If u = 2i + 2j - k and v = 6i - 3j + 2k,then | vedclass

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(B) Let $c$ be a vector perpendicular to both $u$ and $v$. Then $c = u 	imes v$.$c = \begin{vmatrix} i & j & k \ 2 & 2 & -1 \ 6 & -3 & 2 \end{vmatr

Vedclass · Accepted Answer

$\frac{1}{\sqrt{17}} \left( \frac{1}{5}i - 2j - \frac{18}{5}k \right)$

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(B) Let $c$ be a vector perpendicular to both $u$ and $v$. Then $c = u 	imes v$.$c = \begin{vmatrix} i & j & k \ 2 & 2 & -1 \ 6 & -3 & 2 \end{vmatrix}$$c = i(4 - 3) - j(4 - (-6)) + k(-6 - 12)$$c = i(1) - j(10) + k(-18) = i - 10j - 18k$.Now,the magnitude of $c$ is $|c| = \sqrt{1^2 + (-10)^2 + (-18)^2} = \sqrt{1 + 100 + 324} = \sqrt{425} = \sqrt{25 	imes 17} = 5\sqrt{17}$.The unit vector perpendicular to both $u$ and $v$ is $\pm \frac{c}{|c|}$.$\frac{c}{|c|} = \frac{i - 10j - 18k}{5\sqrt{17}} = \frac{1}{\sqrt{17}} \left( \frac{1}{5}i - 2j - \frac{18}{5}k ight)$.Thus,option $B$ is correct.

If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k$,then find the unit vector perpendicular to both $u$ and $v$.

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