Finding Parametric equations for the line of intersection of two plane

In summary, the parametric equations for the line of intersection of two planes are x= 6s, y= s- 1/6, z= 7s- 1/6, where s is a parameter. The equations were found by eliminating variables using the elimination method.
  • #1
Jaqsan
17
0

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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  • #2
Also, it would be helpful if you solved it with the elimination method because that's the way we were shown in class.
 
  • #3
Jaqsan said:
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
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
Jaqsan said:
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
Keep scrolling...
 
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  • #7
HallsofIvy said:
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
Jaqsan said:
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! :-).
 

1. What does it mean to find parametric equations for the line of intersection of two planes?

Finding parametric equations for the line of intersection of two planes means determining the equations that represent the points where the two planes intersect in three-dimensional space. These equations are in the form of parameters, which are variables that describe the position of a point on the line relative to a fixed point.

2. Why is it important to find parametric equations for the line of intersection of two planes?

Finding parametric equations for the line of intersection of two planes is important because it allows us to understand the relationship between the two planes and how they intersect in three-dimensional space. This information is essential in various fields such as engineering, physics, and computer graphics.

3. What is the process for finding parametric equations for the line of intersection of two planes?

The process for finding parametric equations for the line of intersection of two planes involves solving the system of equations formed by the two planes. This can be done by using techniques such as substitution or elimination. Once the system is solved, the equations for the line of intersection can be written in parametric form using the parameters determined from the system.

4. Can parametric equations for the line of intersection of two planes have multiple solutions?

Yes, parametric equations for the line of intersection of two planes can have multiple solutions. This can occur when the two planes are parallel or coincident, resulting in an infinite number of solutions. In this case, the parameters in the equations will have different values, but the equations will still represent the same line of intersection.

5. Are there any real-life applications of finding parametric equations for the line of intersection of two planes?

Yes, there are many real-life applications of finding parametric equations for the line of intersection of two planes. For example, in engineering, these equations are used to determine the angle between two intersecting planes, which is crucial in constructing buildings, bridges, and other structures. In computer graphics, parametric equations are used to create 3D models and animations by representing the movement of objects along the line of intersection of two planes.

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