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

1 Answer
Nov 8, 2016

Answer:

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〉#