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Addition of Vectors

For any two v...

Question

For any two vectors $\vec{a} and \vec{b}, find (\vec{a} \times \vec{b}) . \vec{b} .$

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Solution

$Let: \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} \vec{a} \times \vec{b} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}| = \hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1}) (\vec{a} \times \vec{b}) . \vec{b} = [\hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1})] . (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = b_{1} (a_{2} b_{3} - a_{3} b_{2}) - b_{2} (a_{1} b_{3} - a_{3} b_{1}) + b_{3} (a_{1} b_{2} - a_{2} b_{1}) = a_{2} b_{1} b_{3} - a_{3} b_{1} b_{2} - a_{1} b_{2} b_{3} + a_{3} b_{1} b_{2} + a_{1} b_{2} b_{3} - a_{2} b_{1} b_{3} = 0$

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