Posted by sania on .
prove that the lines through a(0,1,1) and b(4,5,1) intersects the line through c (3,9,4) and d (4,4,4). also find their point of intersection

maths plse help me.. 
Steve,
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 = 37s
1+6t = 95s
1+2t = 4
(t,s) = (5/2, 1) is the solution, so
(x,y,z) = (10,14,4) 
maths plse help me.. 
Reiny,
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
3t1 = 5k+9
3t  5k = 10 (#2)
using the z's :
t1 = 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) 
maths plse help me.. 
MathMate,
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
2u1 = 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(37v, 95v, 4+0v)
=>
37v=10 => v=1
or
95v=14 => v=1
or
4+0v=4 => v=1 (works).
Therefore they intersect at P, since P(10,14,4) lies on both lines.