JEE MAIN - Mathematics (2022 - 24th June Morning Shift - No. 12)
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 :
$$6\left( {3 - \sqrt 3 } \right)$$
$$3 + \sqrt 3 $$
$$6\left( {3 + \sqrt 3 } \right)$$
$$6\left( {\sqrt 3 + 1} \right)$$
Explanation
$\because \quad \hat{b}=\vec{c}+2(\vec{c} \times \hat{a})$
$$ \begin{aligned} &\Rightarrow \hat{b} \cdot \vec{c}=|\vec{c}|^{2} \\\\ &\therefore \hat{b}-\vec{c}=2(\vec{c} \times \vec{a}) \end{aligned} $$
$$ \begin{aligned} &\Rightarrow|\hat{b}|^{2}+|\vec{c}|^{2}-2 \hat{b} \cdot \vec{c}=4|\vec{c}|^{2}|\vec{a}|^{2} \sin ^{2} \frac{\pi}{12} \\\\ &\Rightarrow 1+|\vec{c}|^{2}-2|c|^{2}=4|\vec{c}|^{2}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^{2} \\\\ &\Rightarrow 1=|\vec{c}|^{2}(3-\sqrt{3}) \\\\ &\Rightarrow 36|\vec{c}|^{2}=\frac{36}{3-\sqrt{3}}=6(3+\sqrt{3}) \end{aligned} $$
$$ \begin{aligned} &\Rightarrow \hat{b} \cdot \vec{c}=|\vec{c}|^{2} \\\\ &\therefore \hat{b}-\vec{c}=2(\vec{c} \times \vec{a}) \end{aligned} $$
$$ \begin{aligned} &\Rightarrow|\hat{b}|^{2}+|\vec{c}|^{2}-2 \hat{b} \cdot \vec{c}=4|\vec{c}|^{2}|\vec{a}|^{2} \sin ^{2} \frac{\pi}{12} \\\\ &\Rightarrow 1+|\vec{c}|^{2}-2|c|^{2}=4|\vec{c}|^{2}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^{2} \\\\ &\Rightarrow 1=|\vec{c}|^{2}(3-\sqrt{3}) \\\\ &\Rightarrow 36|\vec{c}|^{2}=\frac{36}{3-\sqrt{3}}=6(3+\sqrt{3}) \end{aligned} $$
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