Find all points where surface normal is perpendicular to plane

In summary: This would give you a single equation to solve for x and y, which should be easier than finding two vectors in the surface.
  • #1
Addez123
199
21
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?
 
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  • #2
Addez123 said:
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.
 
  • #3
Addez123 said:
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?
 
  • #4
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))
 
  • #5
Addez123 said:
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?
 

1. What is the definition of a surface normal?

A surface normal is an imaginary line that is perpendicular to a surface at a specific point. It is used to determine the orientation of a surface and is often represented by a vector.

2. How can I find the surface normal at a specific point on a plane?

To find the surface normal at a specific point on a plane, you will need to know the equation of the plane and the coordinates of the point. Using the equation of the plane, you can find the normal vector and then normalize it to get the surface normal.

3. What does it mean for a surface normal to be perpendicular to a plane?

When a surface normal is perpendicular to a plane, it means that the normal vector is at a 90-degree angle to the plane. This indicates that the surface is perfectly flat at that point, with no slope or curvature.

4. Why is it important to find all points where the surface normal is perpendicular to a plane?

Knowing where the surface normal is perpendicular to a plane can be useful in various applications, such as computer graphics, engineering, and physics. It can help determine the orientation and behavior of a surface and aid in calculations and simulations.

5. Are there any other ways to find the surface normal besides using the equation of the plane?

Yes, there are other methods to find the surface normal, such as using partial derivatives or cross products. However, using the equation of the plane is often the most straightforward and efficient way to find the surface normal at a specific point.

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