How do you find a unit vector normal to the plane which passes through the following points (1,1,1),(2,0,2),(1,1,2)?

1 Answer
Aug 29, 2016

Reqd. unit normal is #(-1/sqrt2,-1/sqrt2,0), or, (1/sqrt2,1/sqrt2,0)#.

Explanation:

Let #A(1,1,1), B(2,0,2), and, C(1,1,2)# be the pts. on the plane, say #pi#.

We need a unit normal, call it #hat n#, to #pi#.

#vec(AB)=(2-1,0-1,2-1)=(1,-1,1) in pi#, and,

#vec(AC)=(1-1,1-1,2-1)=(0,0,1) in pi#.

#:.# Fom Vector Geometry, #vec(AB) xx vec(AC)# is a vector #bot#

to #vec(AB) and vec(AC)#, hence, is the normal to the plane #pi#.

Now, #vec(AB)xxvec(AC)=det|(hati,hatj,hatk),(1,-1,1),(0,0,1)|#

#=-1hati-1hatj+0hatk=(-1,-1,0)#, so that,

#||vec(AB)xxvec(AC)||=sqrt2#.

Therefore,

#hatn=[vec(AB)xxvec(AC)]/||vec(AB)xxvec(AC)||#, i.e.,

#hat n=(-1/sqrt2,-1/sqrt2,0), or, (1/sqrt2,1/sqrt2,0)#

Enjoy maths.!