Question

The unit vector in ZOX plane and making angle ${45}^{\circ }$ and ${60}^{\circ }$ respectively with $\overrightarrow{a}=2\hat{i}+2\hat{j}-\hat{k}$ and $\overrightarrow{b}=0\hat{i}+\hat{j}-\hat{k}$

• A. $-\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$
• B. $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$
• C. $\frac{1}{3\sqrt{2}}\hat{i}+\frac{4}{3\sqrt{2}}\hat{j}+\frac{1}{3\sqrt{2}}\hat{k}$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$

Let the required vector be $\overrightarrow{c}=x\hat{i}+y\hat{k}$. As it is a unit vector, therefore
$|\overrightarrow{c}|=1\Rightarrow x^2+y^2=1......\left(i\right)$
$\therefore cos{45}^{\circ }=\frac{2x-y}{\sqrt{4+4+1}}\Rightarrow 2x-y=\frac{3}{\sqrt{2}}....\left(ii\right)$
$\overrightarrow{b}$ and $\overrightarrow{c}$ are inclined at an angle of ${60}^{\circ }$.
$\therefore -\frac{y}{\sqrt{2}}=cos{60}^{\circ }\Rightarrow y=-\frac{1}{\sqrt{2}}.....\left(iii\right)$
From Eqs. (ii) and (iii), we get $x=\frac{1}{\sqrt{2}}$
Hence, the required vector is $\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{k}.$