A vector with magnitude of 3 units,which is p | vedclass

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(A) First,find the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & -4 \ 6 & 5

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$\pm(2 \hat{i}-2 \hat{j}+\hat{k})$

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(A) First,find the cross product $\vec{a} 	imes \vec{b}$:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & -4 \ 6 & 5 & -2 \end{vmatrix} = \hat{i}(-2 - (-20)) - \hat{j}(-6 - (-24)) + \hat{k}(15 - 6) = 18 \hat{i} - 18 \hat{j} + 9 \hat{k}$Next,find the magnitude of the cross product:$|\vec{a} 	imes \vec{b}| = \sqrt{18^2 + (-18)^2 + 9^2} = \sqrt{324 + 324 + 81} = \sqrt{729} = 27$The unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is $\frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|} = \frac{18 \hat{i} - 18 \hat{j} + 9 \hat{k}}{27} = \frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k}$The required vector with magnitude $3$ is $\pm 3 \left( \frac{2}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{1}{3} \hat{k} ight) = \pm(2 \hat{i} - 2 \hat{j} + \hat{k})$.

$A$ vector with magnitude of $3$ units,which is perpendicular to each of the vectors $\vec{a}=3 \hat{i}+\hat{j}-4 \hat{k}$ and $\vec{b}=6 \hat{i}+5 \hat{j}-2 \hat{k}$,is given by

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