A vector of magnitude 9, which is perpendicular to both the vectors $(4\hat{i} - \hat{j} + 8\hat{k})$ and $(-\hat{j} + \hat{k})$ is

A vector perpendicular to both given vectors can be found using the cross product.

The given vectors are:

$\vec{a} = 4\hat{i} - \hat{j} + 8\hat{k}$

$\vec{b} = 0\hat{i} - \hat{j} + \hat{k}$

The cross product $\vec{a} \times \vec{b}$ is:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 8 \\ 0 & -1 & 1 \end{vmatrix}$

$= \hat{i}\begin{vmatrix} -1 & 8 \\ -1 & 1 \end{vmatrix} - \hat{j}\begin{vmatrix} 4 & 8 \\ 0 & 1 \end{vmatrix} + \hat{k}\begin{vmatrix} 4 & -1 \\ 0 & -1 \end{vmatrix}$

$= \hat{i}[(-1)(1) - (8)(-1)] - \hat{j}[(4)(1) - (8)(0)] + \hat{k}[(4)(-1) - (-1)(0)]$

$= \hat{i}[-1 + 8] - \hat{j}[4 - 0] + \hat{k}[-4 - 0]$

$= 7\hat{i} - 4\hat{j} - 4\hat{k}$

The magnitude of this vector is:

$|\vec{v}| = \sqrt{(7)^2 + (-4)^2 + (-4)^2}$

$= \sqrt{49 + 16 + 16}$

$= \sqrt{81}$

$= 9$

The vector $7\hat{i} - 4\hat{j} - 4\hat{k}$ has magnitude 9 and is perpendicular to both given vectors.