Answer
$ \lt \frac{7}{3 \sqrt 6},\frac{2}{3 \sqrt 6},\frac{-1}{3 \sqrt 6}\gt$, and $ \lt \frac{-7}{3 \sqrt 6},\frac{-2}{3 \sqrt 6},\frac{1}{3 \sqrt 6}\gt$
Work Step by Step
As we know the cross product of two vectors is orthogonal.
Then we will have to calculate the cross product of the two vectors.
Let $a=j+2k$ $\implies$ $a= \lt 0,1,1 \gt$
$b=i-2j+3k$ $\implies$ $b= \lt 1,-2,3 \gt$
$a \times b= \lt 0,1,1 \gt \times \lt 1,-2,3 \gt = \lt 7,2,-1 \gt$
$|a \times b| =\sqrt {49+4+1}= 3 \sqrt 6$
The unit vector is given by: $ \lt \frac{7}{3 \sqrt 6},\frac{2}{3 \sqrt 6},\frac{-1}{3 \sqrt 6}\gt$
The second vectors can be found by reversing the direction of the first.
Thus, $ \lt \frac{-7}{3 \sqrt 6},\frac{-2}{3 \sqrt 6},\frac{1}{3 \sqrt 6}\gt$
Hence, $ \lt \frac{7}{3 \sqrt 6},\frac{2}{3 \sqrt 6},\frac{-1}{3 \sqrt 6}\gt$, and $ \lt \frac{-7}{3 \sqrt 6},\frac{-2}{3 \sqrt 6},\frac{1}{3 \sqrt 6}\gt$
