# Find all points where surface normal is perpendicular to plane

Homework Statement:
Given function surface:
$$z = x^2 + y^2$$
a. Find tangent plane in point (1, -1)
b. Find all points where the tangentplane of the surface is parallel to the plane:
$$2x + 3y - z = 135$$
Relevant Equations:
Math and stuff.
a. I solved a but I don't fully understand how it works.
$$z = f_x'(1, -1)(x -1) + f_y'(1, -1)(y+1) = 2(x-1) + 3(y+1)$$

Eitherway it's b that's my issue.
I can find the gradient of both plane and surface, but trying to do "dot-product of both normals = 1" will give an equation involving two complex square roots and I rather just not. Because I remember there was an easier way!

I don't completely remember what it was, but my solution would be find two vectors within the surface and the dot-product of each of them with the plane normal must equal 0. But how would I find two vectors in the surface without doing excrusiating many equations? I suppose it has something to do with the answer in a, but I just can't think of anything!

Basically, I know this can be solved in a single equation given by normal vector and the function surface (or its gradient), I just don't remember which. Can someone give me a hand?

Last edited:

Staff Emeritus
Homework Helper
Gold Member
do "dot-product of both normals = 1"
This will work only if you normalize the normal vectors. Your other method of finding two vectors parallel to the surface would also work. It is just a matter of picking two such vectors.

Staff Emeritus
Homework Helper
Gold Member
Homework Statement:: Given function surface:
$$z = x^2 + y^2$$
a. Find tangent plane in point (1, -1)
b. Find all points where the tangentplane of the surface is parallel to the plane:
$$2x + 3y - z = 135$$
Relevant Equations:: Math and stuff.

Basically, I know this can be solved in a single equation given by normal vector and the function surface (or its gradient), I just don't remember which. Can someone give me a hand?
What is the condition on the gradient for the plane to be parallel to the surface?

The gradient has to be perpendicular to any vector in the plane.
Picking two points in the plane and then calculating their vector is a giant waste of time, I am certain there was a better way.

For one, I tried to set the equation of the surface to equal the plane:
$$2x(x-1) + 3y^2(y + 1) - z = 2x + 3y - z - 135$$

But this does not give me the correct answer (which is the points (1,1) and (1, -1))

docnet
Homework Statement:: Given function surface:
$$z = x^2 + y^2$$
a. Find tangent plane in point (1, -1)
b. Find all points where the tangentplane of the surface is parallel to the plane:
$$2x + 3y - z = 135$$
Relevant Equations:: Math and stuff.

a. I solved a but I don't fully understand how it works.
$z = f_x'(1, -1)(x -1) + f_y'(1, -1)(y+1) = 2(x-1) + 3(y+1)$\$

Eitherway it's b that's my issue.
I can find the gradient of both plane and surface, but trying to do "dot-product of both normals = 1" will give an equation involving two complex square roots and I rather just not. Because I remember there was an easier way!

I don't completely remember what it was, but my solution would be find two vectors within the surface and the dot-product of each of them with the plane normal must equal 0. But how would I find two vectors in the surface without doing excrusiating many equations? I suppose it has something to do with the answer in a, but I just can't think of anything!

Basically, I know this can be solved in a single equation given by normal vector and the function surface (or its gradient), I just don't remember which. Can someone give me a hand?
for b, you could try finding the point(s) where the gradient of the plane is equal to the gradient of the surface?