A unit vector coplanar with vec{a} = hat{i} + | vedclass

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(C) Let the required vector be $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$.Since $\vec{v}$ is perpendicular to $\vec{c} = \hat{i} + \hat{j} + \hat{k}$,

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$\pm \frac{1}{\sqrt{2}} (\hat{j} - \hat{k})$

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(C) Let the required vector be $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$.Since $\vec{v}$ is perpendicular to $\vec{c} = \hat{i} + \hat{j} + \hat{k}$, we have $\vec{v} \cdot \vec{c} = 0 \implies x + y + z = 0 \implies z = -(x + y)$.Since $\vec{v}$ is coplanar with $\vec{a}$ and $\vec{b}$, $\vec{v}$ must be perpendicular to the normal vector $\vec{n} = \vec{a} 	imes \vec{b}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 2 \ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1-4) - \hat{j}(1-2) + \hat{k}(2-1) = -3\hat{i} + \hat{j} + \hat{k}$.Now, $\vec{v} \cdot \vec{n} = 0 \implies -3x + y + z = 0$.Substituting $z = -(x + y)$, we get $-3x + y - x - y = 0 \implies -4x = 0 \implies x = 0$.Then $z = -y$. So $\vec{v} = y\hat{j} - y\hat{k} = y(\hat{j} - \hat{k})$.For a unit vector, $|\vec{v}| = 1 \implies |y|\sqrt{1^2 + (-1)^2} = 1 \implies |y|\sqrt{2} = 1 \implies y = \pm \frac{1}{\sqrt{2}}$.Thus, $\vec{v} = \pm \frac{1}{\sqrt{2}}(\hat{j} - \hat{k})$.

$A$ unit vector coplanar with $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ and perpendicular to $\vec{c} = \hat{i} + \hat{j} + \hat{k}$ is

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