How do you find the vector perpendicular to the plane containing (-2,1,0), (-3,0,0), and (5,2,0)?

1 Answer
Aug 1, 2016

Answer:

the Reqd. #bot# Vector#=(0,0,-6)=-6hatk#.

Explanation:

Let us name the given pts. contained in the plane, say #pi#,

#A(-3,0,0), B(-2,1,0) and C(5,2,0)#

Then, clearly, both #vec(AB) and, vec(AC) in pi#.

From Vector Geometry , we remember that the Cross Product

of two vectors is perpendicular to each of them.

Accordingly, #(vec(AB) xx vec(AC)) bot vec(AB) &, vec (AC)#, and,

hence, #(vec(AB) xx vec(AC)) bot pi#.

so, the Reqd. #bot# Vector#=(vec(AB) xx vec(AC))#

#=(1,1,0)xx(8,2,0)=(0,0,-6)=-6hatk#.

It is note-worthy that all the pts. #A, B, and, C# have #0# as their z-co-ordinate, so they all are in #XY#-plane, [i.e., plane #pi# is #XY#-plane], and, so, the reqd. vector #bot# to the plane #pi# has to be along #Z-axis#, i.e., along the vector#hatk#!