Find the unit vector perpendicular to vec{A} | vedclass

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(B) vector perpendicular to both $\vec{A}$ and $\vec{B}$ is given by their cross product: $\vec{C} = \vec{A} 	imes \vec{B}$.$\vec{C} = \begin{vmatrix

Vedclass · Accepted Answer

$\frac{-2\hat{i} + \hat{j} + 4\hat{k}}{\sqrt{21}}$

Vedclass · Answer

(B) vector perpendicular to both $\vec{A}$ and $\vec{B}$ is given by their cross product: $\vec{C} = \vec{A} 	imes \vec{B}$.$\vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 1 \ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0 - 2) - \hat{j}(0 - 1) + \hat{k}(2 - (-2)) = -2\hat{i} + \hat{j} + 4\hat{k}$.The magnitude of $\vec{C}$ is $|\vec{C}| = \sqrt{(-2)^2 + (1)^2 + (4)^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$.The unit vector $\hat{C}$ is given by $\frac{\vec{C}}{|\vec{C}|} = \frac{-2\hat{i} + \hat{j} + 4\hat{k}}{\sqrt{21}}$.

Find the unit vector perpendicular to $\vec{A}$ and $\vec{B}$ where $\vec{A} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + 2\hat{j}$.

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