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Find a unit vector perpendicular to A = i + j -  k  and B = 2 i + j - 3 k.

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Unit vector perpendicular to \(\vec A\) and \(\vec B\) is \(\frac{\vec A\times\vec B}{|\vec A\times\vec B|}.\) 

Now, \(\vec A\times\vec B=\begin{bmatrix}\hat i&\hat j&\hat k\\ 1&1&-1\\2&1&-3\end{bmatrix}\) = \(\hat i\) (-3 + 1) - \(\hat j\) (-3 + 2) + \(\hat k\) (1 - 2)

= -2\(\hat i\) + \(\hat j\) - \(\hat k\) .

\(|\vec A\times\vec B|=\sqrt{(-2)^2+1^2+(-1)^2}\) = \(\sqrt{4 + 1 +1}=\sqrt6\)

\(\therefore\) unit vector = \(\frac{\vec A\times\vec B}{|\vec A\times\vec B|}\) = \(\frac{1}{\sqrt6}(2\hat i+\hat j-\hat k)\).

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