Question

What is the unit vector that is normal to the plane containing <1,1,1> and <2,0,-1>?

Answer 1

The unit vector is `=1/sqrt14〈-1,3,-2〉`

Explanation:

You must do the cross product of the two vectors to obtain a vector perpendicular to the plane:
The cross product is the deteminant of
`∣((veci,vecj,veck),(1,1,1),(2,0,-1))∣`

`=veci(-1)-vecj(-1-2)+veck(-2)=〈-1,3,-2〉`

We check by doing the dot products.
`〈-1,3,-2〉.〈1,1,1〉=-1+3-2=0`
`〈-1,3,-2〉.〈2,0,-1〉=-2+0+2=0`
As the dots products are `=0`, we conclude that the vector is perpendicular to the plane.
`∥vecv∥=sqrt(1+9+4)=sqrt14`
The unit vector is `hatv=vecv/(∥vecv∥)=1/sqrt14〈-1,3,-2〉`