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If a is a unit vector, then $|\mathbf{a} \times \hat{\mathbf{i}}|^2+|\mathbf{a} \times \hat{\mathbf{j}}|^2+|\mathbf{a} \times \hat{\mathbf{k}}|^2=$
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The correct answer is:
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We have,
$$
\begin{aligned}
& |\mathbf{a} \times \hat{\mathbf{i}}|^2+|\mathbf{a} \times \hat{\mathbf{j}}|^2+|\mathbf{a} \times \hat{\mathbf{k}}|^2 \\
& (|\mathbf{a}||\hat{\mathbf{i}}| \sin \alpha)^2+(|\mathbf{a}||\hat{\mathbf{j}}| \sin \beta)^2+(|\mathbf{a}||\hat{\mathbf{k}}| \sin \gamma)^2 \\
& =\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma \quad[\because|\mathbf{a}|=|\hat{\mathbf{i}}|=|\hat{\mathbf{j}}|=|\hat{\mathbf{k}}|=1] \\
& =1-\cos ^2 \alpha+1-\cos ^2 \beta+1-\cos ^2 \gamma \\
& =3-\left(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma\right) \\
& =3-1=2 \quad\left[\because \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\right]
\end{aligned}
$$
$$
\begin{aligned}
& |\mathbf{a} \times \hat{\mathbf{i}}|^2+|\mathbf{a} \times \hat{\mathbf{j}}|^2+|\mathbf{a} \times \hat{\mathbf{k}}|^2 \\
& (|\mathbf{a}||\hat{\mathbf{i}}| \sin \alpha)^2+(|\mathbf{a}||\hat{\mathbf{j}}| \sin \beta)^2+(|\mathbf{a}||\hat{\mathbf{k}}| \sin \gamma)^2 \\
& =\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma \quad[\because|\mathbf{a}|=|\hat{\mathbf{i}}|=|\hat{\mathbf{j}}|=|\hat{\mathbf{k}}|=1] \\
& =1-\cos ^2 \alpha+1-\cos ^2 \beta+1-\cos ^2 \gamma \\
& =3-\left(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma\right) \\
& =3-1=2 \quad\left[\because \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\right]
\end{aligned}
$$
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