If a=hat{i}+hat{j}+hat{k},b=hat{i}+hat{j}+2ha | vedclass

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(C) Given vectors are $a=\hat{i}+\hat{j}+\hat{k}$,$b=\hat{i}+\hat{j}+2\hat{k}$ and $c=2\hat{i}+3\hat{j}+4\hat{k}$.First,we find the vector perpendicul

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$\frac{1}{\sqrt{58}}$

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(C) Given vectors are $a=\hat{i}+\hat{j}+\hat{k}$,$b=\hat{i}+\hat{j}+2\hat{k}$ and $c=2\hat{i}+3\hat{j}+4\hat{k}$.First,we find the vector perpendicular to both $a$ and $b$ using the cross product:$a 	imes b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & 1 & 2 \end{vmatrix} = \hat{i}(2-1) - \hat{j}(2-1) + \hat{k}(1-1) = \hat{i} - \hat{j}$.The unit vector perpendicular to both $a$ and $b$ is given by $n = \pm \frac{\hat{i} - \hat{j}}{|\hat{i} - \hat{j}|} = \pm \frac{\hat{i} - \hat{j}}{\sqrt{1^2 + (-1)^2}} = \pm \frac{\hat{i} - \hat{j}}{\sqrt{2}}$.The magnitude of the projection of vector $n$ on vector $c$ is given by $|n \cdot \hat{c}|$,where $\hat{c} = \frac{c}{|c|}$.$|c| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.Projection magnitude $= \left| \left( \pm \frac{\hat{i} - \hat{j}}{\sqrt{2}} ight) \cdot \left( \frac{2\hat{i} + 3\hat{j} + 4\hat{k}}{\sqrt{29}} ight) ight| = \left| \frac{\pm(2 - 3)}{\sqrt{2}\sqrt{29}} ight| = \left| \frac{-1}{\sqrt{58}} ight| = \frac{1}{\sqrt{58}}$.Thus,the correct option is $C$.

If $a=\hat{i}+\hat{j}+\hat{k}$,$b=\hat{i}+\hat{j}+2\hat{k}$ and $c=2\hat{i}+3\hat{j}+4\hat{k}$,then the magnitude of the projection on $c$ of a unit vector that is perpendicular to both $a$ and $b$ is

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