Question

Scaling the vector equation (-3,5,3)+(1,-1,-4)t from r(6) to r(9)?

Answer 1

If you think of "t' as time

then, you start at (-3,5,3) and reach (-2,4,-1) in "1" time units, right ?

Answer 2

and the velocity vector would be <1, -1, -4>

Answer 3

Right, I get that. So that means that <1, -1, -4> is on a scale of 1 from what I can understand

Answer 4

do you want to be at point (-3,5,3) at time t=6 ?

Answer 5

Would you have to scale the point accordingly too?

Answer 6

r(t) = (-3,5,3)+(1,-1,-4)(t-6)

Answer 7

if you want to be at point (-3,5,3) at `t=6`,you just need to start late
subtract `6` from time as shown above ^^

Answer 8

at what point you want to be at `t=9` ?

Answer 9

Would it be the same except it would be t-9 this time?

Answer 10

whats the complete exact question ? i feel you're hiding some details

Answer 11

Find a vector parametric equation r(t) for the line through the points P=(−3,5,3) and Q=(−2,4,−1) for each of the given conditions on the parameter t.
If r(6)=P and r(9)=Q, then r(t)=???
This is the full question

Answer 12

that means you want to write the equation such that r(6) gives you point P and r(9) gives you point Q

Answer 13

direction vector from P to Q = <1, -1, -4>

Answer 14

r(t) = (-3,5,3)+(1,-1,-4)(t-6)


gets you to point P at t=6
but it won't get you to point Q at t=9

we need to fix it a bit

Answer 15

since we want to increase the time between PQ by 9-6 = 3,
divide the velocity vector by 3 to slow down the speed

Answer 16

r(t) = (-3,5,3)+(1,-1,-4)(t-6)/3

Answer 17

that should work, check ^^

Answer 18

Oh, I understand the problem now! Thanks for explaining, my book didn't have anything on this so I had a hard time understanding :)

Answer 19

np :)

Answer 20

if it helps you can use the `moving bug` analogy

Answer 21

A bug is moving on that line with some "velocity"
you need to model its position