How Do You Find the Equation of the Intersection of Two Planes?

[pyrosilver]

urgent, intersection of two planes

Homework Statement

Find the equation of the intersection between 2x + 3y - 4z = 12 and 5x + 2y + 3z = 7.

Homework Equations

none. I don't know how to use cross products, but is there another way?

The Attempt at a Solution

I don't know how to do this and it will be on my final tomorrow... I tried various things but right now the closest I have is, set x to 0 and find for y and z. Please help, I have no idea what else to do.

I know it is in this form (or at least, I think it is):

[x, y, z] + t(a, b, c)

but what are those? t is for time. is the x y z a point on the plane, and how would I find that?

 

[rock.freak667]

but what are those? t is for time. is the x y z a point on the plane, and how would I find that?

t is not for time. t is a parameter.

so for 2x + 3y - 4z = 12 and 5x + 2y + 3z = 7, what are the normals for these two planes?
What will you get when you take the cross-product of these two normals?

 

[pyrosilver]

normals are [2, 3, -4] and [5, 2, 3] right?

I don't know, I never learned to cross product. Is that the only way? I used to be able to do this I think like a year ago (can't find the notes for it) and I didn't use cross producting...how do you cross product? I tried googling it but didn't understand it.

 

[rock.freak667]

normals are [2, 3, -4] and [5, 2, 3] right?

I don't know, I never learned to cross product. Is that the only way? I used to be able to do this I think like a year ago (can't find the notes for it) and I didn't use cross producting...how do you cross product? I tried googling it but didn't understand it.

ok, well there is another way to do it.

Take 2x + 3y - 4z = 12 and 5x + 2y + 3z = 7

Eliminate either x,y or z. When you get an equation with two of variables left, then let either of the variables be your parameter (t,a,b, whatever you like) and then find all the other variables in terms of that parameter.

 

[pyrosilver]

Ok is this how you do it?

Let z = 0

2x + 3y = 12 and 5x + 2y = 7

x ends up equaling 1, y = 10/3

I'm not sure what happens next? so is it

[1, 10/3, 0] + t (a, b, c)

How do i find the other variables in terms of that parameter?

 

[rock.freak667]

Ok is this how you do it?

Let z = 0

Nope.

2x + 3y - 4z = 12 is the same as 10x+15y-20z=60 (x5 throughout)

5x + 2y + 3z = 7 is the same as 10x+4y+6z=14 (x2 throughout)



10x+15y-20z=60 ...(1)
10x+4y+6z=14 ...(2)

Find (1)-(2)

 

[pyrosilver]

OH OKAY. So then:

9y - 26z = 46

y = 46/9 + 26z/9

And then you plug that into the equation above?

Okay thanks, now I get how to get x, y, and z, but I'm still fuzzy on how to get it into [x,y,z] + t(a, b, c) form? Does that still apply?

 

[rock.freak667]

OH OKAY. So then:

9y - 26z = 46

y = 46/9 + 26z/9

And then you plug that into the equation above?

Okay thanks, now I get how to get x, y, and z, but I'm still fuzzy on how to get it into [x,y,z] + t(a, b, c) form? Does that still apply?

it's not best to put the parameter as one of your variables, so we'll use t instead

so you let z=t, and you found y in terms of t

y=46/9 +26t/9

So put that into any of the equations of the planes to find x in terms of t.
What do you get then?

 

[pyrosilver]

9y - 26t = 46
9 (46/9 +26t/9) - 26t = 46

like that?

 

[rock.freak667]

9y - 26t = 46
9 (46/9 +26t/9) - 26t = 46

like that?

Not into that equation. Into either of two equations of the planes.

Either put y=46/9 +26t/9 and z=t into 2x + 3y - 4z = 12 or 5x + 2y + 3z = 7.

EDIT:

When you get x in terms of t. Put x,y and z in the vector form as such

\left( \begin{array}{c}<br /> x\\<br /> y \\<br /> z <br /> \end{array}\right)