How do you find the points of intersection of l(t)=(3+2t, 7+8t, -2+t)?

Question

I'm having a lot of trouble with this problem.

Answer 1

That's a parametric equation of a line in 3-space, so "points of intersection" must mean points where the line intersects the xy plane, the yz plane, and the xz plane. It's just like x and y intercepts in 2 dimensions.

All points in the xy plane have z coordinate 0, so let -2+t = 0, t = 2, and plug it in to get x = 3+2(2) = 7, y = 7+8(2) = 23. So l(t) intersects the xy plane at (7,23,0).

Similarly, let y = 0 = 7+8t, t = -7/8. Then x = 3+2(-7/8) = 10/8, z = -2 + (-7/8) = -23/8, so l(t) intersects the xz plane at (10/8, 0, -23/8).

Finally let x = 0 = 3+2t, t = -3/2. Then y = 7+8(-3/2) = -5 and z = -2+(-3/2) = -7/2, so l(t) intersects they yz plane at (0, -5, -7/2).