JEE Main 2024 — 29 January, Shift 2 — Question 20

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

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

$\overrightarrow{\mathrm{p}}_{3}=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}$

Now angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_{1}=\frac{\pi}{2}$ $\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_{1}=0 \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=0$

$\Rightarrow \mathrm{x}+\mathrm{z}=0$

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

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

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

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

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

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow x+y=\frac{-1}{\sqrt{2}}$ from equation (i), (ii) and (iii)

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

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

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

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