Jee Main 2024 30 01 2024 Shift 2 - Question12
Question 12
If $\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k} \quad|\vec{b}|^{2}=6$ and angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$. If $\vec{a} \cdot \vec{b}=3$ then $\left(\alpha^{2}+\beta^{2}\right)|\vec{a} \times \vec{b}|^{2}$ is
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Answer: (18)
Solution:
$\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta=3$
$$ \begin{aligned} & =\sqrt{1+\alpha^{2}+\beta^{2}} \cdot \sqrt{6} \frac{1}{\sqrt{2}}=3 \\ & \Rightarrow 1+\alpha^{2}+\beta^{2}=3 \\ & \Rightarrow \alpha^{2}+\beta^{2}=2 \end{aligned} $$
Also $|\vec{a}|=\sqrt{1+\alpha^{2}+\beta^{2}}=\sqrt{3}$
$\Rightarrow|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$
$=\sqrt{3} \times \sqrt{6} \times \frac{1}{\sqrt{2}}=3 \Rightarrow|\vec{a} \times \vec{b}|^{2}=9$
$\Rightarrow\left(\alpha^{2}+\beta^{2}\right)|\vec{a} \times \vec{b}|^{2}=2 \times 9=18$
BETA
