A vector of magnitude 8 units in the direction perpendicular to both…

The correct answer is Option (3) → $\frac{8}{\sqrt{6}}(\hat i+\hat j - 2\hat k)$

Given vectors:

$\vec a=\hat i+\hat j+\hat k=(1,1,1)$

$\vec b=2\hat i+\hat k=(2,0,1)$

Required direction is along $\vec a\times \vec b$.

$\vec a\times \vec b= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 1\\ 2 & 0 & 1 \end{vmatrix}$

$= \hat i(1\cdot 1-1\cdot 0)-\hat j(1\cdot 1-1\cdot 2)+\hat k(1\cdot 0-1\cdot 2)$

$= \hat i(1)-\hat j(-1)+\hat k(-2)$

$= \hat i+\hat j-2\hat k$

Magnitude:

$|\vec a\times \vec b|=\sqrt{1^{2}+1^{2}+(-2)^{2}}=\sqrt{6}$

Unit vector in this direction:

$\frac{\hat i+\hat j-2\hat k}{\sqrt{6}}$

Vector of magnitude $8$ in this direction:

$\vec d=8\cdot \frac{\hat i+\hat j-2\hat k}{\sqrt{6}}$

$=\frac{8}{\sqrt{6}}(\hat i+\hat j-2\hat k)$

Final answer: $\frac{8}{\sqrt{6}}(\hat i+\hat j-2\hat k)$ is the required vector