If $\vec a,\vec b$ are non-zero and non-collinear vectors, then $[\vec a\,\,\vec b\,\,\hat i]\hat i +[\vec a\,\,\vec b\,\,\hat j]\hat j+[\vec a\,\,\vec b\,\,\hat k]\hat k$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$[\vec a\,\,\vec b\,\,\hat i]\hat i +[\vec a\,\,\vec b\,\,\hat j]\hat j+[\vec a\,\,\vec b\,\,\hat k]\hat k$
$=\{(\vec a×\vec b).\hat i\}\hat i+\{(\vec a×\vec b).\hat j\}\hat j+\{(\vec a×\vec b).\hat k\}\hat k=\vec a×\vec b$