
#1
Sep2009, 06:58 PM

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. [tex]2x2y+z=12, x\frac{1}{2}=y\frac{3}{2}=\frac{x+1}{2}[/tex] 2. Relevant equations [tex]2x2y+z=12[/tex] [tex]x\frac{1}{2}=y\frac{3}{2}=\frac{x+1}{2}[/tex] 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: [tex]x=t+\frac{1}{2}[/tex] [tex]y=t\frac{3}{2}[/tex] [tex]z=2t1[/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})+2t1=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? 



#2
Sep2009, 09:08 PM

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 nonzero it cannot be contained in the plane 



#3
Jun1911, 08:52 PM

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? 



#4
Jun1911, 09:05 PM

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 



#5
Mar1512, 08:36 PM

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 coordinaterotate the entire problem so that the Line L stands perfectly vertical (ZAxisAligned) 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
Mar1612, 10:31 AM

HW Helper
P: 3,309

Hey pierre, welcome to PF
This is a pretty old post, so probably no need to reopen it. 


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