What is the unit vector that is normal to the plane containing 3i+7j-2k and 8i+2j+9k?

1 Answer

Answer:

The unit vector normal to the plane is
#(1/94.01)(67hati-43hatj+50hatk)#.

Explanation:

Let us consider #vecA=3hati+7hatj-2hatk, vecB= 8hati+2hatj+9hatk#
The normal to the plane #vecA, vecB# is nothing but the vector perpendicular i.e., cross product of #vecA, vecB#.
#=>vecAxxvecB= hati(63+4)-hatj(27+16)+hatk(6-56) =67hati-43hatj+50hatk#.
The unit vector normal to the plane is
#+-[vecAxxvecB//(|vecAxxvecB|)]#
So#|vecAxxvecB|=sqrt[(67)^2+(-43)^2+(50)^2]=sqrt8838=94.01~~94#
Now substitute all in above equation, we get unit vector =#+-{[1/(sqrt8838)][67hati-43hatj+50hatk]}#.