Parametric equation of an intersection.

[-Dragoon-]

Homework Statement

Find the parametric equations of the intersection line of two planes 2x - 3y - z + 1 = 0 and
3x - 2y + 3z - 4 = 0

Homework Equations

N/A

The Attempt at a Solution

First I'll label them:
2x - 3y - z + 1 = 0 [1]
3x - 2y + 3z - 4 = 0 [2]
Then I get rid of the z variable for now, and multiply [1] by 3 to do that, then eliminate by adding:
6x - 9y - 3z +3 = 0
3x - 2y + 3z - 4 = 0
_________________
9x - 11y - 1 = 0 [3]
Then I write y in terms of x:
y = (9/11)x - 1/11[4]
Then substitute [4] back into [1]:
2x - 3((9/11)x - 1/11) - z +1 = 0
2x - (27/11)x +3/11 - z + 1 = 0
Then write z in terms of x:
z = (-5/11)x + 14/11
Finally, I set x = t to write the parametric equations:
x = t
y = (9/11)t - 1/11
z = (-5/11)t + 14/11
However, this was the answer my book got:
x = (11/9)t + 1/9
y = t
z = (-5/11)t + 11/9

Can anyone help me figure out what I did wrong? I double checked all the tedious calculations, and they seem correct to me. Thanks in advance.

 

[rock.freak667]

I believe you as well are correct, they just parameterized y and you did it with x.

Alternatively you could use 9x - 11y - 1 = 0 to get a point (x,y) and then find the cross-product of the perpendicular vectors of the first two planes to get the direction of the line.

 

[-Dragoon-]

I believe you as well are correct, they just parameterized y and you did it with x.

I see. Thank you for the help.

Alternatively you could use 9x - 11y - 1 = 0 to get a point (x,y) and then find the cross-product of the perpendicular vectors of the first two planes to get the direction of the line.

Isn't the cross product only used for lines and planes in 3 dimensional space?

 

[mege]

I see. Thank you for the help.


Isn't the cross product only used for lines and planes in 3 dimensional space?

A plane really only has meaning in 3d space ;) (esoteric examples aside)

With the tools you have, I think your approach is fine. When dealing with the gradient in Calc III some of this will come up again, but will be solved using a different approach.

The cross product of two vectors normal (perpindicular) to intersecting planes will result in a vector parallell to the line formed by intersection of the planes. Using this method, your answer checks out to be OK. If you don't get this concept, it's OK - your method works just as well.

 

[-Dragoon-]

A plane really only has meaning in 3d space ;) (esoteric examples aside)

With the tools you have, I think your approach is fine. When dealing with the gradient in Calc III some of this will come up again, but will be solved using a different approach.

The cross product of two vectors normal (perpindicular) to intersecting planes will result in a vector parallell to the line formed by intersection of the planes. Using this method, your answer checks out to be OK. If you don't get this concept, it's OK - your method works just as well.

Interesting. I do know that the normals of two planes are: (2, -3, -1) and (3, -2, 3), but how could I use their cross product to check if my answer is correct? This would be really helpful to do for an upcoming exam, as a question like this was worth quite a lot on a previous test of mine. Thanks in advance.

 

[HallsofIvy]

The simplest way to check your answer is to put the parametric equations of the line back into the equations for the plane and see if they are satisfied for all t.

For example, you have
x = t
y = (9/11)t - 1/11
z = (-5/11)t + 14/11

and the equation of the first plane is 2x - 3y - z + 1 = 0.

Then 2t- 3((9/11)t- 1/11)- ((-5/11)t+ 14/11)+ 1= 2t- (27/11)t+ 3/11+ (5/11)t- 14//11+ 1= (2- 27/11+ 5/11)t+ 3/11- 14/11+ 1= (2- 22/11)t- 11/11+ 1= (2- 2)t- 1+ 1= 0 for all t.

 

[-Dragoon-]

The simplest way to check your answer is to put the parametric equations of the line back into the equations for the plane and see if they are satisfied for all t.

For example, you have
x = t
y = (9/11)t - 1/11
z = (-5/11)t + 14/11

and the equation of the first plane is 2x - 3y - z + 1 = 0.

Then 2t- 3((9/11)t- 1/11)- ((-5/11)t+ 14/11)+ 1= 2t- (27/11)t+ 3/11+ (5/11)t- 14//11+ 1= (2- 27/11+ 5/11)t+ 3/11- 14/11+ 1= (2- 22/11)t- 11/11+ 1= (2- 2)t- 1+ 1= 0 for all t.

Thank you, I am extremely grateful. This will be very helpful for my upcoming exam.