Let a→=a1i^+a2j^+a3k^,b→=b1i^+b2j^+b3k^ and c→=c1i^+c2j^+c3k^ be three non-zero vectors such that c→ is an unit vector perpendicular to both a→ and b→ if the angle between a→ and b→ is π6 then the value of a1 b1 c1a2 b2 c2a3 b3 c32 is

(a→×b→⋅c→)2=|a→|2|b→|2|c→|2sin2⁡θcos2⁡ϕ (θ is the angle between a→ and b→,ϕ=0)=14a12+a22+a32b12+b22+b32

Let a→=a1i^+a2j^+a3k^,b→=b1i^+b2j^+b3k^ and c→=c1i^+c2j^+c3k^ be three non-zero vectors such that c→ is an unit vector perpendicular to both a→ and b→ if the angle between a→ and b→ is π6 then the value of a1 b1 c1a2 b2 c2a3 b3 c32 is

(a→×b→⋅c→)2=|a→|2|b→|2|c→|2sin2⁡θcos2⁡ϕ (θ is the angle between a→ and b→,ϕ=0)=14a12+a22+a32b12+b22+b32

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Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} and \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ be three non-zero vectors such that $\vec{c}$ is an unit vector perpendicular to both $\vec{a} and \vec{b}$ if the angle between $\vec{a} and \vec{b}$ is $\frac{π}{6}$ then the value of ${|\begin{array}{lll} a_{1} b_{1} c_{1} \\ a_{2} b_{2} c_{2} \\ a_{3} b_{3} c_{3} \end{array}|}^{2}$ is

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$\frac{3}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})$

$\frac{1}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})$

answer is C.

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Detailed Solution

$\begin{array}{l} (\vec{a} \times \vec{b} \cdot \vec{c})^{2} = | \vec{a} |^{2} | \vec{b} |^{2} | \vec{c} |^{2} \sin^{2} {θcos}^{2} ϕ \\ (θ is the angle between \vec{a} and \vec{b}, ϕ = 0) \\ = \frac{1}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2}) \end{array}$