Plane determined by intersecting lines
1. The problem statement, all variables and given/known data
Find the point of intersection of the lines: x=2t+1, y=3t+2, z=4t+3, and x=s+2, y=2s+4, z=4s1, and then find the plane determined by these lines. 2. Relevant equations How do i find the plane determined by these lines? 3. The attempt at a solution Ive read through the text, and i figured out the first part about where they intersect: v=<2,3,4> Pt. A=(1,2,3) 2(x1)+3(y2)+4(z3)=0 2x+3y+4z=20 then i substituted the 2nd parametric equation into the x,y,z variables and solved for s. s=1 then i plugged s=1 back into the parametric equation to find x,y,z for intersection the equations intersect at (1,2,3) Now i'm stuck...how do i find the planes determined by these lines? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 
Re: Plane determined by intersecting lines
Quote:
But v = <2, 3, 4> is a vector pointing in the direction of the first line it is NOT perpendicular to the plane which is what you need. (In fact, since the lines lie in the plane, <2, 3, 4> is a vector in the plane, not perpendicular to it.) Quote:
Quote:

Re: Plane determined by intersecting lines
The standard equation for a plane is [tex] a(x  x_0) + b(y  y_0) + c(z  z_0) = 0, [/tex] where [tex] \vec{n} = <a, b, c> [/tex] is the normal vector to the plane. Now, if you know two vectors (the direction vectors of your 2 lines) that are already on the plane, can you think of any operation between two vectors that gives you a normal vector (thus giving you a normal vector to your plane)? Can you get the rest?

Re: Plane determined by intersecting lines
darn you HallsofIvy!!...beat me by a minute :P

Re: Plane determined by intersecting lines
Great! Thanks a lot guys!

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