Find the intersection of three planes (a line)?

[lillybeans]

Homework Statement

The following system of equations represents three planes that intersect in a line.

1. 2x+y+z=4
2. x-y+z=p
3. 4x+qy+z=2

Determine p and q

2. The attempt at a solution

The problem I have with this question is that you are solving 5 variables with only 3 equations. I attempted at this question for a long time, to no avail.

What I did was I tried to convert everything into parametric form, so to make z a parameter (s) and express x and y in terms of s. The steps are as follows:

1. Let z=s

In terms of p and s
2. From ① + ②, we get ④ x=(4+p-2s)/3 ---> "y" eliminated
3. From ① - ②x2, we get ⑤ y=(4-2p+s)/3 ---> "x" eliminated

In terms of q and s
4. From ①x2 - ③, we get ⑥ y=(6-2)/(2-q) ---> "x" eliminated
5. From ①xq - ③, we get ⑦ x=(3q-2+s)/2(q-2) ---> "y" eliminated

I did not do operations with ② and ③, because that will re-introduce the 5 variables (x,y,z,p,q), and after elimination there will still be 4, which is not what we want. We want as little as variables as possible and eliminate as many as possible.

Since they all intersect at the same line, I can make ④=⑦, ⑤=⑥. That's 2 equations. Since we are now down to 3 variables (p,q,s), we need a 3rd equation containing these 3 variables to solve. I was thinking of bringing back ② and ③, but that doesn't work because I will be re-introducing either x or y after elimination.

Please help me and correct my thought process if anywhere along the way I didn't seem to make much sense.

Thank you all so much!

 

[HallsofIvy]

Treat it as three equations in three unknowns, x, y, and z, letting s and q be the parameters. I can't see any reason to think that z= s.

Just go ahead and solve the equations for x, y, and z, leaving s and q as numbers in the solution. Then choose s and q so that there exist a solution (the planes are not parallel) but there is no unique solution (the planes do not all cross at a single point).

For example, it is easy see that we an eliminate y by adding the first two equations:
3x+ 2z= p+ 4. We can also eliminate y by multiplying the second equation by q and adding to the third equation: (q+ 4)x+ (q+ 1)z= pq+ 2.

Now, say, multiply the first equation by q+ 1, the second equation by 2, and subtract to eliminate z. The gives a solution for x depending upon q and s. For what values of q and s can you not get a single solution?

Remember that 0x= a has no solution but that 0x= 0 has many solutions.

 

[lillybeans]

Treat it as three equations in three unknowns, x, y, and z, letting s and q be the parameters. I can't see any reason to think that z= s.

Just go ahead and solve the equations for x, y, and z, leaving s and q as numbers in the solution. Then choose s and q so that there exist a solution (the planes are not parallel) but there is no unique solution (the planes do not all cross at a single point).

For example, it is easy see that we an eliminate y by adding the first two equations:
3x+ 2z= p+ 4. We can also eliminate y by multiplying the second equation by q and adding to the third equation: (q+ 4)x+ (q+ 1)z= pq+ 2.

Now, say, multiply the first equation by q+ 1, the second equation by 2, and subtract to eliminate z. The gives a solution for x depending upon q and s. For what values of q and s can you not get a single solution?

Remember that 0x= a has no solution but that 0x= 0 has many solutions.

Thank you so much for your help! I got it. (q=5)