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

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(B) Given vectors are $\bar{a} = \bar{i} + 2\bar{j} + 2\bar{k}$ and $\bar{b} = 2\bar{i} - \bar{j} + p\bar{k}$.Magnitude of $\bar{a}$ is $|\bar{a}| = \

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$\frac{3\sqrt{5}}{\sqrt{7}}$

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(B) Given vectors are $\bar{a} = \bar{i} + 2\bar{j} + 2\bar{k}$ and $\bar{b} = 2\bar{i} - \bar{j} + p\bar{k}$.Magnitude of $\bar{a}$ is $|\bar{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.Magnitude of $\bar{b}$ is $|\bar{b}| = \sqrt{2^2 + (-1)^2 + p^2} = \sqrt{4 + 1 + p^2} = \sqrt{5 + p^2}$.The dot product is $\bar{a} \cdot \bar{b} = (1)(2) + (2)(-1) + (2)(p) = 2 - 2 + 2p = 2p$.We know that $\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos(	heta)$, where $	heta = 60^{\circ}$.So, $2p = 3 	imes \sqrt{5 + p^2} 	imes \cos(60^{\circ})$.Since $\cos(60^{\circ}) = \frac{1}{2}$, we have $2p = 3 	imes \sqrt{5 + p^2} 	imes \frac{1}{2}$.$4p = 3\sqrt{5 + p^2}$.Squaring both sides, $16p^2 = 9(5 + p^2) = 45 + 9p^2$.$7p^2 = 45 \implies p^2 = \frac{45}{7} \implies p = \sqrt{\frac{45}{7}} = \frac{3\sqrt{5}}{\sqrt{7}}$.Thus, the correct option is $B$.

Let $\bar{a} = \bar{i} + 2\bar{j} + 2\bar{k}$ and $\bar{b} = 2\bar{i} - \bar{j} + p\bar{k}$ be two vectors. If the angle between $\bar{a}$ and $\bar{b}$ is $60^{\circ}$, then $p =$

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