use the parametric equations for the lines to see whether the equations are consistent
You need to solve
1+3t = 3-3s
2+4t = 5-3s
3-6t = 3+s
Solving the first two yields
t=1, s = -1/3
If that fits the 3rd equation as well, then the two lines intersect.
Determine if the line through A(1, 2, 3) and B(4, 8, -3) and the line through C(3, 5, 3) and D(0, 2, 4) intersect.
Having a hard time figuring out this problem.
6 answers
Does my answer have to be expressed as a point, or just what the parameters are that satisfy the condition that they make both lines intersect?
Also, I think you've got the wrong answer posted. I don't know how you got -1/3, but when I use both equations for the third equation, I get different results.
Steve, I still don't understand how those values for t and s are the same, when I do substitute the numbers in, it gives me different values.
direction vector of AB = (3, 6, -6) <------ Steve had (3,4,-6), a slight arithmetic error.
x = 1 + 3t
y = 2 + 6t
z = 3 - 6t
direction vector CD = (-3,-3,1)
x = 3 - 3s
y = 5 - 3s
z = 3 + s
setting the x's equal ---> 1+3t = 3-3s or 3t + 3s = 2
setting the y's equal ----> 2+6t = 5-3s or 6t + 3s = 3
subtract them: 3t = 1, t = 1/3
sub into 3t + 3s = 2
1 + 3s = 2
s = 1
does that work in the z's ??
or : 3-6t = 3+s ??
LS = 3 - 2 = 1
RS = 3+1 = 4 ≠ LS
So they do NOT intersect.
x = 1 + 3t
y = 2 + 6t
z = 3 - 6t
direction vector CD = (-3,-3,1)
x = 3 - 3s
y = 5 - 3s
z = 3 + s
setting the x's equal ---> 1+3t = 3-3s or 3t + 3s = 2
setting the y's equal ----> 2+6t = 5-3s or 6t + 3s = 3
subtract them: 3t = 1, t = 1/3
sub into 3t + 3s = 2
1 + 3s = 2
s = 1
does that work in the z's ??
or : 3-6t = 3+s ??
LS = 3 - 2 = 1
RS = 3+1 = 4 ≠ LS
So they do NOT intersect.
correct.
1 answer