A vector in the direction of v = 2hat{i} + 3h | vedclass

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(C) Given vector $v = 2\hat{i} + 3\hat{j} + \hat{k}$.First,calculate the magnitude of vector $v$:$|v| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \s

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$\frac{2}{\sqrt{2}}\hat{i} + \frac{3}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$

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(C) Given vector $v = 2\hat{i} + 3\hat{j} + \hat{k}$.First,calculate the magnitude of vector $v$:$|v| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.The unit vector in the direction of $v$ is $\hat{v} = \frac{v}{|v|} = \frac{2\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{14}}$.$A$ vector with magnitude $\sqrt{7}$ in the direction of $v$ is given by $\sqrt{7} 	imes \hat{v}$:$= \sqrt{7} 	imes \left( \frac{2\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{14}} ight) = \frac{\sqrt{7}}{\sqrt{14}}(2\hat{i} + 3\hat{j} + \hat{k}) = \frac{1}{\sqrt{2}}(2\hat{i} + 3\hat{j} + \hat{k}) = \frac{2}{\sqrt{2}}\hat{i} + \frac{3}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$.Thus,the correct option is $C$.

$A$ vector in the direction of $v = 2\hat{i} + 3\hat{j} + \hat{k}$ with magnitude $\sqrt{7}$ is

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