Finding Parametric equations for the line of intersection of two plane

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  • #1
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Homework Statement



Find the parametric equations for the line of intersection of two planes

Homework Equations



Equations for the two planes...
z=x+y,-------(1)
2x-5y-z=1 -----(2)

The Attempt at a Solution



My answers are not correct so I guess I'm going about it the wrong way. Someone please walk me through it from the top.
 
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Answers and Replies

  • #2
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Also, it would be helpful if you solved it with the elimination method because that's the way we were shown in class.
 
  • #3
Dick
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Also, it would be helpful if you solved it with the elimination method because that's the way we were shown in class.
You really need to show your attempt before anyone can help you, please?
 
  • #4
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First thing I did was eliminate x.
Rearranged the first equation to get, x+y-z=0 -------(3)

Multiplied (3) through by 2 to get, 2x+2y-2z=0 -------(4)
Subtracted (2) from (4) to get, 7y-z=-1 or y=(z-1)/7 ----(5)
Let z=t, so I got y=(t-1)/7.

Did the same thing for y, multiplying the first equation through by -5. And I got my final answer to be
x=(1+6t)/(-7) then z=t

The answers are x=6t, y=(-1/6)+t, z=(-1/6)+7t
My process could be completely wrong, which is why I need someone to help me out from the top. I'm just so confused.
 
  • #5
HallsofIvy
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First thing I did was eliminate x.
Rearranged the first equation to get, x+y-z=0 -------(3)

Multiplied (3) through by 2 to get, 2x+2y-2z=0 -------(4)
Subtracted (2) from (4) to get, 7y-z=-1 or y=(z-1)/7 ----(5)
Let z=t, so I got y=(t-1)/7.
Yes, this is correct.

Did the same thing for y, multiplying the first equation through by -5. And I got my final answer to be
x=(1+6t)/(-7) then z=t
This is incorrect. Multiplying x+y- z= 0 by 5 gives 5x+ 5y- 5z= 0. Adding that to 2x- 5y- z= 1 gives 7x- 6z= 1 so that 7x= 1+ 6z, x= (1+ 6z)/7, not "-7".

The answers are x=6t, y=(-1/6)+t, z=(-1/6)+7t
You understand that there can be many different parameters for the same line don't you?
From your "x= (1+ 6t)/7, y= (t- 1)/7, z= t" (that negative sign corrected), we can let x= 6s and have, from the first equation, 6s= (1+6t)/7 so 1+ 6t= 42s, 6t= 42s- 1, t= 7s- 1/6. Then y= (t- 1)/7= (7s- 1/6- 1)/7= (7s- 7/6)/7= s- 1/6. That is x= 6s, y= (-1/6)+ s, z= (-1/6)+ 7s, just what you give, except with "s" instead of "t".

My process could be completely wrong, which is why I need someone to help me out from the top. I'm just so confused.
 
  • #6
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Keep scrolling...
 
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  • #7
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Adding that to 2x- 5y- z= 1 gives 7x- 6z= 1 so that 7x= 1+ 6z, x= (1+ 6z)/7, not "-7".
Okay, Why did you add the two equations? instead of subtract like in the first one. Aren't you supposed to do the same thing to both parts?

Hmmmm....I didn't know that about the final answer. Must have missed it in class. Now let's just clarify the middle part. Thanks and thanks in advance! :-).
 
  • #8
HallsofIvy
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Okay, Why did you add the two equations?
I added 5x+ 5y- 5z= 0 to 2x- 5y- z= 1 in order to eliminate "y"": 5y- 5y= 0.

instead of subtract like in the first one. Aren't you supposed to do the same thing to both parts?
What two parts are you talking about? I did add both parts of the equations: on the left (5x+ 5y- 5z)+ (2x-5y- z)= (5x+ 2x)+ (5y- 5y)+ (-5z- z)= 7x- 6z and on the right 0+ 1= 1
7x- 6z= 1.

Hmmmm....I didn't know that about the final answer. Must have missed it in class. Now let's just clarify the middle part. Thanks and thanks in advance! :-).
 

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