Posted by sania on Wednesday, December 12, 2012 at 11:13am.
L1:
x = 0 + 4t
y = -1 + 6t
z = -1 + 2t
L2:
x = 3 - 7s
y = 9 - 5s
z = 4 + 0s
L1 intersects L2 if
4t = 3-7s
-1+6t = 9-5s
-1+2t = 4
(t,s) = (5/2, -1) is the solution, so
(x,y,z) = (10,14,4)
direction vector of 1st line = (4,6,2) or reduced to (2,3,1)
parametric equation of 1st:
x = 0 + 2t
y = -1 + 3t
z = -1 + t
direction vector of 2nd line = (7, 5, 0)
parametric equation of 2nd:
x = 3 + 7k
y = 9 + 5k
z = 4
solving for x's
0+2t = 3 + 7k
2t = 3+ 7k (#1)
solving for y's
3t-1 = 5k+9
3t - 5k = 10 (#2)
using the z's :
t-1 = 4
t = 5
sub that into #1:
10 = 3 + 7k
k = 1
so t=5 and k = 1
check for consistency in #2 (the equation not used to solve)
3t - 5k = 10
LS = 15 - 5 = 10 = RS
so the point of intersection is: (Using the 1st, could use either one)
x = 0+2t = 10
y = -1+3t = 14
z = -1 + t = 4
They intersect at (10,14,4)
L(ab): (0,-1,-1)+(4u,6u,2u)
L(cd): (3,9,4)+(-7v,-5v,0)
If they intersect at P, then corresponding x,y,z values match, hence
2u-1 = 0 => u=2.5
where
P=(0+4*2.5, -1+6*2.5, -1+2*2.5)=(10,14,4)
For line L(cd),
we have P(3-7v, 9-5v, 4+0v)
=>
3-7v=10 => v=-1
or
9-5v=14 => v=-1
or
4+0v=4 => v=-1 (works).
Therefore they intersect at P, since P(10,14,4) lies on both lines.