Let a, b, c be three mutually perpendicular vectors of the same m… - Vidyadip

MCQ

Let $\vec{a}, \vec{b}, \vec{c}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$, with the vector $\vec{a}+\vec{b}+\vec{c}$. Then $36 \cos ^{2} 2 \theta$ is equal to $.....$

A
$1$
B
$2$
C
$3$
✓
$4$

✓

Answer

Correct option: D.

$4$

d
$|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\bar{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\overrightarrow{a \cdot c}+\vec{b} \cdot \vec{c})=3$

$\Rightarrow|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}$

$\vec{a}(\vec{a}+\vec{b}+\vec{c})=|\vec{a}|+|\vec{a}+\vec{b}+\vec{c}| \cos \theta$

$\Rightarrow 1=\sqrt{3} \cos \theta$

$\Rightarrow \cos 2 \theta=-\frac{1}{3}$

$\Rightarrow 36 \cos ^{2} 2 \theta=4$

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