Question

A unit vector coplanar with $ \hat{i}+\text{ }\hat{j}+2\hat{k} $ and $ \hat{i}+2\text{ }\hat{j}+\hat{k}, $ and perpendicular to $ \hat{i}+\text{ }\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 $ \bot $ 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{matrix} a & b & c \\ 1 & 1 & 2 \\ 1 & 2 & 1 \\ \end{matrix} \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}} $