Let a be a unit vector and b is a nonzero vector not parallel to … - Vidyadip

MCQ

Let $\vec{a}$ be a unit vector and $\vec{b}$ is a nonzero vector not parallel to $\vec{a}$ . The angles of the triangle, two of whose sides are represented by $\sqrt 3 (\vec{a} \times \vec{b} )$ and $\vec{b} - (\vec{a} . \vec{b}) \vec{a}$ are

A
$\frac{\pi }{4}$ , $\frac{\pi }{4}$ , $\frac{\pi }{2}$
B
$\frac{\pi }{4}$ , $\frac{\pi }{3}$ , $\frac{5\pi }{12}$
✓
$\frac{\pi }{6}$ , $\frac{\pi }{3}$ , $\frac{\pi }{2}$
D
$\frac{\pi }{3}$ , $\frac{\pi }{3}$ , $\frac{\pi }{3}$

✓

Answer

Correct option: C.

$\frac{\pi }{6}$ , $\frac{\pi }{3}$ , $\frac{\pi }{2}$

c
$\vec b - (\vec a \cdot \vec b)\vec a = (\vec a \cdot \vec a)\vec b - (\vec a \cdot \vec b)\vec a = \vec a \times (\vec b \times \vec a)$

here $\vec a \times \vec b \bot \vec a \times (\vec b \times \vec a)$

$\angle \mathrm{ABC}=90^{\circ}$

$\tan \theta = \frac{{\sqrt 3 |\vec a \times \vec b|}}{{|\vec a||\vec b \times \vec a|}} = \sqrt 3 $

$\theta=\pi / 3$

$\angle \mathrm{ABC}=\pi / 2: \angle \mathrm{BAC}=\pi / 6$

$\angle \mathrm{ACB}=\pi / 3$

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