Let $\vec a =\hat i +4\hat j,\vec b=4\hat j+\hat k$ and $\vec c=\hat i-2\hat k$. If $\vec d$ is a vector perpendicular to both $\vec a$ and $\vec b$ such that $\vec c.\vec d=16$, then $|\vec d|$ is equal to

$4\sqrt{33}$

The correct answer is Option (4) → $4\sqrt{33}$

$\vec a=(1,4,0)$

$\vec b=(0,4,1)$

$\vec c=(1,0,-2)$

$\vec d$ is perpendicular to both $\vec a$ and $\vec b$, so $\vec d$ is parallel to $\vec a\times \vec b$.

$\vec a\times \vec b= \begin{vmatrix} i & j & k\\ 1 & 4 & 0\\ 0 & 4 & 1 \end{vmatrix} $

$\vec a\times \vec b=(4,-1,4)$

Hence $\vec d=t(4,-1,4)$.

Given: $\vec c\cdot \vec d=16$.

$\vec c\cdot \vec d=(1,0,-2)\cdot t(4,-1,4)$

$t(4+0-8)=16$

$t(-4)=16$

$t=-4$

$|\vec d|=|-4|\sqrt{4^2+(-1)^2+4^2}$

$|\vec d|=4\sqrt{16+1+16}$

$|\vec d|=4\sqrt{33}$

Final answer: $4\sqrt{33}$