Consider the two lines l1: x=−2t, y=1+2t, z=3tl1: x=−2t, y=1+2t, z=3t and l2: x=−8+4s, y=0+5s, z=5+1sl2: x=−8+4s, y=0+5s, z=5+1s find the point of intersection of the two lines.

At the point of intersection, the coordinates are the same:

... (x, y, z) = (-2t, 1+2t, 3t) = (-8+4s, 5s, 5+s)

We only need two of the coordinates to solve for the values of s and t. Using the x- and y-coordinates, we have

... 4s + 2t = 8 . . . . . . x2 - x1 = 0, in standard form

... 5s -2t = 1 . . . . . . . y2 - y1 = 0, in standard form

Adding these equations gives 9s=9, so s=1. Substituting into either equation gives t=2.

Using the expression for l1 with t=2, the point of intersection is

... (-2·2, 1+2·2, 3·2) = (-4, 5, 6).

We could have stopped after finding the value of s, because that defines the point. By finding the value of t, we can check the solution to make sure that l1 and l2 both give the same point for the respective values of s and t. (They do.)

45 minutes

step-by-step explanation: