Question
Let $$\overrightarrow A = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k,\,\overrightarrow B = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k$$ and $$\overrightarrow C = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k$$ be three non-zero vectors such that $$\overrightarrow C $$ is a unit vector perpendicular to both the vectors $$\overrightarrow A $$ and $$\overrightarrow B .$$ If the angle between $$\overrightarrow A $$ and $$\overrightarrow B $$ is $$\frac{\pi }{6},$$ then \[{\left| \begin{array}{l}
{a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\
{b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\
{c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3}
\end{array} \right|^2}\] is equal to :
A.
$$0$$
B.
$$1$$
C.
$$\frac{1}{4}\left( {a_1^2 + a_2^2 + a_3^2} \right)\left( {b_1^2 + b_2^2 + b_3^2} \right)$$
D.
$$\frac{3}{4}\left( {a_1^2 + a_2^2 + a_3^2} \right)\left( {b_1^2 + b_2^2 + b_3^2} \right)\left( {c_1^2 + c_2^2 + c_3^2} \right)$$
Answer :
$$\frac{1}{4}\left( {a_1^2 + a_2^2 + a_3^2} \right)\left( {b_1^2 + b_2^2 + b_3^2} \right)$$
Solution :
\[\begin{array}{l}
{\left| \begin{array}{l}
{a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\
{b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\
{c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3}
\end{array} \right|^2} = {\left[ {\overrightarrow A \overrightarrow B \overrightarrow C } \right]^2}\\
\Rightarrow \left| \begin{array}{l}
{a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\
{b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\
{c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3}
\end{array} \right| = {\left( {\left( {\overrightarrow A \times \overrightarrow B } \right).\overrightarrow C } \right)^2}\\
\Rightarrow \left| \begin{array}{l}
{a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\
{b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\
{c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3}
\end{array} \right| = {\left\{ {\left| {\overrightarrow A } \right|\,\left| {\overrightarrow B } \right|\sin \frac{\pi }{6}\left( {\overrightarrow C } \right).\overrightarrow C } \right\}^2}\\
\Rightarrow \left| \begin{array}{l}
{a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\
{b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\
{c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3}
\end{array} \right| = {\left| {\overrightarrow A } \right|^2}{\left| {\overrightarrow B } \right|^2}{\left( {\frac{1}{2}} \right)^2}{\left| {\overrightarrow C } \right|^4}\\
\Rightarrow \left| \begin{array}{l}
{a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\
{b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\
{c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3}
\end{array} \right| = \frac{1}{4}{\left| {\overrightarrow A } \right|^2}{\left| {\overrightarrow B } \right|^2}\\
\Rightarrow \left| \begin{array}{l}
{a_1}\,\,\,\,{a_2}\,\,\,\,{a_3}\\
{b_1}\,\,\,\,\,{b_2}\,\,\,\,{b_3}\\
{c_1}\,\,\,\,\,{c_2}\,\,\,\,{c_3}
\end{array} \right| = \frac{1}{4}\left( {a_1^2 + a_2^2 + a_3^2} \right)\left( {b_1^2 + b_2^2 + b_3^2} \right)
\end{array}\]