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

$7\hat i-4\hat j-4\hat k$

The correct answer is Option (1) → $7\hat i-4\hat j-4\hat k$

Let the required vector be $\vec{r}$.

Given vectors:

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

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

Since $\vec{r}$ is perpendicular to both $\vec{a}$ and $\vec{b}$,

$\vec{r} \propto \vec{a} \times \vec{b}$

Compute cross product:

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

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

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

Magnitude of $\vec{a} \times \vec{b}$:

$|\vec{a} \times \vec{b}| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{81} = 9$

Hence, a unit vector along $\vec{a} \times \vec{b}$ is $\frac{1}{9}(7\hat{i} - 4\hat{j} - 4\hat{k})$.

Since the required vector has magnitude 9,

$\vec{r} = 9 \times \frac{1}{9}(7\hat{i} - 4\hat{j} - 4\hat{k}) = 7\hat{i} - 4\hat{j} - 4\hat{k}$

Therefore, the required vector is $7\hat{i} - 4\hat{j} - 4\hat{k}$.