Question

Find the normal vector to the plane 3x+2y+6z=6.

Answer 1

I think I remember this from Calc 3. From what I recall you take the magnitude of all the values. So
sqrt(9 + 4 + 36) = 7
Then you divide the x, y, and z by the magnitude.

The normal vector would then be 3/7x + 2/7x + 6/7x = 6/7.

I am going to take a look at my old notes and will get back to you to tell you if this was correct or not.

Answer 2

@ajmayberry, the normal vector should have three components. What you suggest doesn't give a vector.

The normal vector is orthogonal to any two vectors in the plane, so taking the cross product of any two vectors in the plane, say \(\vec{a}\) and \(\vec{b}\) will give you an orthogonal vector, which you would then normalize.

Answer 3

Yeah.. I just realized what I wrote isn't making sense. The equation he gave is a PLANE. The normal vector would just be 3i + 2j + 6k

Answer 4

But the answer is (-3, -2, -6). How do I get there?

Answer 5

The vector in the answer is technically the same, just the opposite direction.

Answer 6

In this pic, both \(\vec{n}\) and \(-\vec{n}\) are normal unit vectors.