Finding line of intersection of two planes

  • #1
Spacec0wboy
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Homework Statement



Find parametric equations for the line of intersection of the
planes x + y + z = 1 and
r = (1, 0, 0) + [itex]\lambda[/itex](2, 1, 0) + [itex]\mu[/itex](0, 1, 1) where [itex]\lambda[/itex], [itex]\mu[/itex] [itex]\in[/itex] R

Homework Equations





The Attempt at a Solution



I attempted to convert the 2nd plane equation to scalar form by finding the normal: (2, -1, 0) cross product with (0, 1, -1). That ended up being i + 2j + 2k, meaning that the plane had equation x + 2y + 2z = d, subbing in the point in the vector equation (1, 0, 0) gives d =1 therefore x + 2y + 2z = 1. I'm not sure what to do from here or if what I've done is either correct or necessary.

Help please?
 

Answers and Replies

  • #2
kushan
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the normal vector would be (-1,2,-2).
And Cartesian equation would be x-2y+2z=1
 
  • #3
HallsofIvy
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The parametric equations for the second plane are saying that [itex]x= 1+ 2\lambda+ 0\mu[/itex], [itex]y= 0+ 1\lambda+ 1\mu[/itex], and [itex]z= 0+ 0\lambda+ 1\mu[/itex]. Putting those into the equation of the first plane, x+ y+ z= 1 gives [itex](1+ 2\lambda)+ (\lambda+ \mu)+ (\mu)= 1+ 3\lambda+ 2\mu= 1[/itex] or [itex]3\lambda+ 2\mu= 0[/itex]. Solve that for either [itex]\lambda[/itex] or [itex]\mu[/itex] in terms of the other and replace in the parametric equation, getting linar equations in a single parameter- a line.

I don't know if it was just a typo or copying mistake but the normal to the second plane is NOT the cross product of (2, -1, 0) and (0, 1, -1). I'm not sure where you got those vectors. Two vectors in the plane are (1, 0, 0)+ (2, 1, 0)= (3, 1, 0) (taking [itex]\lambda= 1[/itex] and [itex]\mu= 0[/itex] and another is (1, 0, 0)+ (0, 1, 1)= (1, 1, 1) (taking [itex]\lambda= 0[/itex] and [itex]\mu= 1[/itex]. The cross product of those vectors is normal to the plane.
 
  • #4
kushan
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HallsofIvy
I don't think cross product of 3,1,0 and 1,1,1 would be normal to the 2nd plane
as the cross product of above is 1,-3,2 , but the normal vector should be 1,-2,2
 
Last edited:
  • #5
ehild
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The Attempt at a Solution



I attempted to convert the 2nd plane equation to scalar form by finding the normal: (2, -1, 0) cross product with (0, 1, -1).

(2,1,0) and (0,1,1) are vectors which lay in the plane. You mixed the signs a bit.

Kushan is right, (-1, 2,-2) is a normal vector of the plane.

ehild
 
  • #6
kushan
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ehild would it be -1,2,-2 or 1,-2,2 ?
 
  • #7
ehild
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Two vectors [STRIKE]in the plane[/STRIKE] of the plane are (1, 0, 0)+ (2, 1, 0)= (3, 1, 0) (taking [itex]\lambda= 1[/itex] and [itex]\mu= 0[/itex] and another is (1, 0, 0)+ (0, 1, 1)= (1, 1, 1) (taking [itex]\lambda= 0[/itex] and [itex]\mu= 1[/itex]. The cross product of those vectors is normal to the plane.

The normal of a plane is the cross product of two vectors laying in the plane, your vectors are position vectors of two points in the plane.

ehild
 
Last edited:
  • #8
ehild
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ehild would it be -1,2,-2 or 1,-2,2 ?

They are parallel vectors, both normal to the plane, one is -1 times the other.

ehild
 
  • #9
kushan
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Direction doesn't matter?
 
  • #10
Bacle2
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Direction doesn't matter?

Not for normality: if a vector v is normal to the surface, then so is the vector -v.
 
  • #11
kushan
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Ok I used to imagine like the one originating from center and piercing though the plane perpendicularly
So it is both way .
Thank Ehild and Bacle2
 
  • #12
Bacle2
Science Advisor
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Actually,

maybe thinking of it this way would help: imagine you just have , say the vector

1,-2,2 on the plane. Now, give the plane a 180 turn,and you get the vector

-1,2,-2 --which is the same as 1,-2,2, but with opposite orientation.

Give 99% of the tanks to Ehild, he did most of the work.
 
  • #13
kushan
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1/100 thanks Bacle
 
  • #14
Spacec0wboy
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Wow yep I can't even read the question properly, the normal is indeed (-1, 2,-2). So if I cross product (-1, 2, -2) with (1, 1, 1) I get 4i + 3j - 3k, which is the direction vector of the line. So that's half of what I need for the parametric equations, the other half is a point on the line, how do I get that?
 
  • #15
kushan
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find the equation of line Spacec , using the equation of planes .
 
  • #16
ehild
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Wow yep I can't even read the question properly, the normal is indeed (-1, 2,-2). So if I cross product (-1, 2, -2) with (1, 1, 1) I get 4i + 3j - 3k, which is the direction vector of the line. So that's half of what I need for the parametric equations, the other half is a point on the line, how do I get that?

You do not need that cross product really. You need to give the parametric equation of the line. Follow HallsofIvy's hint.

ehild
 
  • #17
Spacec0wboy
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Ok so using HallsofIvy's method I get:

λ = -2/3*μ

x = 1 - 4/3*μ
y = 5/3*μ
z = -μ

Is this right? Thanks for helping me out and that method was pretty easy but I would still like to know how to do it using the direction vector of the line and finding a point on it, just so I know another way to do it.
 
  • #18
kushan
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You mean other method for solving this?
 
  • #19
Spacec0wboy
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You mean other method for solving this?

Yea, by finding the vector equation of the line then putting it into parametric form. We got the direction vector of the line before from the cross product of the 2 plane normals. Now we need a point on the line but I'm not sure how to get that.
 
  • #20
kushan
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put some x=0 in both plane then solve the two equation .
 
  • #21
Spacec0wboy
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Ok, but why do we put x = 0? Does that mean that both planes share a point with an x value of 0?
 
  • #22
kushan
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Its like we are looking at the line , and assuming it passes through x=0 , and if it does what would be y ,z . In some cases putting x=0 would give two unsolvable equations .
And it is not necessary to put x=0 , you can try other things like y=0 or z=0 aswell.
 
  • #23
Spacec0wboy
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Ah ok. But I don't get how we can assume that the line goes through x = 0, I mean most questions like this are rigged but with 2 random planes the chances of that happening are quite low aren't they? Hypothetically, what if it doesn't go through x = 0 y = 0 or z = 0? What happens then? Unless I have the wrong idea of how this works..

EDIT: I just realized how this would work haha. The line extends forever in a give direction meaning that it has to cross an axis once, meaning at some point one of x, y, z will equal 0.
 
Last edited:
  • #24
kushan
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there is just one case when line is parallel to one of the axis , rest is taken care of .
 
  • #25
HallsofIvy
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The normal of a plane is the cross product of two vectors laying in the plane, your vectors are position vectors of two points in the plane.

ehild
Ah! Thanks. I should have used <3, 1, 0>, and <1, 1, 1>.
 
  • #26
ehild
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Ah! Thanks. I should have used <3, 1, 0>, and <1, 1, 1>.

You did use <3, 1, 0> and <1, 1, 1>, which are points of the plane but the vectors <3, 1, 0> and <1, 1, 1> are not in the plane. :smile:

ehild
 

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