Find all points where surface normal is perpendicular to plane

[Addez123]

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?

 

[Orodruin]

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.

 

[Orodruin]

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?

 

[Addez123]

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?