# Equation of line intersecting two planes

1. Jun 11, 2015

### BigFlorida

1. The problem statement, all variables and given/known data

Find the cross product equation for the line L that forms the intersection of the planes:

A: 2x - y + 2z = 1 and B: x + y - 2z = 1.

2. Relevant equations

General equations for planes A and B:

(a, b, and r are vectors).

A: a dot r = 1

B: b dot r = 1

r = xi + yj + zk ; i, j, k here are the unit vectors.

r X (b X a) = b - a

alternatively, (b X a) X r = a - b

3. The attempt at a solution

I only know one way in which to solve for the equation of a line intersecting two planes and that is to use the coefficients of the variables of x, y, and z to define two normal vectors, cross those normal vectors to get a parallel line to the intersecting line, then set x, y, or z equal to zero and solve the equations for x and y to define a point. Then, having a vector in the direction of the line (the cross product of the norms) and a point on the line, put it in the form r(t) = (position vector of point) + t(cross product of the norms). I solved the equation for the line in this way, but my teacher wants it as a "cross product equation." I have no clue how to get a and b to stick into the equations for the planes listed above. The only thing I could think of was using the coefficients of x, y, and z; but that makes no sense because that would be dotting a vector with its normal vector and the product would be 0, not 1. Does anyone have any ideas? The parametric equation I got for the line was r(t) = < 2/3, 1/3, 0, 6, 3 >, but this is not in the r X (b X a) form he wants it in. Thanks in advance.

Last edited: Jun 11, 2015
2. Jun 11, 2015

### HallsofIvy

Staff Emeritus
Here is a simpler method: Any point on the intersection of the two planes must lie on both planes. And that means that if we designate the point (x, y, z), x, y, and z must satisfy both equations. Now, if we had three equations, we could, generally, solve for specific values of x, y, and z: three planes intersect in a single point.

With two equations, we can solve for two of the variables in terms of the third, using that third as parameter in parametric equations for a line.

Here, notice that one equation has "-y+ 2z" and the other has "y- 2z". Adding the equations, both y and z cancel, leaving a single equation in x only. Solve that for a specific value for x. The line is parallel to the yz-plane, x is that constant all along the line. Putting that value of x into the equations gives you one equation in y and z. Solve for y in terms of z and use z as parameter or solve for z in terms of y and use y as parameter.

Last edited: Jun 12, 2015
3. Jun 11, 2015

### vela

Staff Emeritus
Actually, your idea is right. Take the first plane, for example. You have $2x-y+2z=1$, which can be written as $(2,-1,2)\cdot(x,y,z) = 1$, which is the form you said you want.

An equation of a plane of the form $\vec{n}\cdot \vec{r} = 0$ corresponds to a plane that passes through the origin. For one that doesn't go through the origin, its equation can be expressed in the form $\vec{n}\cdot (\vec{r} - \vec{r}_0) = 0$, where $\vec{r}_0$ is a point that lies in the plane.

Last edited: Jun 11, 2015
4. Jun 11, 2015

### BigFlorida

@HallsofIvy I did not think of doing that, but I will definitely give it a shot. Thank you!

@vela I never even considered the case of n⃗ ⋅(r⃗ − r⃗ 0) = 0, thank you so much. That has just helped me accomplish what I set out to do!