Question

Determine whether the lines L1:→r(t)=〈0,−2,3〉t+〈0,3,1〉 and L2:→p(s)=〈3,−1,−2〉s+〈−3,−4,15〉 intersect. If they do, find the point of intersection a. They intersect at the point b. They are skew lines c. They are parallel or equal

Answer 1

To determine if the lines intersect, we need to set the parametric equations equal to each other and solve for t and s:

0t + 0 = 3s - 3
-2t + 3 = -1s - 4
3t + 1 = -2s + 15

From the first equation, we can solve for s:
3s = 3
s = 1

Substituting s = 1 into the second equation, we can solve for t:
-2t + 3 = -1(1) - 4
-2t + 3 = -1 - 4
-2t + 3 = -5
-2t = -8
t = 4

Now that we have values for t and s, we can substitute them back into the parametric equations to find the point of intersection:

For L1:
r(t) = <0, -2, 3>(4) + <0, 3, 1>
= <0, -8, 12> + <0, 3, 1>
= <0, -5, 13>

For L2:
p(s) = <3, -1, -2>(1) + <-3, -4, 15>
= <3, -1, -2> + <-3, -4, 15>
= <0, -5, 13>

Since the point <0, -5, 13> is common to both lines, the lines L1 and L2 intersect at that point.

Therefore, the correct answer is a. They intersect at the point.