The unit vector perpendicular to the vectors | vedclass

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(C) Let the given vectors be $\vec{a} = i - j + k$ and $\vec{b} = 2i + 3j - k$.The vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by th

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$\frac{-2i+3j+5k}{\sqrt{38}}$

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(C) Let the given vectors be $\vec{a} = i - j + k$ and $\vec{b} = 2i + 3j - k$.The vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by the cross product $\vec{a} 	imes \vec{b}$.$\vec{a} 	imes \vec{b} = \begin{vmatrix} i & j & k \ 1 & -1 & 1 \ 2 & 3 & -1 \end{vmatrix}$$= i((-1)(-1) - (1)(3)) - j((1)(-1) - (1)(2)) + k((1)(3) - (-1)(2))$$= i(1 - 3) - j(-1 - 2) + k(3 + 2)$$= -2i + 3j + 5k$The magnitude of the cross product is $|\vec{a} 	imes \vec{b}| = \sqrt{(-2)^2 + (3)^2 + (5)^2} = \sqrt{4 + 9 + 25} = \sqrt{38}$.The unit vector perpendicular to both vectors is $\frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|} = \frac{-2i + 3j + 5k}{\sqrt{38}}$.

The unit vector perpendicular to the vectors $i - j + k$ and $2i + 3j - k$ is

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