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
Solution
<p>$$\begin{aligned}
& |\overrightarrow{\mathrm{b}}|^2=6 ;|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=3 \sqrt{2} \\
& |\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{b}}|^2 \cos ^2 \theta=18 \\
& |\overrightarrow{\mathrm{a}}|^2=6
\end{aligned}$$</p>
<p>Also $1+\alpha^2+\beta^2=6$</p>
<p>$\alpha^2+\beta^2=5$</p>
<p>to find</p>
<p>$$\begin{aligned}
& \left(\alpha^2+\beta^2\right)|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta \\
& =(5)(6)(6)\left(\frac{1}{2}\right) \\
& =90
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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