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

by raytrace
Tags: determine, intersection, lies, line, plane, point
 P: 9 1. The problem statement, all variables and given/known data Find the point(s) of intersection (if any) of the plane and the line. Also determine whether the line lies in the plane. $$2x-2y+z=12, x-\frac{1}{2}=-y-\frac{3}{2}=\frac{x+1}{2}$$ 2. Relevant equations $$2x-2y+z=12$$ $$x-\frac{1}{2}=-y-\frac{3}{2}=\frac{x+1}{2}$$ 3. 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?
 HW Helper P: 3,309 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
 P: 1 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?
HW Helper
P: 3,309

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

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
 P: 2 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 )
 HW Helper P: 3,309 Hey pierre, welcome to PF This is a pretty old post, so probably no need to re-open it.

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