Question 14 Marks

For any vector $\vec{a}$, prove that :
$
|\overrightarrow{ a } \times \hat{ i }|^2+|\overrightarrow{ a } \times \hat{ j }|^2+|\overrightarrow{ a } \times \hat{ k }|^2= 2 |\overrightarrow{ a }|^2
$

Answer

Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$
$
\begin{array}{llrl}
& \therefore & |\vec{a}| & =\sqrt{a_1^2+a_2^2+a_3^2} \\
\text { or } & & |\vec{a}|^2 & =a_1^2+a_2^2+a_3^2
\end{array}
$
Therefore, $\vec{a} \times \hat{i}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \times \hat{i}$
$
\begin{array}{l}
=a_1(\hat{i} \times \hat{i})+a_2(\hat{j} \times \hat{i})+a_3(\hat{k} \times \hat{i}) \\
=a_1 \times 0+a_2(-\hat{k})+a_3 \hat{j} \\
=-a_2 \hat{k}+a_3 \hat{j}
\end{array}
$
$
\begin{array}{l}
\Rightarrow \quad|\vec{a} \times \hat{i}|=\sqrt{\left(-a_2\right)^2+\left(a_3\right)^2}=\sqrt{a_2^2+a_3^2} \\
\text { or } \quad|\vec{a} \times \hat{i}|^2=a_2^2+a_3^2 \quad \quad \ldots \ldots(1)\\
\text { Similarly } \quad \vec{a} \times \hat{j}=\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \times \hat{j} \\
=a_1 \hat{k}-a_3 \hat{i}
\end{array}
$

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