Let vec{a} = a_1hat{i} + a_2hat{j} + a_3hat{k | vedclass

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(C) Given that $\vec{c}$ is a unit vector,$|\vec{c}| = 1$,so $c_1^2 + c_2^2 + c_3^2 = 1 \dots (i)$.Since $\vec{c} \perp \vec{a}$ and $\vec{c} \perp \v

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$\frac{1}{4} |\vec{a}|^2 |\vec{b}|^2$

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(C) Given that $\vec{c}$ is a unit vector,$|\vec{c}| = 1$,so $c_1^2 + c_2^2 + c_3^2 = 1 \dots (i)$.Since $\vec{c} \perp \vec{a}$ and $\vec{c} \perp \vec{b}$,we have $\vec{c} \cdot \vec{a} = 0 \Rightarrow a_1c_1 + a_2c_2 + a_3c_3 = 0 \dots (ii)$ and $\vec{c} \cdot \vec{b} = 0 \Rightarrow b_1c_1 + b_2c_2 + b_3c_3 = 0 \dots (iii)$.The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,so $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} |\vec{a}| |\vec{b}|$.Squaring both sides,$(\vec{a} \cdot \vec{b})^2 = \frac{3}{4} |\vec{a}|^2 |\vec{b}|^2 \dots (iv)$.Now,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 $D^2 = \left| \begin{array}{ccc} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{array} ight| \left| \begin{array}{ccc} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{array} ight|$.Using the property of the determinant of the product of matrices,$D^2 = \left| \begin{array}{ccc} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{array} ight| = \left| \begin{array}{ccc} |\vec{a}|^2 & \vec{a} \cdot \vec{b} & 0 \ \vec{a} \cdot \vec{b} & |\vec{b}|^2 & 0 \ 0 & 0 & 1 \end{array} ight|$.Expanding along the third row,$D^2 = 1 \cdot (|\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2) = |\vec{a}|^2 |\vec{b}|^2 - \frac{3}{4} |\vec{a}|^2 |\vec{b}|^2 = \frac{1}{4} |\vec{a}|^2 |\vec{b}|^2$.

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