- #1

jaejoon89

- 195

- 0

L1: R = 2i + 3j + 3k + t(i - 2j + 5k)

L2: (x + 3) / 2 = (y + 1) / 2 = -z

(I assume the plural in points is wrong... since that would be impossible)

R, i, j, k are vectors; x, y, z are not

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x = 2 + t

y = 3 - 2t

z = 3 + 5t

for some value(s) of t. If this point also lies on L2, then (x + 3)/2 = -z, so that:

(2 + t + 3)/2 = -(3 + 5t)

5 + t = -6 - 10t

11t = -11

t = -1

For this to really be on L2, you also need to have (y + 1)/2 = -z, or:

(3 - 2t + 1)/2 = -(3 + 5t)

4 - 2t = -6 - 10t

8t = -10

t = -5/4

The two t's don't equal each other. What am I doing wrong?