Let bar{a} = 2hat{i} + hat{j} + hat{k},bar{b} | vedclass

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(D) Since $\bar{c}$ is coplanar with $\bar{a}$ and $\bar{b}$,we can write $\bar{c} = x\bar{a} + y\bar{b}$ for some scalars $x$ and $y$.$\bar{c} = x(2\

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$-\hat{j} + \hat{k}$

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(D) Since $\bar{c}$ is coplanar with $\bar{a}$ and $\bar{b}$,we can write $\bar{c} = x\bar{a} + y\bar{b}$ for some scalars $x$ and $y$.$\bar{c} = x(2\hat{i} + \hat{j} + \hat{k}) + y(\hat{i} + 2\hat{j} - \hat{k}) = (2x + y)\hat{i} + (x + 2y)\hat{j} + (x - y)\hat{k}$.Given that $\bar{c} \perp \bar{a}$,the dot product $\bar{c} \cdot \bar{a} = 0$.$(2x + y)(2) + (x + 2y)(1) + (x - y)(1) = 0$.$4x + 2y + x + 2y + x - y = 0 \implies 6x + 3y = 0 \implies y = -2x$.Substituting $y = -2x$ into the expression for $\bar{c}$:$\bar{c} = x(2\hat{i} + \hat{j} + \hat{k}) - 2x(\hat{i} + 2\hat{j} - \hat{k}) = x(2\hat{i} + \hat{j} + \hat{k} - 2\hat{i} - 4\hat{j} + 2\hat{k}) = x(-3\hat{j} + 3\hat{k}) = 3x(-\hat{j} + \hat{k})$.For $x = 1/3$,we get $\bar{c} = -\hat{j} + \hat{k}$. Thus,option $D$ is correct.

Let $\bar{a} = 2\hat{i} + \hat{j} + \hat{k}$,$\bar{b} = \hat{i} + 2\hat{j} - \hat{k}$,and vector $\bar{c}$ be coplanar with $\bar{a}$ and $\bar{b}$. If $\bar{c}$ is perpendicular to $\bar{a}$,then $\bar{c}$ is:

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