+b\hat{j}+c\hat{k}=\frac{1}{\sqrt{2}}\hat{j}-\frac{1}{\sqrt{2}}\hat{k}$
$=\frac{\hat{j}-\hat{k}}{\sqrt{2}}$