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

1 Answer
SaiGanesh Gotetti
Nov 7, 2016

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]}.

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