Vector Algebra Question 11
Question 11 - 2024 (30 Jan Shift 2)
Let $\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k}, \alpha, \beta \in R$. 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
(1) 90
(2) 75
(3) 95
(4) 85
Show Answer
Answer (1)
Solution
$|\vec{b}|^{2}=6 ;|\vec{a}||\vec{b}| \cos \theta=3 \sqrt{2}$
$|\overrightarrow{\mathrm{a}}|^{2}|\overrightarrow{\mathrm{b}}|^{2} \cos ^{2} \theta=18$
$|\vec{a}|^{2}=6$
Also $1+\alpha^{2}+\beta^{2}=6$
$\alpha^{2}+\beta^{2}=5$
to find
$$ \begin{aligned} & \left(\alpha^{2}+\beta^{2}\right)|\overrightarrow{\mathrm{a}}|^{2}|\overrightarrow{\mathrm{b}}|^{2} \sin ^{2} \theta \ & =(5)(6)(6)\left(\frac{1}{2}\right) \ & =90 \end{aligned} $$
