Find the point of intersection of the plane and line. Determine if line lies in plane

  • Thread starter raytrace
  • Start date
  • #1
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Homework Statement


Find the point(s) of intersection (if any) of the plane and the line. Also determine whether the line lies in the plane.

[tex]2x-2y+z=12, x-\frac{1}{2}=-y-\frac{3}{2}=\frac{x+1}{2}[/tex]


Homework Equations


[tex]2x-2y+z=12[/tex]

[tex]x-\frac{1}{2}=-y-\frac{3}{2}=\frac{x+1}{2}[/tex]


The Attempt at a Solution



I can seem to find the point of intersection just fine. What I'm having difficulty figuring out is how to determine whether the line lies in the plane or not.

I solved all the x, y, and z equations to t:

[tex]x=t+\frac{1}{2}[/tex]
[tex]y=-t-\frac{3}{2}[/tex]
[tex]z=2t-1[/tex]

I put the x, y, and z into the equation for the plane and solved for t:

[tex]2(t+\frac{1}{2})-2(-t-\frac{3}{2})+2t-1=12[/tex]

[tex]t=\frac{3}{2}[/tex]

Plug t=3/2 back into the equations for x, y, and z and I end up with the point of intersection at (2, -3, 2)

Now here is where I get stuck.

Shoot, while typing this I may have just came up with the solution but I'll ask and see if anyone can confirm.

Do I calculate the vector of the line and do a dot product to the normal of the plane. If the result is 0 then that means that the line lies on the plane?

Therefore
[tex]\vec{l}\bullet\vec{n}= <2,-2,1>\bullet<1,-1,2> = 2(1) + (-2)(-1) + 1(2) = 6[/tex]

So the line does not lie in the plane.

Is this assumption and my math correct? If not, where did I go wrong?
 
Last edited:

Answers and Replies

  • #2
lanedance
Homework Helper
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if the dot product of the normal to the plane & line is zero, you know the line is parallel to the plane, but you must also show it contains a point of the plane, to be contained by the plane

if the dot product is non-zero it cannot be contained in the plane
 
  • #3


i have the same problem but i dont understand how to tell if the line lies in the plane?
where does the <1,-1,2> come from?
 
  • #4
lanedance
Homework Helper
3,304
2


hey jcreed this post is 2 yrs old so you should probably open a new thread, you'll definitely get more answers that way as well

to show a line is contained is a plane you need to show
- there is a point on the line contained in the plane
- the line is parallel to the plane
This shows all points in the line are contained in the plane
 
  • #5
9
3


Algorithmic Geometry approach. This problem is straightforward if you know how to form generalized coordinate rotations (3D matrix rotators). The trick is to coordinate-rotate the entire problem so that the Line L stands perfectly vertical (Z-Axis-Aligned) in rotated space.

Inputs:
Line: defined by 3D points p1, p2
Plane: PL defined by equation: p • o == L, where: p is any point on the plane
o is the plane's orientation (normalized "normal")
L is the signed distance of plane from origin along direction o

Result: i is the intersection point


Steps to solve algorithmically:
0) bail if Line is parallel to plane
1) compute the run direction d of Line L = (p2 - p1)norm (normalized vector diff)
2) construct a Rotator with [ newXaxis, newYaxis, d ] where the software chooses the newXaxis, newYaxis dirVec pair arbitrarily
3) compute the invariantX'Y' coords of the rotated line L' by coordRotating p1 --> p1'
3) rotate the entire problem by the ZAlignRotator in 2)
a) computationally coordRotate PL by ZAlignRotator --> PL'
4) Use the plane equation of PL' : i' • PL'.o == PL'.L to solve for the z' coord of i'
5) coordUnrotate i' --> i

To determine if a Line lies within a Plane, simply generate 2 points on the line, and then test both to see if they lie in the plane (if they both satisfy the plane equation p • o == L )
 
  • #6
lanedance
Homework Helper
3,304
2


Hey pierre, welcome to PF

This is a pretty old post, so probably no need to re-open it.
 

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