Let $\vec u=\hat i+\hat j,\,\vec v=\hat i-\hat j$ and $\vec w=\hat i+2\hat j+3\hat k$. If $\hat n$ is a unit vector such that $\vec u.\hat n=0$ and $\vec v.\hat n=0$, then $\vec w.\hat n$ is equal to:

3

Given $\vec u=\hat i+\hat j,\,\vec v=\hat i-\hat j$

$\vec w=\hat i+2\hat j+3\hat k$

$\vec u.\hat n=0$ and $\vec v.\hat n=0$

$\hat n=\frac{\vec u×\vec v}{|\vec u×\vec v|}$

$⇒\vec u×\vec v=\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&0\\1&-1&0\end{vmatrix}=0\hat i-0\hat j-2\hat k=-2\hat k$

$\vec w.\hat n=\frac{|\vec w.\vec u×\vec v|}{|\vec u×\vec v|}=\frac{|-6\hat k|}{|-2\hat k|}=3$

Since $\vec w.(\vec u×\vec v)=(\hat i+2\hat j+3\hat k).(-2\hat k)=-6\hat k$

$⇒|\vec w.\hat n|=3$