Question

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

विकल्प

• \[\left| \vec{a} \right|^2 \left| \vec{b} \right|^2\]

• \[\left| \vec{a} + \vec{b} \right|^2\]

• \[\left| \vec{a} \right|^2 + \left| \vec{b} \right|^2\]

• \[2 \left| \vec{a} \right|^2 \left| \vec{b} \right|^2\]

Answer 1

\[\left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \]

We have

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

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

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

\[ = \left( \left| \vec{a} \right|\left| \vec{b} \right| \sin \theta \right)^2 + \left( \left| \vec{a} \right| \left| \vec{b} \right|^{} \cos \theta \right)^2 \]

\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \sin^2 \theta + \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \cos^2 \theta\]

\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \left( \sin^2 \theta + \cos^2 \theta \right)\]

\[ = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \]