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=-4s-1, 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(x-1)+3(y-2)+4(z-3)=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
The equations of your lines are x= 2t+ 1, y= 3t+ 2, and z= 4t+ 3. If x= 2t+ 1= 1, then t= 0 so y= 2 and z= 3. Also x= s+ 2= 1 for s= -1 and then y= 2(-1)+ 4= 2, z= -4(-1)- 1= 3. Yes, the two lines intersect at that point. 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.) The coefficients in the parametric equations give vectors <2, 3, 4> and <1, 2, -4> which point in the directions the lines and so are two vectors in the plane. You want a vector perpendicular (normal) to the plane. Take the cross product of those two vectors.
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?