Question

For any two vectors \[\vec{a} \text { and } \vec{b}\] of magnitudes 3 and 4 respectively, write the value of \[\left[ \vec{a} \vec{b} \vec{a} \times \vec{b} \right] + \left( \vec{a} \cdot \vec{b} \right)^2 .\]

Answer 1

We have 

\[\left[ \vec{a} \vec{b} \vec{a} \times \vec{b} \right] + \left( \vec{a} . \vec{b} \right)^2 \]

\[ = \left[ \left( \vec{a} \times \vec{b} \right) . \left( \vec{a} \times \vec{b} \right) \right] + \left( \vec{a} . \vec{b} \right)^2 \left( \text { By definition of scalar triple product } \right)\]

\[ = \left| \left( \vec{a} \times \vec{b} \right) \right|^2 + \left( \vec{a} . \vec{b} \right)^2 \]

\[ = \left( \begin{vmatrix}\vec{a}\end{vmatrix}\begin{vmatrix}\vec{b}\end{vmatrix} \text { sin }\theta \right)^2 + \left( \begin{vmatrix}\vec{a}\end{vmatrix}\begin{vmatrix}\vec{b}\end{vmatrix}cos\theta \right)^2 \]

\[ = \left( \begin{vmatrix}\vec{a}\end{vmatrix}\begin{vmatrix}\vec{b}\end{vmatrix} \right)^2 s {in}^2 \theta + \left( \begin{vmatrix}\vec{a}\end{vmatrix}\begin{vmatrix}\vec{b}\end{vmatrix} \right)^2 \cos^2 \theta\]

\[ = \left( \begin{vmatrix}\vec{a}\end{vmatrix}\begin{vmatrix}\vec{b}\end{vmatrix} \right)^2 \left( \text { sin }^2 \theta + c {os}^2 \theta \right)\]

\[ = \left( \begin{vmatrix}\vec{a}\end{vmatrix}\begin{vmatrix}\vec{b}\end{vmatrix} \right)^2 \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \left( \because \sin^2 \theta + c {os}^2 \theta = 1 \right)\]

\[ = \left( 3 \times 4 \right)^2 \left( \text { Given }: \begin{vmatrix}\vec{a}\end{vmatrix} = 3 \text { and } \begin{vmatrix}\vec{b}\end{vmatrix} = 4 \right)\]

\[ = 144\]