- #1

GreenPrint

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## 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.