$\begin{aligned}
& \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \
1 & -2 & 3 \
2 & 3 & -1
\end{array}\right|=\hat{i}(-7)+7 \hat{j}+7 \hat{k} \
& \hat{a}= \pm \frac{(-7 \hat{i}+7 \hat{j}+7 \hat{k})}{\sqrt{7^2+7^2+7^2}}= \pm\left(\frac{-\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\right)
\end{aligned}\begin{aligned}
& \text { Now, } \cos \theta= \pm \frac{(-1+1+1)}{\sqrt{3} \cdot \sqrt{3}}= \pm \frac{1}{3} \
& \Rightarrow \cos ^{-1}\left(\frac{-1}{3}\right) \Rightarrow \hat{a}=\frac{-(-\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}} \
& \hat{a}=\frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}}
\end{aligned}\begin{aligned}
& \cos \frac{\pi}{3}=\frac{1-\alpha-1}{\sqrt{3} \cdot \sqrt{\alpha^2+2}} \
& \frac{1}{2}=\frac{-\alpha}{\sqrt{3} \cdot \sqrt{\alpha^2+2}} \rightarrow \alpha < 0 \
& 3\left(\alpha^2+2\right)=4 \alpha^2 \
& 6=\alpha^2 \
& \alpha= \pm \sqrt{6}
\end{aligned}Clearly,\alpha=-\sqrt{6}$
Let a be a unit vector perpendicular to the vectors overrightarrow b= i-2 j+3 k and overrightarrow =2 i+3 j- k, and makes an angle of cos ^-1 (- 1 3…
JEE Main 2025 — Mathematics Vectors & 3D Geometry
Let a^ be a unit vector perpendicular to the vectors b=i^−2j^+3k^ and c=2i^+3j^−k^, and makes an angle of cos−1(−31) with the vector i^+j^+k^. If a^ makes an angle of 3π with the vector i^+αj^+k^, then the value of α is :
Held on 29 Jan 2025 · Verified 6 Jul 2026.
Options
- A
6
- B
−6
- C
−3
- D
3
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