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Change of Variables

If a → is any...

Question

If $\vec{a}$ is any vector, then ${(\vec{a} \times \hat{i})}^{2} + {(\vec{a} \times \hat{j})}^{2} + {(\vec{a} \times \hat{k})}^{2} =$
(a) ${\vec{a}}^{2}$

(b) $2 {\vec{a}}^{2}$

(c) $3 {\vec{a}}^{2}$

(d) $4 {\vec{a}}^{2}$

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Solution

(b) $2 {\vec{a}}^{2}$

$Let \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \vec{a} \times \hat{i} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 1 & 0 & 0 \end{matrix}| = a_{3} \hat{j} - a_{2} \hat{k} \Rightarrow {(\vec{a} \times \hat{i})}^{2} = {(a_{3} \hat{j} - a_{2} \hat{k})}^{2} = {a_{3}}^{2} {|\hat{j}|}^{2} + {a_{2}}^{2} {|\hat{k}|}^{2} - 2 a_{3} a_{2} (\hat{j} . \hat{k}) = {a_{3}}^{2} + {a_{2}}^{2} (∵ \hat{j} . \hat{k} =0) . . . (1) ∴ \vec{a} \times \hat{j} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 0 & 1 & 0 \end{matrix}| = - a_{3} \hat{i} + a_{1} \hat{k} \Rightarrow {(\vec{a} \times \hat{j})}^{2} = {(- a_{3} \hat{i} + a_{1} \hat{k})}^{2} = {a_{3}}^{2} {|\hat{i}|}^{2} + {a_{1}}^{2} {|\hat{k}|}^{2} - 2 a_{3} a_{2} (\hat{i} . \hat{k}) = {a_{3}}^{2} + {a_{1}}^{2} (∵ \hat{i} . \hat{k} =0) . . . (2) ∴ \vec{a} \times \hat{k} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 0 & 0 & 1 \end{matrix}| = a_{2} \hat{i} - a_{1} \hat{j} \Rightarrow {(\vec{a} \times k)}^{2} = {(a_{2} \hat{i} - a_{1} \hat{j})}^{2} = {a_{2}}^{2} {|\hat{i}|}^{2} + {a_{1}}^{2} {|j|}^{2} + 2 a_{1} a_{2} (\hat{i} . \hat{j}) = {a_{2}}^{2} + {a_{1}}^{2} (∵ \hat{i} . \hat{j} =0) . . . (3) Adding (1), (2) and (3), we get {(\vec{a} \times \hat{i})}^{2} + {(\vec{a} \times \hat{j})}^{2} + {(\vec{a} \times k)}^{2} = {a_{3}}^{2} + {a_{2}}^{2} + {a_{3}}^{2} + {a_{1}}^{2} + {a_{2}}^{2} + {a_{1}}^{2} = 2 ({a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}) = 2 {\vec{a}}^{2} (∵ |\vec{a}| = \sqrt{{a_{1}}^{2} + {a_{2}}^{2} + {a_{3}}^{2}})$

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