If $\vec{a}$ is any vector, then $|\vec a×\hat i|^2 +|\vec a×\hat j|^2 + |\vec a×\hat k|^2$ is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $2|\vec a|^2$
Let $\vec{a} = a_{1}\hat{i} + a_{2}\hat{j} + a_{3}\hat{k}$
$|\vec{a} \times \hat{i}|^{2} = |(a_{1}\hat{i} + a_{2}\hat{j} + a_{3}\hat{k}) \times \hat{i}|^{2}$
$= |(a_{2}\hat{j} + a_{3}\hat{k}) \times \hat{i}|^{2} = |a_{2}(\hat{j}\times\hat{i}) + a_{3}(\hat{k}\times\hat{i})|^{2}$
$= |(-a_{2}\hat{k} + a_{3}\hat{j})|^{2} = a_{2}^{2} + a_{3}^{2}$
Similarly,
$|\vec{a} \times \hat{j}|^{2} = a_{3}^{2} + a_{1}^{2}$
$|\vec{a} \times \hat{k}|^{2} = a_{1}^{2} + a_{2}^{2}$
Add all three:
$|\vec{a} \times \hat{i}|^{2} + |\vec{a} \times \hat{j}|^{2} + |\vec{a} \times \hat{k}|^{2} = (a_{2}^{2} + a_{3}^{2}) + (a_{3}^{2} + a_{1}^{2}) + (a_{1}^{2} + a_{2}^{2})$
$= 2(a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) = 2|\vec{a}|^{2}$
$|\vec{a} \times \hat{i}|^{2} + |\vec{a} \times \hat{j}|^{2} + |\vec{a} \times \hat{k}|^{2} = 2|\vec{a}|^{2}$