# Orthogonal Projection onto XY plane

1. Mar 12, 2009

### brendan

I have an assignment question to find an equation of the orthogonal projection onto the XY plane of the curve of intersection of twp particular functions.
If some one knows of a good web page that might explain this to me I would be greatly appreciate it.

regards
Brendan

2. Mar 12, 2009

### HallsofIvy

Staff Emeritus
If you can write the equation of the curve of intersection as parametric equations, just set z= 0. That is, if the curve is given by x= f(t), y= g(t), z= h(t), its projection onto the xy-plane is x= f(t), y= g(t), z= 0.

3. Mar 12, 2009

### brendan

The two functions that intersect are:

z = y^2 and z = 4-X^2 -y^2

z = 4-X^2 -y^2

As parametric

x = 4-cos(t) y = -sin(t)

z = y^2

As parametric

z = sin(t)

Do I have to combine the two Equations?

4. Mar 13, 2009

### HallsofIvy

Staff Emeritus
Those are NOT parametric equations for the two surfaces. Since they involve only a single parameter they are equations of curves, not surfaces.

The two surfaces are $z= y^2$ and $z= 4- x^2- y^2$. In this particular case, you don't need parametric equations at all. You can just eliminate z from the two equations to get the projection onto the xy plane.

5. Mar 13, 2009

### WWGD

Just to understand Ivy's answer better, check that the vector in the projection

is actually orthogonal to the xy-plane: describe the xy-plane in "coordinates"

(i.e., in the same way as describing the x-axis as the set of points (x,0) ) , and then

find the coordinates of your vector : coordinates (x,y,z) of intersection point -

coordinates (x,y,z) of projection . You can use dot product to show that the

projection is perpendicular : xy-plane is perp. to the vector going from :

intersection point to image of orth. projection. (It would help if I could draw pictures.)

6. Mar 13, 2009

### brendan

The two surfaces are z = y^2 and z = 4-X^2 -y^2

surface 1 (xy plane) = (0,y,0)
surface 2 (xy plane)= (x,y,0)

They both intersect at (0,y,0)

is Y the projection?

7. Mar 13, 2009

### brendan

Sorry, I'm getting confused.

The two surfaces are z = y^2 and z = 4-X^2 -y^2

4-x^2 - y^2 =z
y^2 = z

Subtract one equation from the other gives

4-x^2 -2y^2 = 0

Is that the orthogonal projection?

8. Mar 13, 2009

### owlpride

Yes, that is correct.

In your previous post you projected the surfaces first (you should have gotten (x,y,0) for both surfaces btw) and then intersected their projection. That doesn't work. You have to intersect first and then project.

That's exactly what you did in your latest post. Usually you would have to set z = 0 to get the projection. However, since we already have a description of just the x- and y-coordinates of the intersection that does not depend on z, we can skip this step.

By the way, what does the answer look like geometrically?

9. Mar 13, 2009

### brendan

It looks like a convex Elliptic paraboloid

10. Mar 14, 2009

### HallsofIvy

Staff Emeritus
No, the projection of the curve of intersection of two surfaces into the xy-plane is a curve in the xy-plane, not a surface.

Your first surface, $z= y^2$ is a "parabolic cylinder", your second surface is, $z= 4- x^2- y^2$ is an "elliptic paraboloid" (or, more correctly, a "circular paraboloid". The intersection of the two is given by $y^2= 4- x^2- y^2$ or $x^2+ 2y^2= 4$, an ellipse.