For any vector $\vec a$, $|\vec a×\hat i|^2+|\vec a×\hat j|^2+|\vec a×\hat k|^2$ is equal to
Answer & explanation
Correct answer: option 2
Let $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$. Then,
$\vec a×\hat i=-a_2\hat k+a_3\hat j,\vec a×\hat j=a_1\hat k-a_3\hat i$ and $\vec a×\hat k=-a_1\hat j+a_2\hat i$
$∴\vec a$, $|\vec a×\hat i|^2+|\vec a×\hat j|^2+|\vec a×\hat k|^2$
$=({a_2}^2 + {a_3}^2) + ({a_1}^2 + {a_3}^2) + ({a_1}^2+ {a_2}^2)$
$=2({a_1}^2+ {a_2}^2+{a_3}^2)=2|\vec a|^2$