Question

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

Answer 1

`(-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)`.