Point A (1, -1, 2)
Line s = 2i - j + t(3i -j +k)
The Attempt at a Solution
Ordinarily these are pretty obvious, but in this case the line is also a parametrized vector.
So if I consider r = r0 + st
And sub in s as I would do normally, I'd end up getting t^2's, and that's not the correct answer.
The answer is r = (1, -1, 2) + t(3, -1, 1)
Which looks like the the t part of the original s line was kept with a new position vector (given point). Why is this?