# Intersection of three planes

• Mar 24th 2008, 09:44 AM
Hasan1
Intersection of three planes
Solve each of the following systems of equations. Give a geometrical interpretation of each system and its solution.

1. x - y + 3z = 4
2. x + y + 2z = 2
3. 3x + y + 7z = 9

OK so the first thing I did was check to see if any of the equations were scalar multiples, to see if any of the planes were coincident, which they weren't. Then I checked their normals to see if they were scalar multiples to see if the planes were parallel, they weren't.
Then I took the triple scalar product to see if the normals were coplanar. They were. From this I said the planes either have one line of intersection, or they have no solution, because they have three separate intersections as a triangular prism.

I do not know how to determine which possibility it is, the answer says it has no solution, it forms a triangular prism.

Anybody know?(Yes)
• Mar 24th 2008, 10:31 AM
iknowone
One way to see this is to manually calculate the solution of these three equations. I can use the coefficient matrix to determine if any solutions exist.

$\displaystyle \left( \begin{array}{ccc} 1 & -1 & 3 \\ 1 & 1 & 2 \\ 3 & 1 & 7 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 4 \\ 2 \\ 9 \end{array} \right)$

For this equation to have a solution it be true that the first matrix (call it $\displaystyle A$) has an inverse $\displaystyle A^{-1}$. Since then:

$\displaystyle A^{-1}A\left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array} \right)= A^{-1}\left( \begin{array}{c} 4 \\ 2 \\ 9 \end{array} \right)$

But if $\displaystyle A^{-1}$ does not exist then there is no solution. So when doesn't $\displaystyle A^{-1}$ exist? $\displaystyle A$ having a determinant equal to zero would do it. And if this case the determinant does equal zero.
• Mar 24th 2008, 11:27 AM
mathceleb
You can also use Cramers' Rule for 3 unknowns as well.

3 Unknown Calculator Cramer's Rule

The math work involved is all there if you wish to see how a no solution is developed. If you already know Cramers' Rule, then disregard this message.
• Mar 24th 2008, 12:52 PM
Hasan1
Thanks, but we never learned about determinants or cramer's rule in class, the most we learned about matrices was gaussian and gauss-jhordan elimination, so i don't really understand those solutions. Is there a way to determine if it has no solution manually?
• Mar 24th 2008, 01:04 PM
iknowone
Add the first and second equations. Then add the first and third equations. You will have 2 equations in x and z:

2x+5z = 6 and
4x+10z = 13

This is an impossibility. To see this more clearly observe that equation 2 can be written

2x+5z = 6.5

We know there are no values of x and y that make 2x+5z = 6.5 AND 6. If there were solutions you could use substitution or elimination to solve for x,y, and z.

For example:
x+y-2z = 2
x-y+z = 6
2x-2y-3z=2

we eliminate y from 2 of the equations like we did before (adding 2 equations)

2x-z = 8
4x-7z=6

now multiply the first equation by 2 and subtract the second from it.
This gives:

0+5z = 10 which implies that z=2.

Plug this into either of the above 2 equations:
2x-z = 2x - 2 = 8 which gives x = 5.

Plug z=2 and x=5 into one of the original equations and solve for y:
x+y-2z = 5 + y - 4 = 2, y=1.

So the solution is a single point (5,1,2).