# Solving 2 planes with a line?

1. Nov 15, 2004

### Physics is Phun

How would you go about finding cartesian equations of intersecting planes when you have the vector equation of the line they intersect at. THere in obviously an infinate number of solution. The catch is that all the x,y,z values of the planes normals must not be 0.
So I was thinking you but the lines equation into parametric for then from that you can put it into a matrix, then you can somehow "unsolve" it (at the moment in our geometry class we have learned how to solve systems using the matrix but this seems to be the reverse.) One of my friends did this problem that way and it seems to work but I was wondering if there is something else that I should be doing.

thanks

2. Nov 15, 2004

### Physics is Phun

anyone? I can give you the specific example, but I wanted to keep this general as not to put it in the homework section.
here it is. r= (2,-1,0) + t(4,5,1) remember, all the components of the normals are non-zero values.

3. Nov 16, 2004

### Muzza

In the abscense of further requirements... Choose any two non-parallel vectors. Take the cross product of each with a vector that is parallel to the line. The resulting two vectors can be used as plane normals. You can then find the equations of the planes if you know a point on the line.

For the example you gave, you could choose the first vector as (1, 2, 3), giving a plane normal (4,5,1) x (1,2,3) = (13, -11, 3). The equation for a plane with (13,-11,3) as a normal is 13x - 11y + 3z + d = 0, this plane must include the point (2, -1, 0), so we take d = -37.

Take the second vector to be (1, 2, 4), giving the normal (4,5,1) x (1,2,4) = (18, -15, 3). The plane's equation is 18x - 15y + 3z - 51 = 0.

4. Nov 16, 2004

### Physics is Phun

but how do you know the vectors (1,2,3) and (1,2,4) are on the plane?

5. Nov 16, 2004

### NateTG

Let's say you've got some nonzero vector $$<A_1,B_1,C_1>$$ then take some non-parralel, non-zero but otherwise arbitrary vector $$<A_2,B_2,C_2>$$.

Then let
$$\vec{v}_1=<A_1,B_1,C_1>\times<A_2,B_2,C_2>$$
and
$$\vec{v}_2=<A_1,B_1,C_1>\times \vec{v}_1$$

From the properties of the cross product those two vectors are perpendicular to each other, and perpendicular to your original vector. Then if
$$\vec_{v}_1=<x_1,y_1,z_1>$$
the equation
$$?=x_1x+y_1y+z_1z$$
represents a plane parralel to the line since it is normal to $$\vec{v}_1$$(you'll have to solve to get the appropriate "?" and use a similar technique for finding the other plane.

6. Nov 17, 2004

### Muzza

I don't - they're not on the plane, they're just parallel to their respective planes.