Question

Find a vector parametric equation $$\vec{r}(t)$$ for the line through the points P=(−1,−4,2) and Q=(−1,1,−3) for the given conditions on the parameter t.

If $$\vec{r}(0)=⟨−1,−4,2⟩$$ and $$\vec{r}(3)=⟨−1,1,−3⟩$$

Answer 1

position vector given (-1,-4,2) is <-1,-4,2> @shamim

Answer 2

i think

Answer 3

what does Op mean?

Answer 4

where are you stuck ?

Answer 5

you need a `point` and a `direction vector` to write equation of line

Answer 6

so i have a line <-1, -4, 2> + t<-2, 5, -5> ?

Answer 7

looks there is a mistake in diraction vector, check once

Answer 8

ah, <-1, -4, 2> +t<0, 5, -5>

Answer 9

Yes! that does give you the equation of line but there is still a problem

Answer 10

does it satisfy the given conditions ?

Answer 11

plugin t = 0, what do you get ?

Answer 12

plugin t = 3, what do you get ?

Answer 13

t = 0, <-1, -4, 2> + <0,0,0> = <-1, -4, 2>

Answer 14

t = 3, <-1, -4, 2> + <-6, 15, -15> = <-7, 11, -13>

Answer 15

does not satisfy t = 3

Answer 16

Right, how do we fix it

Answer 17

hmm

Answer 18

plugin t = 1, what do you get ?

Answer 19

oh wait my t=3 was way off i think

Answer 20

Lets slow down the speed by a factor of 3 : \[r(t) = \langle -1, -4, 2\rangle +\frac{t}{3} \langle 0, 5, -5\rangle \]

Answer 21

well the x component was anyways

Answer 22

see if this this satisfies both the conditions now

Answer 23

hmm, t = 0 is same

Answer 24

and t = 3 is..

Answer 25

<-1, -4, 2> + <0, 5, -5> = <-1, 1, -3> wow, nice

Answer 26

i would have never thought to do that, ty @ganeshie8

Answer 27

im finally starting to understand this stuff a little bit

Answer 28

still need lots of work though :)

Answer 29

you're good :)