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.
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.
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.
correct.
To determine if the two lines intersect, we can begin by setting up a system of equations representing the parametric equations for each line.
Line AB:
x = 1 + t(4 - 1)
y = 2 + t(8 - 2)
z = 3 + t(-3 - 3)
Line CD:
x = 3 + s(0 - 3)
y = 5 + s(2 - 5)
z = 3 + s(4 - 3)
To find the point of intersection between the two lines, we can set up a system of equations by equating the corresponding components:
1 + t(4 - 1) = 3 + s(0 - 3) (Equation 1)
2 + t(8 - 2) = 5 + s(2 - 5) (Equation 2)
3 + t(-3 - 3) = 3 + s(4 - 3) (Equation 3)
Simplifying these equations, we get:
3t - 3s = 2 (Equation 4)
6t + 3s = 3 (Equation 5)
-6t + 6s = 0 (Equation 6)
By adding Equation 4 and Equation 6, we can eliminate the variable t:
-3s + 6s = 2
3s = 2
s = 2/3
Substituting the value of s into Equation 6, we can find the value of t:
-6t + 6(2/3) = 0
-6t + 4 = 0
-6t = -4
t = 2/3
Now that we have the values of t and s, we can substitute them back into the parametric equations for the lines to find the point of intersection. Let's use the values of t = 2/3 and s = 2/3:
For Line AB:
x = 1 + (2/3)(4 - 1) = 3
y = 2 + (2/3)(8 - 2) = 6
z = 3 + (2/3)(-3 - 3) = -1
For Line CD:
x = 3 + (2/3)(0 - 3) = 1
y = 5 + (2/3)(2 - 5) = 3
z = 3 + (2/3)(4 - 3) = 7/3
The point of intersection is (3, 6, -1) for Line AB and (1, 3, 7/3) for Line CD.
Therefore, the two lines do not intersect since the point of intersection (3, 6, -1) on Line AB does not lie on Line CD (1, 3, 7/3).