Find the unit vector perpendicular to the two | vedclass

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(B) Let $\vec{A} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + \hat{j} + \hat{k}$.To find a vector perpendicular to both $\vec{A}$ and $\vec

Vedclass · Accepted Answer

$\frac{-\hat{i} + \hat{k}}{\sqrt{2}}$

Vedclass · Answer

(B) Let $\vec{A} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + \hat{j} + \hat{k}$.To find a vector perpendicular to both $\vec{A}$ and $\vec{B}$,we calculate their cross product $\vec{C} = \vec{A} 	imes \vec{B}$.$\vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 1) - \hat{j}(1 - 1) + \hat{k}(1 - (-1)) = -2\hat{i} + 2\hat{k}$.The magnitude of $\vec{C}$ is $|\vec{C}| = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.The unit vector $\hat{n}$ perpendicular to both is $\hat{n} = \pm \frac{\vec{C}}{|\vec{C}|} = \pm \frac{-2\hat{i} + 2\hat{k}}{2\sqrt{2}} = \pm \frac{-\hat{i} + \hat{k}}{\sqrt{2}}$.Comparing this with the given options,the correct choice is $\frac{-\hat{i} + \hat{k}}{\sqrt{2}}$.

Find the unit vector perpendicular to the two vectors $\vec{A} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + \hat{j} + \hat{k}$.

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