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

1 Answer
Narad T.
Nov 8, 2016

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〉

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