Let $\vec V=2\hat i+\hat j-\hat k$ and $\vec W =\hat i +3\hat k$. It $\vec U$ is a unit vector, then the maximum value of the scalar triple product $[\vec U\,\,\vec V\,\,\vec W]$, is
Answer & explanation
Correct answer: option 3
We have,
$[\vec U\,\,\vec V\,\,\vec W]=\vec U.(\vec V×\vec W)$
$⇒[\vec U\,\,\vec V\,\,\vec W]≤|\vec U||\vec V×\vec W|$ $[∵\vec a.\vec b≤|\vec a||\vec b|]$
$⇒[\vec U\,\,\vec V\,\,\vec W]≤|\vec V×\vec W|$
Now,
$\vec V×\vec W=\begin{vmatrix}\hat i&\hat j&\hat k\\2&1&-1\\1&0&3\end{vmatrix}=3\hat i-7\hat j-\hat k$
$∴|\vec V×\vec W|=\sqrt{9+49+1}=\sqrt{59}$
Hence, $[\vec U\,\,\vec V\,\,\vec W]≤\sqrt{59}$