Medium MCQ +4 / -1 PYQ · JEE Mains 2022

Let $\widehat a$, $\widehat b$ be unit vectors. If $\overrightarrow c$ be a vector such that the angle between $\widehat a$ and $\overrightarrow c$ is ${\pi \over {12}}$, and $$\widehat b = \overrightarrow c + 2\left( {\overrightarrow c \times \widehat a} \right)$$, then ${\left| {6\overrightarrow c } \right|^2}$ is equal to :

  1. A $6\left( {3 - \sqrt 3 } \right)$
  2. B $3 + \sqrt 3$
  3. C $6\left( {3 + \sqrt 3 } \right)$ Correct answer
  4. D $6\left( {\sqrt 3 + 1} \right)$

Solution

$\because \quad \hat{b}=\vec{c}+2(\vec{c} \times \hat{a})$ <br/><br/> $$ \begin{aligned} &\Rightarrow \hat{b} \cdot \vec{c}=|\vec{c}|^{2} \\\\ &\therefore \hat{b}-\vec{c}=2(\vec{c} \times \vec{a}) \end{aligned} $$ <br/><br/> $$ \begin{aligned} &\Rightarrow|\hat{b}|^{2}+|\vec{c}|^{2}-2 \hat{b} \cdot \vec{c}=4|\vec{c}|^{2}|\vec{a}|^{2} \sin ^{2} \frac{\pi}{12} \\\\ &\Rightarrow 1+|\vec{c}|^{2}-2|c|^{2}=4|\vec{c}|^{2}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^{2} \\\\ &\Rightarrow 1=|\vec{c}|^{2}(3-\sqrt{3}) \\\\ &\Rightarrow 36|\vec{c}|^{2}=\frac{36}{3-\sqrt{3}}=6(3+\sqrt{3}) \end{aligned} $$

About this question

Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors

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