Thread: point of intersection of two vector

1. point of intersection of two vector

Hi all,

I have the lines:

A: x = 1 + 2t, y = 2+ 3t, z = 3 + 4t

B: x-2 = (y-4)/2 = (-z -1)/4

I am having trouble determining the coordinates of the point of intersection of A and B.

Well i found B in parametric form first:

x = 2 + t, y = 4 +2t, Z = -4 -4t

and then i equated each component....

e.g 1 + 2t = 2 + t; 2 +3t = 4 +2t

the t values differed for each component and thus i thought there was no point of intersection. However, in the answer there is a point of intersection at (1,2,3). So i am lost here...

ArTiCk

2. Originally Posted by ArTiCK
I have the lines:
A: x = 1 + 2t, y = 2+ 3t, z = 3 + 4t
B: x-2 = (y-4)/2 = (-z -1)/4
x = 2 + t, y = 4 +2t, Z = -4 -4t
Mistake #1. Always use different parameters for different lines.
So B: x=2+s, y=4+2s, & z=-4-4s.

Solve for t & s.
1+2t=2+s
2+3t=4+2s
(Be sure to check to see if the solution works for the z value!)