## Calculus: Early Transcendentals 8th Edition

$\lt \frac{7}{3 \sqrt 6},\frac{2}{3 \sqrt 6},\frac{-1}{3 \sqrt 6}\gt$, and $\lt \frac{-7}{3 \sqrt 6},\frac{-2}{3 \sqrt 6},\frac{1}{3 \sqrt 6}\gt$
As we know, the cross product of two vectors is orthogonal. 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 $\lt \frac{-7}{3 \sqrt 6},\frac{-2}{3 \sqrt 6},\frac{1}{3 \sqrt 6}\gt$