Find Intersection of Plane & Line, Does Line Lie in Plane?

[raytrace]

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.

$$2x-2y+z=12, x-\frac{1}{2}=-y-\frac{3}{2}=\frac{x+1}{2}$$

Homework Equations

$$2x-2y+z=12$$

$$x-\frac{1}{2}=-y-\frac{3}{2}=\frac{x+1}{2}$$

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:

$$x=t+\frac{1}{2}$$
$$y=-t-\frac{3}{2}$$
$$z=2t-1$$

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

$$2(t+\frac{1}{2})-2(-t-\frac{3}{2})+2t-1=12$$

$$t=\frac{3}{2}$$

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
$$\vec{l}\bullet\vec{n}= <2,-2,1>\bullet<1,-1,2> = 2(1) + (-2)(-1) + 1(2) = 6$$

So the line does not lie in the plane.

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

 

[lanedance]

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

 

[jcreed2323]

i have the same problem but i don't understand how to tell if the line lies in the plane?
where does the <1,-1,2> come from?

 

[lanedance]

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

 

[pbierre]

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 )

 

[lanedance]

Hey pierre, welcome to PF

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