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(C) The vector $m$ is coplanar with $(2 \hat{i}+\hat{j})$ and $(\hat{j}-\hat{k})$.So,$m = x(2 \hat{i}+\hat{j}) + y(\hat{j}-\hat{k}) = 2x \hat{i} + (x+

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$\frac{5}{\sqrt{26}}$

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(C) The vector $m$ is coplanar with $(2 \hat{i}+\hat{j})$ and $(\hat{j}-\hat{k})$.So,$m = x(2 \hat{i}+\hat{j}) + y(\hat{j}-\hat{k}) = 2x \hat{i} + (x+y) \hat{j} - y \hat{k}$.Since $m$ is orthogonal to $\hat{i}-\hat{j}+\hat{k}$,we have:$2x - (x+y) - y = 0 \Rightarrow x - 2y = 0 \Rightarrow x = 2y$.Substituting $x = 2y$ into the expression for $m$:$m = 2(2y) \hat{i} + (2y+y) \hat{j} - y \hat{k} = 4y \hat{i} + 3y \hat{j} - y \hat{k}$.Since $m$ is a unit vector,$|m| = 1$:$\sqrt{(4y)^2 + (3y)^2 + (-y)^2} = 1 \Rightarrow \sqrt{16y^2 + 9y^2 + y^2} = 1 \Rightarrow \sqrt{26y^2} = 1 \Rightarrow y = \pm \frac{1}{\sqrt{26}}$.Thus,$m = \pm \frac{1}{\sqrt{26}}(4 \hat{i} + 3 \hat{j} - \hat{k})$.The length of the perpendicular from the origin to the plane $r \cdot m = a \cdot m$ is $|a \cdot m|$.$|a \cdot m| = |(\hat{i}-\hat{k}) \cdot \pm \frac{1}{\sqrt{26}}(4 \hat{i} + 3 \hat{j} - \hat{k})| = \frac{1}{\sqrt{26}} |(1)(4) + (0)(3) + (-1)(-1)| = \frac{1}{\sqrt{26}} |4+1| = \frac{5}{\sqrt{26}}$ units.

Let $m$ be the unit vector orthogonal to the vector $\hat{i}-\hat{j}+\hat{k}$ and coplanar with the vectors $2 \hat{i}+\hat{j}$ and $\hat{j}-\hat{k}$. If $a=\hat{i}-\hat{k}$,then the length of the perpendicular from the origin to the plane $r \cdot m=a \cdot m$ is

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