Let a unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$ make angles $\frac{\pi}{2}, \frac{\pi}{3}$ and $\frac{2 \pi}{3}$ with the vectors $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$$ and $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ respectively. If $$\vec{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$$ then $|\hat{u}-\vec{v}|^2$ is equal to

<p>Unit vector $$\hat{\mathrm{u}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$$</p> <p>$$\begin{aligned} & \overrightarrow{\mathrm{p}}_1=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2=\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{p}}_3=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \end{aligned}$$</p> <p>Now angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_1=\frac{\pi}{2}$</p> <p>$$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_1=0 \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=0$$</p> <p>$\Rightarrow \mathrm{x}+\mathrm{z}=0$ ...... (i)</p> <p>Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_2=\frac{\pi}{3}$</p> <p>$$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_2=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_2\right| \cos \frac{\pi}{3}$$</p> <p>$$\Rightarrow \frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2} \Rightarrow y+z=\frac{1}{\sqrt{2}}$$ ...... (ii)</p> <p>Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_3=\frac{2 \pi}{3}$</p> <p>$$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_3=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_3\right| \cos \frac{2 \pi}{3}$$</p> <p>$$\Rightarrow \frac{x}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow x+y=\frac{-1}{\sqrt{2}}$$ ..... (iii)</p> <p>from equation (i), (ii) and (iii) we get</p> <p>$x=\frac{-1}{\sqrt{2}} \quad y=0 \quad z=\frac{1}{\sqrt{2}}$</p> <p>$$\begin{aligned} & \text { Thus } \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}-\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\ & \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-2}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \\ & \therefore|\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}|^2=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^2=\frac{5}{2} \end{aligned}$$</p>