Question

\[\text{ If }  \left| \vec{a} \right| = 10, \left| \vec{b} \right| = 2 \text{ and }  \left| \vec{a} \times \vec{b} \right| = 16, \text{ find }  \vec{a} . \vec{b} .\]

 

Answer 1

\[\text{ We know }  \]
\[ \left( \vec{a} . \vec{b} \right)^2 + \left| \vec{a} \times \vec{b} \right|^2 = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \]
\[ \Rightarrow \left( \vec{a} . \vec{b} \right)^2 + \left( 16 \right)^2 = \left( 10 \right)^2 \times 2^2 ( \because \left| \vec{a} \times \vec{b} \right| = 16, \left| \vec{a} \right| = 10 \text{ and  } \left| \vec{b} \right| = 2)\]
\[ \Rightarrow \left( \vec{a} . \vec{b} \right)^2 + 256 = 400\]
\[ \Rightarrow \left( \vec{a} . \vec{b} \right)^2 = 144\]
\[ \Rightarrow \left( \vec{a} . \vec{b} \right) = \pm 12\]