- #1
lillybeans
- 68
- 1
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!