Medium MCQ +4 / -1 PYQ · JEE Mains 2024

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

  1. A 85
  2. B 90 Correct answer
  3. C 75
  4. D 95

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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