JEE Main 2025 — 29 January, Evening Shift — Question 60

Let $\overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$

$$\begin{aligned} & \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{array}\right| \\ & =-7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & =-7(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}) \end{aligned}$$

Now $\hat{a}=\frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}}$ or $\frac{-\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$

$\cos \theta=\frac{\hat{\mathrm{a}} \cdot \overrightarrow{\mathrm{v}}}{|\overrightarrow{\mathrm{v}}|}=\frac{1-1-1}{\sqrt{3} \sqrt{3}}=\frac{-1}{3} \quad \cos \theta=\frac{\hat{\mathrm{a}} \cdot \overrightarrow{\mathrm{v}}}{|\overrightarrow{\mathrm{v}}|}=\frac{-1+1+1}{3}=\frac{1}{3}$ (rejected)

$\Rightarrow \hat{\mathrm{a}}=\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{3}}$ Now $\cos \frac{\pi}{3}=\frac{\hat{\mathrm{a}} \cdot(\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}})}{\sqrt{1+\alpha^{2}+1}}$

$\Rightarrow \frac{1}{2}=\frac{1-\alpha-1}{\sqrt{3} \sqrt{\alpha^{2}+2}}$

$\Rightarrow \frac{\sqrt{3}}{2} \sqrt{\alpha^{2}+2}=-\alpha \quad(\therefore \alpha<0)$

$3 \alpha^{2}+6=4 \alpha^{2}$

$\Rightarrow \alpha=-\sqrt{6}$