Intersection of two planes

  • #1

Homework Statement


Consider the intersection between the following two planes given in
parametric form:

P1 : x = [2, 4. 3] + s1[1, 2, 1] + s2[2, 5, 4]
P2 : x = [1, 0, -5] + t1[3, 8, 7] + t2[2, 1, -5]

Find the intersection of the two planes as a line in parametric form.



Homework Equations





The Attempt at a Solution


I don't understand how to approach this question at all. At first I thought just to approach it as a system of linear equations and leave one variable as free. But it went nowhere for me.
 

Answers and Replies

  • #2
Dick
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Why don't you show us how you approached it as a system of linear equations and tell us what went wrong. You've got 3 equations, one for each coordinate, and 4 unknowns {t1,t2,s1,s2}. You should be able to eliminate all but one.
 
  • #3
I made P1 and P2 equal and got myself a system of 3 equations with four variables (s1 s2 t1 and t2).
s1 + 2s2 - 3t1 - 2t2 = -1
2s1 + 5s2 - 8t1 - t2 = -4
s1 + 4s2 - 7t1 + 5t2 = -8


Then I put the equation into a matrix and used gaussian elimination to get it to echelon form. t2 was the free variable so I got:

[s1 s2 t1 t2] = t2[8 -17 -8 1] + [8 -12 -5 0]

After that, I'm not sure what to do.
But I think my method is totally wrong. I think the solution isn't as complicated as I made it.
 
  • #4
Dick
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What you should be able to do now is find expressions for s1, s2 and t1 in terms of t2 and substitute into the original line equations. They both should give you the same line. But I think there's a mistake in what you did. I tried it and I can't use t2 as the free parameter. One of my equations turns out to be t2=(-3). I needed to use t1.
 
  • #5
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0
i got up until solving for t2 = -3 but i have no idea where to go from there :S
what do you guys mean by free parameter
 
  • #6
Dick
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i got up until solving for t2 = -3 but i have no idea where to go from there :S
what do you guys mean by free parameter
A free parameter is one that your system of equations don't determine, unlike t2. You can solve for s1 and s2 in terms of t1. That makes t1 a 'free parameter'.
 
  • #7
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this is how i got t2 = -3

1 2 -3 -2 -1
0 1 -2 3 -2
0 0 0 1 -3

that gave me t2 = -3

when i plug that back into the equations and get another augmented matrix it doesnt work...am i doing it wrong
 
  • #8
Dick
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this is how i got t2 = -3

1 2 -3 -2 -1
0 1 -2 3 -2
0 0 0 1 -3

that gave me t2 = -3

when i plug that back into the equations and get another augmented matrix it doesnt work...am i doing it wrong
Why are you putting it into another augmented matrix? You should be basically done. Now write out the other equations and back substitute t2=(-3) to find s1 and s2 in terms of t1.
 
  • #9
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so I would need to isolate t1 in each equation after plugging the t2 value in?
 
  • #10
Dick
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so I would need to isolate t1 in each equation after plugging the t2 value in?
Substitute t2=(-3) into both of the equations above it. Solve the second equation for s2 in terms of t1. Put that into into the first equation. Solve that for s1 in terms of t1 as well. Isn't that the usual sort of thing you do once you've got a reduced echelon form? You could also try to get some more zeros in the upper rows using the lower rows first. But it's still the same procedure.
 
  • #11
12
0
sorry i meant s1 & s2
i did that and i get
s2 = 7 + 2t1
s1 = -21 - t1
 
  • #12
Dick
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sorry i meant s1 & s2
i did that and i get
s2 = 7 + 2t1
s1 = -21 - t1
That's funny. It's the same thing I get.
 

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