Let a¯=a1i¯+a2j¯+a3k¯,b¯=b1i¯+b2j¯+b3k¯, c¯=c1i¯+c2j¯+c3k¯ be three non zero vectors such that |c¯|=1 angle between a¯ and b¯ is π4 and c¯ is perpendicular to a¯ and b¯ then |a1b1c1a2b2c2a3b3c3|2= α(a12+a22+a32) (b12+b22+b32) where α=

|a1b1c1a2b2c2a3b3c3|2=|a1a2a3b1b2b3c1c2c3|2=|a¯.(b¯×c¯)|2 =|(a¯×b¯) . c¯|2 =|a¯×b¯|2|c¯|2 =|a¯|2|b¯|2sin2π/4 =12(a12+a22+a32)(b12+b22+b32) ∴α=12

Let a¯=a1i¯+a2j¯+a3k¯,b¯=b1i¯+b2j¯+b3k¯, c¯=c1i¯+c2j¯+c3k¯ be three non zero vectors such that |c¯|=1 angle between a¯ and b¯ is π4 and c¯ is perpendicular to a¯ and b¯ then |a1b1c1a2b2c2a3b3c3|2= α(a12+a22+a32) (b12+b22+b32) where α=

|a1b1c1a2b2c2a3b3c3|2=|a1a2a3b1b2b3c1c2c3|2=|a¯.(b¯×c¯)|2 =|(a¯×b¯) . c¯|2 =|a¯×b¯|2|c¯|2 =|a¯|2|b¯|2sin2π/4 =12(a12+a22+a32)(b12+b22+b32) ∴α=12

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Let $\bar{a} = a_{1} \bar{i} + a_{2} \bar{j} + a_{3} \bar{k}, \bar{b} = b_{1} \bar{i} + b_{2} \bar{j} + b_{3} \bar{k},$ $\bar{c} = c_{1} \bar{i} + c_{2} \bar{j} + c_{3} \bar{k}$ be three non zero vectors such that $| \bar{c} | = 1$ angle between $\bar{a}$ and $\bar{b}$ is $\frac{π}{4}$ and $\bar{c}$ is perpendicular to $\bar{a}$ and $\bar{b}$ then ${| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} |}^{2} = α (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})$ where $α =$

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$\frac{1}{2}$

$\frac{1}{4}$

answer is A.

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Detailed Solution

${| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} |}^{2} = {| \begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} |}^{2} = {| \bar{a} . (\bar{b} \times \bar{c}) |}^{2} = {| (\bar{a} \times \bar{b}) . \bar{c} |}^{2} = {| \bar{a} \times \bar{b} |}^{2} {| \bar{c} |}^{2} = {| \bar{a} |}^{2} {| \bar{b} |}^{2} \sin^{2} π / 4 = \frac{1}{2} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2}) ∴ α = \frac{1}{2}$