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Q.

If $\vec{a}, \vec{b} a n d \vec{c}$ are three non – coplanar unit vectors inclined at an angle of $30^{0}$ to each other, then volume tetrahedron whose edges are $\vec{a}, \vec{b} a n d \vec{c}$ is

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a

$\frac{\sqrt{3 \sqrt{3} + 5}}{12}$

b

$\frac{\sqrt{3 \sqrt{3} - 5}}{12}$

c

$\frac{5 \sqrt{2} + 3}{12}$

d

$\frac{5 \sqrt{2} - 3}{12}$

answer is B.

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Detailed Solution

${[\vec{a} \vec{b} \vec{c}]}^{2} = | \begin{matrix} \vec{a} . \vec{a} & \vec{a} . \vec{b} & \vec{a} . \vec{c} \\ \vec{b} . \vec{a} & \vec{b} . \vec{b} & \vec{b} . \vec{c} \\ \vec{c} . \vec{a} & \vec{c} . \vec{b} & \vec{c} . \vec{c} \end{matrix} |$

$| \begin{matrix} 1 & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & 1 & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} & 1 \end{matrix} | = \frac{3 \sqrt{3} - 5}{4}$

$∴ V = \frac{1}{6} | [\vec{a} \vec{b} \vec{c}] | = \frac{\sqrt{3 \sqrt{3} - 5}}{12}$

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