Mathematics - Vector Question with Solution | TestHub

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

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

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

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

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

$\Rightarrow x+z=0 \dots \left(\mathrm{i}\right)$

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

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

$\hat{\mathrm{u}}·{\overrightarrow{\mathrm{p}}}_2= \frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2} \dots \left(\mathrm{ii}\right)$

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

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

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{-1}{2}\Rightarrow x+y=\frac{-1}{\sqrt{2}} \dots \left(\mathrm{iii}\right)$

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

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

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

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

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