Given overrightarrow A = 3hat i + hat j + 2ha | vedclass

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(C) The unit vector $\hat n$ perpendicular to both $\overrightarrow A$ and $\overrightarrow B$ is given by $\hat n = \pm \frac{\overrightarrow A 	ime

Vedclass · Accepted Answer

Both $(a)$ and $(b)$

Vedclass · Answer

(C) The unit vector $\hat n$ perpendicular to both $\overrightarrow A$ and $\overrightarrow B$ is given by $\hat n = \pm \frac{\overrightarrow A 	imes \overrightarrow B}{|\overrightarrow A 	imes \overrightarrow B|}$.First,calculate the cross product $\overrightarrow A 	imes \overrightarrow B$:$\overrightarrow A 	imes \overrightarrow B = \begin{vmatrix} \hat i & \hat j & \hat k \ 3 & 1 & 2 \ 2 & -2 & 4 \end{vmatrix} = \hat i(4 - (-4)) - \hat j(12 - 4) + \hat k(-6 - 2) = 8\hat i - 8\hat j - 8\hat k$.Next,calculate the magnitude $|\overrightarrow A 	imes \overrightarrow B|$:$|\overrightarrow A 	imes \overrightarrow B| = \sqrt{8^2 + (-8)^2 + (-8)^2} = \sqrt{64 + 64 + 64} = \sqrt{192} = 8\sqrt{3}$.Finally,the unit vector is $\hat n = \pm \frac{8\hat i - 8\hat j - 8\hat k}{8\sqrt{3}} = \pm \frac{1}{\sqrt{3}}(\hat i - \hat j - \hat k)$.Thus,both options $(a)$ and $(b)$ are correct.

Given $\overrightarrow A = 3\hat i + \hat j + 2\hat k$ and $\overrightarrow B = 2\hat i - 2\hat j + 4\hat k$,find the unit vector perpendicular to both.

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