If a→,b→,c→ are unit vectors, b→ and c→ are inclined to a→at an angle π6 and the angle between b→ and c→ is π3, then (a→×b→)⋅(a→×c→)=

a→⋅b→=a→⋅c→=cos⁡π6=32 and b→⋅c→=cos⁡π3=12(a→×b→)⋅(a→×c→)=a→⋅a→ a→⋅c→b→⋅a→ b→⋅c→=1323212=12−34=−14

If a→,b→,c→ are unit vectors, b→ and c→ are inclined to a→at an angle π6 and the angle between b→ and c→ is π3, then (a→×b→)⋅(a→×c→)=

a→⋅b→=a→⋅c→=cos⁡π6=32 and b→⋅c→=cos⁡π3=12(a→×b→)⋅(a→×c→)=a→⋅a→ a→⋅c→b→⋅a→ b→⋅c→=1323212=12−34=−14

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If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, $\vec{b}$ and $\vec{c}$ are inclined to $\vec{a}$ at an angle $\frac{π}{6}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{π}{3}$ , then $(\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c}) =$

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$\frac{1}{\sqrt{2}}$

$- \frac{1}{4}$

answer is D.

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Detailed Solution

$\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = \cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\vec{b} \cdot \vec{c} = \cos \frac{π}{3} = \frac{1}{2}$

$(\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c}) = |\begin{array}{ll} \vec{a} \cdot \vec{a} \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} \vec{b} \cdot \vec{c} \end{array}| = |\begin{array}{cc} 1 & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}| = \frac{1}{2} - \frac{3}{4} = - \frac{1}{4}$