Find symmetric equations for the line of intersection of the planes

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Homework Statement



Find symmetric equations for the line of intersection of the planes
The planes:
5x - 2y - 2z = 1
4x + y + z = 6

Homework Equations



r = r0 + tv
x = x0 + at
y = y0 + bt
z = z0 + ct

The Attempt at a Solution



I have attempted this in many different manners and would like to find a way to the solution. The answer is:

x = 1, y - 2 = - z

The book doesn't have a good example for this problem, and as classes haven't started for me, I have no notes from lectures (i need to learn this for another class...taking 2 classes in which 1 is a prereq for the other). Any help would be appreciated.
 
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  • #2
gabbagabbahey
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Homework Statement



The planes:
5x - 2y - 2z = 1
4x + y + z = 6
I know you put what you were looking for in your thread title, but you really should also include the full problem statement in the first section of your post in future threads. Just writing two equations, does not qualify as a complete problem statement.

I have attempted this in many different manners and would like to find a way to the solution. The answer is:

x = 1, y - 2 = - z
I'm not sure exactly what makes the equations in the given answer symmetric (as implied by your thread title), but realize that any point [itex](x,y,z)[/itex] that lies on the line of intersection of two planes will simultaneously satisfy the equations of both planes. How do you normally solve a system of simultaneous equations?
 
  • #3
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I know you put what you were looking for in your thread title, but you really should also include the full problem statement in the first section of your post in future threads. Just writing two equations, does not qualify as a complete problem statement.
I fixed it, thank-you.

I'm not sure exactly what makes the equations in the given answer symmetric (as implied by your thread title), but realize that any point [itex](x,y,z)[/itex] that lies on the line of intersection of two planes will simultaneously satisfy the equations of both planes. How do you normally solve a system of simultaneous equations?
We're apparently supposed to be using the cross product of the two normal vectors of the planes (which gives the same vector as the intersection line, just parallel). It doesn't give me a point, and I can't figure out how they got that specific answer.

Would that be the only answer? Or is there more than one answer (i spent 2 hours on this problem today after spending only 20-30 minutes on the other 20 problems in the lesson).
 
  • #4
gabbagabbahey
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We're apparently supposed to be using the cross product of the two normal vectors of the planes
Why do you say that? Does the full problem statement tell you to use that method?

I can't figure out how they got that specific answer.
If I gave you the system of equations [itex]x+2y = 1[/itex] and [itex]2x+6y = 4[/itex], could you solve it for [itex]x[/itex] and [itex]y[/itex]?
 
  • #5
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Why do you say that? Does the full problem statement tell you to use that method?
It's what the chapter & lesson is about.


If I gave you the system of equations [itex]x+2y = 1[/itex] and [itex]2x+6y = 4[/itex], could you solve it for [itex]x[/itex] and [itex]y[/itex]?
x = -1
y = 1
 
  • #6
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5x - 2y - 2z = 1
4x + y + z = 6

So, for z = 0,

5x - 2y = 1
4x + y = 6

5x - 2y = 1
+ 8x + 2y = 12

13x = 13

x = 1

4 + y + z = 6
y + z = 2
y - 2 = -z

Thanks =D So just set up a simple system of equations? I thought I tried that -.- I must have set it up wrong. Thank-you though!
 
  • #7
gabbagabbahey
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5x - 2y - 2z = 1
4x + y + z = 6

So, for z = 0,

5x - 2y = 1
4x + y = 6
Careful, when you do this you are assuming that z=0 is on the line of intersection (it is in this case, but you have no reason to assume it beforehand). Instead, just add 2 times the second equation to the first equation if you want to get rid of [itex]z[/itex].
 
  • #8
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I assumed it because of how y and z cancel each other out. Plus i kinda knew that x held constant and y & z covers all integers =P (oops)
 

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