Answer (2)

Solution

Unit vector $\hat{\mathbf{u}}=x \hat{i}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$

$\overrightarrow{p} _1=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \overrightarrow{p} _2=\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$

$\overrightarrow{p} _3=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$

Now angle between $\hat{u}$ and $\overrightarrow{p} _1=\frac{\pi}{2}$

$\hat{u} \cdot \overrightarrow{p} _1=0 \Rightarrow \frac{x}{\sqrt{2}}+\frac{z}{\sqrt{2}}=0$

$\Rightarrow x+z=0 \ldots(i)$

Angle between $\hat{u}$ and $\overrightarrow{p} _2=\frac{\pi}{3}$

$\hat{u} \cdot \overrightarrow{p} _2=|\hat{u}| \cdot\left|\overrightarrow{p} _2\right| \cos \frac{\pi}{3}$

$\Rightarrow \frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2} \Rightarrow y+z=\frac{1}{\sqrt{2}}$

Angle between $\hat{u}$ and $\overrightarrow{p} _3=\frac{2 \pi}{3}$

$\hat{u} \cdot \overrightarrow{p} _3=|\hat{u}| \cdot\left|\overrightarrow{p} _3\right| \cos \frac{2 \pi}{3}$

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow x+y=\frac{-1}{\sqrt{2}} \ldots$

from equation (i), (ii) and (iii) we get

$x=\frac{-1}{\sqrt{2}} \quad y=0 \quad z=\frac{1}{\sqrt{2}}$

Thus $\hat{u}-\overrightarrow{v}=\frac{-1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}-\frac{1}{\sqrt{2}} \hat{i}-\frac{1}{\sqrt{2}} \hat{j}-\frac{1}{\sqrt{2}} \hat{k}$

$\hat{u}-\overrightarrow{v}=\frac{-2}{\sqrt{2}} \hat{i}-\frac{1}{\sqrt{2}} \hat{j}$

$\therefore|\hat{u}-\overrightarrow{v}|^{2}=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^{2}=\frac{5}{2}$