Q27 / 75 · JEE Main 2025 (29 Jan Shift 2) · Mathematics
Let $\hat{a}$ be a unit vector perpendicular to the vectors $\overrightarrow{\mathrm{b}}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\overrightarrow{\mathrm{c}}=2 \hat{i}+3 \hat{j}-\hat{k}$, and makes an angle of $\cos ^{-1}\left(-\frac{1}{3}\right)$ with the vector $\hat{i}+\hat{j}+\hat{k}$. If $\hat{\mathrm{a}}$ makes an angle of $\frac{\pi}{3}$ with the vector $\hat{i}+\alpha \hat{j}+\hat{k}$, then the value of $\alpha$ is :
- A$\sqrt{6}$
- Correct answerB$-\sqrt{6}$
- C$-\sqrt{3}$
- D$\sqrt{3}$
Solution
$\begin{aligned}
& \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
2 & 3 & -1
\end{array}\right|=\hat{i}(-7)+7 \hat{j}+7 \hat{k} \\
& \hat{a}= \pm \frac{(-7 \hat{i}+7 \hat{j}+7 \hat{k})}{\sqrt{7^2+7^2+7^2}}= \pm\left(\frac{-\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\right)
\end{aligned}$
$\begin{aligned}
& \text { Now, } \cos \theta= \pm \frac{(-1+1+1)}{\sqrt{3} \cdot \sqrt{3}}= \pm \frac{1}{3} \\
& \Rightarrow \cos ^{-1}\left(\frac{-1}{3}\right) \Rightarrow \hat{a}=\frac{-(-\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}} \\
& \hat{a}=\frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}}
\end{aligned}$
$\begin{aligned}
& \cos \frac{\pi}{3}=\frac{1-\alpha-1}{\sqrt{3} \cdot \sqrt{\alpha^2+2}} \\
& \frac{1}{2}=\frac{-\alpha}{\sqrt{3} \cdot \sqrt{\alpha^2+2}} \rightarrow \alpha < 0 \\
& 3\left(\alpha^2+2\right)=4 \alpha^2 \\
& 6=\alpha^2 \\
& \alpha= \pm \sqrt{6}
\end{aligned}$
Clearly, $\alpha=-\sqrt{6}$
Hence, the correct answer is option: 'B'.