# Thread: Intersection of three planes

1. ## 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?

2. 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.

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

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

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

But if $A^{-1}$ does not exist then there is no solution. So when doesn't $A^{-1}$ exist? $A$ having a determinant equal to zero would do it. And if this case the determinant does equal zero.

3. 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.

4. 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?

5. 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).