# Find two planes that intersect at a given line.

1. Sep 30, 2011

### ElijahRockers

1. The problem statement, all variables and given/known data
Find two different planes whose intersection is the line $x=-7+t, y=9-t, z=6+2t$. Write equations for each plane in the form $Ax+By+Cz=D$.

2. Relevant equations
The solutions are
x+y=2;
2y+z=24;
but I do not know how to obtain these answers.
(This is not for class work, I am just trying to teach myself.)

3. The attempt at a solution

When t = 0, the point on the line (and also a point on both planes) is P=(-7,9,6).

I understand that any point on the line must also be a point on both planes. What I really need is to find the 'tilt' or normal vector of each plane, though when I visualize it, I can see an infinite number of intersecting planes that could share this line. Am I visualizing it wrong?

Understanding that a vector parallel to the line is $L=<i-j+2k>$ and is also the result of the cross product between the normal vectors of each plane, I tried to work the cross product backwards to obtain the two normal vectors. This didn't seem to get me anywhere.

$n_{1}=<A_{1}i+B_{1}j+C_{1}k>$
$n_{2}=<A_{2}i+B_{2}j+C_{2}k>$
$n_{1} \times n_{2} = <i-j+2k>$

I wound up with three equations and 6 variables.

$B_{1}C_{2}-B_{2}C_{1}=1$
$A_{2}C_{1}-A_{1}C_{2}=-1$
$A_{1}B_{2}-A_{2}B_{1}=2$

Looking back on the process of finding the line when given the two planes, I notice that I set z to 0 so solve for x in terms of y, and did some substitution to eventually find a point on the line, but I am unsure of what the analogous 'backwards' process of this would be.

Any insight would be appreciated.

Last edited: Sep 30, 2011
2. Sep 30, 2011

### SammyS

Staff Emeritus
I think you mean:

x+y=2;

2y+z=24;

3. Sep 30, 2011

### ElijahRockers

Ooops. Thanks. It's quite possible I copied the solutions wrong. That really is besides the point though, whatever the solutions may be, I don't know how to get them.

4. Sep 30, 2011

### HallsofIvy

Do you understant that there are an infinite number of pairs of planes that intersect at that line? There is no "the" solution- there are many solutions of which the given solution is one.

The line is given by x=−7+t,y=9−t,z=6+2t and has "direction vector" <1,-1,2>. Also (-7, 9, 6) is a point on that line. Choose any other vector, u, and take the cross product to get a vector perpendicular to both. Find the equation of the plane having that vector as normal vector and containing point (-7, 9, 6). This plane will contain the given line.

Choose any vector, v, other than u and do the same to get a second plane also containing that line. Now you have two planes whose intersection is that plane.

5. Sep 30, 2011

### ElijahRockers

Yes actually, that's what was throwing me off I guess. The computer generated question was multiple choice, so it kind of led me to believe that there was only one solution, when I knew from visualizing the problem (as I said in my original post) that there should be infinitely many solutions. I have a hard time understanding how they would expect me to arrive at one of their 4 multiple choice answers if I need to just choose any vector u at random. Perhaps they expected me to just find the intersecting lines of each possible answer until I found one that matched the given line.

However, I think I do understand the process now, I will take another look at a similar question to see if I can arrive at one of their solutions.

Last edited: Sep 30, 2011
6. Sep 30, 2011

### SammyS

Staff Emeritus
FYI

The plane given by x+y=2 is parallel to the z-axis.

The plane given by 2y+z=24 is parallel to the x-axis.

7. Sep 30, 2011

### ElijahRockers

Yea, though I'm uncertain if that information is supposed to be able to help me choose the correct pair of planes just by looking at the parametrization of the line. I am so used to doing calculus in 2 dimensions, there is a much bigger playing field that I'm not used to now.