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

1 Answer
Topscooter
Jan 7, 2016

(-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).

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