If bar{a} = bar{i} + sqrt{11} bar{j} - 2 bar{ | vedclass

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(C) The component of vector $\bar{b}$ perpendicular to $\bar{a}$ is given by $\bar{b} - 	ext{proj}_{\bar{a}} \bar{b} = \bar{b} - \left( \frac{\bar{b}

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$-(\bar{i} + \sqrt{11} \bar{j} + 6 \bar{k})$

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(C) The component of vector $\bar{b}$ perpendicular to $\bar{a}$ is given by $\bar{b} - 	ext{proj}_{\bar{a}} \bar{b} = \bar{b} - \left( \frac{\bar{b} \cdot \bar{a}}{|\bar{a}|^2} ight) \bar{a}$.First, calculate the dot product $\bar{b} \cdot \bar{a} = (1)(1) + (\sqrt{11})(\sqrt{11}) + (-10)(-2) = 1 + 11 + 20 = 32$.Next, calculate the magnitude squared $|\bar{a}|^2 = (1)^2 + (\sqrt{11})^2 + (-2)^2 = 1 + 11 + 4 = 16$.Now, find the projection: $\frac{\bar{b} \cdot \bar{a}}{|\bar{a}|^2} \bar{a} = \frac{32}{16} \bar{a} = 2 \bar{a} = 2(\bar{i} + \sqrt{11} \bar{j} - 2 \bar{k}) = 2 \bar{i} + 2\sqrt{11} \bar{j} - 4 \bar{k}$.Finally, the perpendicular component is $\bar{b} - 2 \bar{a} = (\bar{i} + \sqrt{11} \bar{j} - 10 \bar{k}) - (2 \bar{i} + 2\sqrt{11} \bar{j} - 4 \bar{k}) = (1-2) \bar{i} + (\sqrt{11} - 2\sqrt{11}) \bar{j} + (-10 + 4) \bar{k} = -\bar{i} - \sqrt{11} \bar{j} - 6 \bar{k} = -(\bar{i} + \sqrt{11} \bar{j} + 6 \bar{k})$.Thus, the correct option is $C$.

If $\bar{a} = \bar{i} + \sqrt{11} \bar{j} - 2 \bar{k}$ and $\bar{b} = \bar{i} + \sqrt{11} \bar{j} - 10 \bar{k}$ are two vectors, then the component of $\bar{b}$ perpendicular to $\bar{a}$ is:

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