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Addition of Vectors

For any two v...

Question

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

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Solution

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

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