If three non-zero vectors are a = a_1 i + a_2 | vedclass

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(D) Given that $c$ is a unit vector,$|c| = 1,$ so $|c|^2 = c_1^2 + c_2^2 + c_3^2 = 1$ .....$(i)$Since $c \perp a$ and $c \perp b,$ we have $c \cdot a

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$\frac{(\Sigma a_1^2)(\Sigma b_1^2)}{4}$

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(D) Given that $c$ is a unit vector,$|c| = 1,$ so $|c|^2 = c_1^2 + c_2^2 + c_3^2 = 1$ .....$(i)$Since $c \perp a$ and $c \perp b,$ we have $c \cdot a = 0$ and $c \cdot b = 0.$This implies $a_1 c_1 + a_2 c_2 + a_3 c_3 = 0$ .....$(ii)$and $b_1 c_1 + b_2 c_2 + b_3 c_3 = 0$ .....$(iii)$The angle between $a$ and $b$ is $\frac{\pi}{6},$ so $a \cdot b = |a||b| \cos(\frac{\pi}{6}) = |a||b| \frac{\sqrt{3}}{2}.$Squaring both sides,$(a \cdot b)^2 = |a|^2 |b|^2 \frac{3}{4} \Rightarrow (a_1 b_1 + a_2 b_2 + a_3 b_3)^2 = \frac{3}{4} (\Sigma a_1^2)(\Sigma b_1^2)$ .....$(iv)$Let $D = \left| \begin{array}{ccc} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{array} ight|.$ Then $D^2 = \left| \begin{array}{ccc} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{array} ight| \left| \begin{array}{ccc} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{array} ight| = \left| \begin{array}{ccc} a \cdot a & a \cdot b & a \cdot c \ b \cdot a & b \cdot b & b \cdot c \ c \cdot a & c \cdot b & c \cdot c \end{array} ight|.$Using the conditions $a \cdot c = 0,$ $b \cdot c = 0,$ and $c \cdot c = 1,$ we get $D^2 = \left| \begin{array}{ccc} |a|^2 & a \cdot b & 0 \ a \cdot b & |b|^2 & 0 \ 0 & 0 & 1 \end{array} ight| = |a|^2 |b|^2 - (a \cdot b)^2.$Substituting $(a \cdot b)^2 = \frac{3}{4} |a|^2 |b|^2,$ we get $D^2 = |a|^2 |b|^2 - \frac{3}{4} |a|^2 |b|^2 = \frac{1}{4} |a|^2 |b|^2 = \frac{(\Sigma a_1^2)(\Sigma b_1^2)}{4}.$

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