Multivariable Calculus

  • Thread starter GreenPrint
  • Start date
  • #1
1,196
0

Homework Statement



Hi,

Find the line L of the intersection of the two planes
x+y+z=1
z-2y+3z=1

What I did was use Gaussian reduction on the augmented matrix. It was easy

[x y z] = [3/2 -1/2 0] + z[-5/2 3/2 1]
or in equation form or whatever it's called
x = 3/2 -5/2 z
y = -1/2 + 3/2 z
z is free variable

What my professor did was take the cross product of the normal vectors of the planes which I guess is just the coefficient matrix with the unit vectors

i j k
1 1 1
1 -2 3

let n1=<1,1,1>
and n2=<1,-2,3>

and he got
5i - 2j - 3k
I agree that this is correct
but then he set z = 0 and said you could set it equal to anything but he recommended zero because it would make the problem easier
got
x+y=1
x-2y=1
solved for x and y and got
x=1, y=0
so he used the point (1,0,0)
and said that the symmetric equation of the line was
(x-1)/5 = - y/2 = -z/3

so I figured out that the parametric equation of the line is
x = 5t+1
y=-2t
z=-3t

OK so I've studied some linear algebra and solved it the way I learned in linear algebra and got a different answer. The professor assumes that we haven't taken linear algebra yet. Yet I find it weird that we would parametric the line in terms of t when z is a free variable from what I learned in linear algebra, so it makes more sense to parametric the line in terms of z, at least that's what I would think.

I'm also not exactly sure why n1Xn2 gives you the direction of the line. Wouldn't n1xn2 give you the direction of the vector perpendicular to both n1 and n2. I don't see why this would be the direction of the line of the intersection of the planes. I also don't see why the components of n1xn2 would be the slope of the of the parametric line.

thanks for any help understanding this.

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
OK so I've studied some linear algebra and solved it the way I learned in linear algebra and got a different answer. The professor assumes that we haven't taken linear algebra yet. Yet I find it weird that we would parametric the line in terms of t when z is a free variable from what I learned in linear algebra, so it makes more sense to parametric the line in terms of z, at least that's what I would think.
you could paramterise in terms of either, it shouldn;t make a difference

I'm also not exactly sure why n1Xn2 gives you the direction of the line. Wouldn't n1xn2 give you the direction of the vector perpendicular to both n1 and n2. I don't see why this would be the direction of the line of the intersection of the planes. I also don't see why the components of n1xn2 would be the slope of the of the parametric line.
The line of intersection is parallel to both plane one and plane 2, so is normal to both n1 and n2. Upto a multiplicative scaling, there is only one direction that satisfie sthis constraint and is given by n1xn2
 

Related Threads on Multivariable Calculus

  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
0
Views
4K
Top