How do you find all unit vectors normal to the plane which contains the points #(0,1,1),(1,−1,0)#, and #(1,0,2)#?

1 Answer
Jul 4, 2016

#hat n = {-3/sqrt[14], -sqrt[2/7], 1/sqrt[14]}#

Explanation:

Given three non aligned points there is an unique plane which contains them.

#p_1={0,1,1}#
#p_2={1,-1,0}#
#p_3 ={1,0,2}#

#p_1,p_2,p_2# define two segments

#p_2-p_1# and #p_3-p_1# parallel to the plane which contains #p_1,p_2,p_3#.

The normal to them is also the normal to the plane so

#hat n = ((p_2-p_1) xx (p_3-p_1))/abs( (p_2-p_1) xx (p_3-p_1)) = {-3/sqrt[14], -sqrt[2/7], 1/sqrt[14]}#