What is the unit vector that is orthogonal to the plane containing # <3, -6, 2> # and # <1, 1, 1> #?

1 Answer
Jan 7, 2016

Answer:

#(-8/sqrt146,-1/sqrt146,9/sqrt146)#

Explanation:

First take the cross product of both the vectors. Let's say #u = (3,-6,2)# and #v = (1,1,1)#.

The cross product of 2 vectors can be seen as the calculus of 3 determinants, and the vector you will create with the cross product is always orthogonal to the 2 first vectors, that's why we have to begin with it.

#u ^^ v = (1*(-6) - 2*1,2*1 - 3*1, 3*1 -1*(-6)) = (-8,-1,9)#

Now that we have #u ^^ v = w#, we now need to normalize it, it means we have to multiply the vector #w# by #1/(||w||)# in order to make it a unit vector.

#||w|| = sqrt((-8)^2 + (-1)^2 + 9^2) = sqrt(146)#. So the unit vector orthogonal to the plan containing #(3,-6,2)# and #(1,1,1)# is #w/||w|| = (-8/sqrt146,-1/sqrt146,9/sqrt146)#.