A vector of magnitude 8 units in the direction perpendicular to both the vectors $\hat{i} + \hat{j} + \hat{k}$ and $2\hat{i} + \hat{k}$ is

A vector perpendicular to two given vectors is found using the cross product.

Let:

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

$\vec{b} = 2\hat{i} + \hat{k}$

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

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

$= \hat{i}\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} - \hat{j}\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} + \hat{k}\begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix}$

$= \hat{i}(1 \times 1 - 1 \times 0) - \hat{j}(1 \times 1 - 1 \times 2) + \hat{k}(1 \times 0 - 1 \times 2)$

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

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

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The magnitude of this vector is:

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

$= \sqrt{1 + 1 + 4}$

$= \sqrt{6}$

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To obtain a vector of magnitude 8 in the same direction:

Required Vector $= \frac{8}{|\vec{a} \times \vec{b}|} \times (\vec{a} \times \vec{b})$

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

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

Therefore, the required vector is $\frac{8}{\sqrt{6}}(\hat{i} + \hat{j} - 2\hat{k})$.