# Interection of a plane and a 3D curve

1. Dec 20, 2008

### soandos

is there a way to find out what a given "cross-section" of a curve looks like?
assuming that the curve itself is known and can be easily plotted?
is there a way to define a plane in a way that the points of intersection form the cross section?
thanks.

2. Dec 21, 2008

### HallsofIvy

Strictly, a "curve" is a one-dimensional object and so does not have any "cross section". Do you mean the cross section where a plane intersects a region of space? If so, how is the region given?

3. Dec 21, 2008

### soandos

so i guess a "curve" is the wrong way to describe it.
lets say that i want to find the points of intersection between Sin[x*y]=z with the xy plane.
i.e. find an equation that can be plotted on two axis that will describe such a 1D curve.
is there a way to do this?

4. Dec 21, 2008

### HallsofIvy

Okay, that's the intersection of a surface and a plane. Of course, the xy-plane is defined by z= 0 so just put z= 0 in sin(xy)= z to get sin(xy)= 0. Since sin(u)= 0 is satified for $u= n\pi$ for n any integer, that is $xy= n\pi$ or $y= n\pi/x$, a hyperbola, in the z= 0 plane: The intersection is a series of hyperbolae $(x, n\pi/x, 0)$ for n any integer.

5. Dec 22, 2008

### soandos

and if i wanted a plane in general, not one that is parallel to any axis?

6. Dec 22, 2008

### adriank

For a general plane (ax + by + cz = d), you could try solving for one of the coordinates in the equation for the plane, and inserting it into the equation for your surface.

7. Dec 22, 2008

### soandos

1) if i wanted to define a plane, can i just pick three points on the plane and then solve for a,b,c?
2) would solving the above equation and finding the points of intersection with the surface in the form y= yield the "cross section" that i want?

8. Dec 22, 2008

### adriank

1) Yes you can. There's a nice formula for that as well: given three non-collinear points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) on a plane, a point (x, y, z) lies on the plane iff
$$\begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x & y & z & 1 \end{vmatrix} = 0.$$
That is, the point (x, y, z) is an affine combination of the three given points.

Obviously, a nice choice of your three points would simplify these calculations significantly; for instance:
$$\begin{vmatrix} x_1 & 0 & 0 & 1 \\ 0 & y_2 & 0 & 1 \\ 0 & 0 & z_3 & 1 \\ x & y & z & 1 \end{vmatrix} = 0 \Longleftrightarrow \frac{x}{x_1} + \frac{y}{y_2} + \frac{z}{z_3} = 1,$$
provided that $$x_1 y_2 z_3 \ne 0$$.

2) Yes.

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