JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 12)
Let $$\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k}, \alpha, \beta \in \mathbb{R}$$. Let a vector $$\vec{b}$$ be such that the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\frac{\pi}{4}$$ and $$|\vec{b}|^2=6$$. If $$\vec{a} \cdot \vec{b}=3 \sqrt{2}$$, then the value of $$\left(\alpha^2+\beta^2\right)|\vec{a} \times \vec{b}|^2$$ is equal to
85
90
75
95
Explanation
$$\begin{aligned} & |\overrightarrow{\mathrm{b}}|^2=6 ;|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=3 \sqrt{2} \\ & |\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{b}}|^2 \cos ^2 \theta=18 \\ & |\overrightarrow{\mathrm{a}}|^2=6 \end{aligned}$$
Also $$1+\alpha^2+\beta^2=6$$
$$\alpha^2+\beta^2=5$$
to find
$$\begin{aligned} & \left(\alpha^2+\beta^2\right)|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta \\ & =(5)(6)(6)\left(\frac{1}{2}\right) \\ & =90 \end{aligned}$$
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