# Finding the equation of a plane given 2 lines and a point

1. Feb 19, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
Find the equation for the plane containing the lines:
x-y+2z=4
2x+y+3z=6
and the point: (1,-2,4)

2. Relevant equations
A(x-xi)+B(y-yi)+C(z-zi)=0

3. The attempt at a solution
First I wrote down the vectors from the line equations,
V1=<1,-1,2>
V2=<2,1,3>

After this I did the cross product to get the perpendicular vector and I ended up with
Vp=<-5,1,3>
now using the equation and the point given at the start
-5(x-1)+1(y+2)+3(z-4)=0
-5x+y+3z-5=0

2. Feb 19, 2015

### Orodruin

Staff Emeritus
These are planes, not lines. Coulld you check the problem formulation?

3. Feb 19, 2015

### Potatochip911

I rechecked it and it says "containing the line x-y+2z=4 and 2x+y+3z=6 and the point (1,-2,4)" so apparently they're both the same line?

4. Feb 19, 2015

### Orodruin

Staff Emeritus
They are both planes. However, planes cross in a line (or they are parallel, which is not the case here). Thus, there is only one line in the problem. The vectors V1 and V2 are normal to the two planes and so they are both orthogonal to the line. Taking the cross product gives you a vector which is therefore the tangent vector of the line and using that as a normal vector of your plane gives you a plane which is orthogonal to both of the original planes, i.e., it is orthogonal to your line and does not contain it.

I suggest you try to figure out how you can end up with a vector which is orthogonal to both the line and to the displacement between any point on the line and the point (1,-2,4).

You can check whether or not your final equation is correct by inserting some points from the line into your formula and see if it is fulfilled or not.

5. Feb 19, 2015

### Potatochip911

It seems as though I have to solve for the parametric equations, is it not possible to do this with the cross product?

6. Feb 19, 2015

### HallsofIvy

You are determined to use the cross product, aren't you? You will but not to find the equation of the line. You are given the two planes x-y+2z=4 and 2x+y+3z=6. You have two equations in three unknowns. You can solve for any two of them in terms of the third. Use that third unknown (or some multiple of it) as parameter to get parametric equations for the line of intersection of the two planes. Take the vector in the direction of that line and the vector from some point on that line to the initially given line and find the cross product of them.

7. Feb 19, 2015

### Potatochip911

Okay I managed to get the answer but now I am just really confused by the theory of all this. So because the point given to me (1,-2,4) is not on either of the planes I need to solve for a point that is on both planes and then find the vector that will move my point onto the plane?

8. Feb 20, 2015

### HallsofIvy

I'm not sure what you mean by "move my point onto the plane". There is no "moving" in this problem. Two planes intersect in a line. There exist a unique plane containing any given line and point not on the line. Equivalently, you could choose any two points that lie on the line. A plane will contain a line if and only if it contains two points on the line. Those two points, together with the given point give you three points on the plane. There exist a unique plane containing any three non-collinear points.