Question

Let L be the line with parametric equations: x=5-2t
y=-10+t
z=-2+2t
Find the vector equation for a line that passes through the point P=(2, 9, 9) and intersects L at a point that is distance 3 from the point Q=(5, −10, −2). Note that there are two possible correct answers.

Answer 1

x= 2+t(?)
y=9+t(?)
z=9+t(?)
This is all i know so far

Answer 2

well you need to find those two points on the line that are 3 units away from point Q

Answer 3

essentially you have:
3^2 = (x - 5)^2 + (y + 10)^2 + (z + 2)^2

and you know what x,y,z are

Answer 4

solve for t. Then use it find the two points on the line. Then find the vectors of the line and you'll have the answer

Answer 5

Answers were:
x=2+5t
y=9-20t
z=9-13t
and
x=2+t
y=9-18t
z=9-9t

Still don't understand it though :(

Answer 6

did you solve for t?

Answer 7

3^2 = (5-2t - 5)^2 + (-10+t + 10)^2 + (-2+2t + 2)^2

^ this is what I meant when I said solve for t

Answer 8

https://www.wolframalpha.com/input/?i=3%5E2+%3D+%285-2t+-+5%29%5E2+%2B+%28-10%2Bt+%2B+10%29%5E2+%2B+%28-2%2B2t+%2B+2%29%5E2

so t = +/-1

when t = 1
x=5-2(1) = 3
y=-10+1 = -9
z=-2+2(1) = 0

when t = -1
x=5-2(-1) = 7
y=-10+(-1) = -11
z=-2+2(-1) = -4

so the two points that are 3 units away from Q are (3,-9,0) and (7,-11,-4)

Answer 9

the two vectors are:
(3-2,-9-9,0-9) = (1,-18,-9)
and
(7-2,-11-2,-4-2) = (5,-13,-6)

Answer 10

oops typo
(7-2,-11-9,-4-9) = (5,-20,-13)

Answer 11

r1(t) = t(1,-18,-9) + (2,9,9)

r2(t) = t(5,-20,-13) + (2,9,9)

which match the solution

Answer 12

Cool, makes sense. Thanks!