let a→,b→ and c→ be three unit vectors such that a→×(b→×c→)=32(b→+c→).if b→ is not parallel to c¯, then angle between a→ and c→ is

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let $\vec{a}, \vec{b} and \vec{c}$ be three unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2} (\vec{b} + \vec{c})$ .if $\vec{b}$ is not parallel to $\bar{c}$ , then angle between $\vec{a} and \vec{c}$ is

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$π$

$\frac{5 π}{6}$

$\frac{π}{6}$

answer is D.

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Detailed Solution

given $| \vec{a} | = | \vec{b} | = | \vec{c} | = 1$
and $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2} (\vec{b} + \vec{c})$
$\Rightarrow (\vec{a} . \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{\sqrt{3}}{2} \vec{b} + \frac{\sqrt{3}}{2} \vec{c}$
on comparing, we get
$\begin{aligned} \vec{a} . \vec{c} = \frac{\sqrt{3}}{2} \Rightarrow | \vec{a} ∥ \vec{c} | \cos θ = \frac{\sqrt{3}}{2} \\ \Rightarrow \cos θ = \cos \frac{π}{6} \Rightarrow θ = \frac{π}{6} \end{aligned}$