Question

Consider u=[2,3] and v=[-4,-6]. Find 2 unit vectors parallel to u.

Answer 1

I THINKING OF treating it as a parallel thing, but that's not correct, is it?

Answer 2

\[{\vec u\over||\vec u||}\]makes it a unit vector, did you know that?

Answer 3

what's that.

Answer 4

dividing by the norm/magnitude of the vector

Answer 5

Unit vector parallel to \(\vec{u}\) is : \( \frac{1}{\sqrt{13}} <2,3> \)

Answer 6

ok...
I don't understand, but I don't think I really need further explaination @TuringTest

Answer 7

Also, what did you type FFM? That looks...strange

Answer 8

But there is one one unique vector (parallel) isn't ?

Answer 9

uhhh...i dunno. Was just gunna treat this as collinear. NEver saw how to deal with parallels in my txtbook

Answer 10

FFM I think is pointing out that the other, \(\vec v\), is antiparallel I suppose...
I just see it as
i.e. magnitude is the length of the vector\[\vec u=<2,3>\] and the magnitude is\[||\vec u||=\sqrt{2^2+3^2}\]\(\vec v\) is parallel to \(\vec u\) so you should be able to do the same trick there

Answer 11

I think the equation editor is messing around with us. I see squares everywhere.
Also, maybe I should skip to the chase. How would I get [0.55,0,83],[-0.55,-0.83]?

Answer 12

what is the magnitude of u ?

Answer 13

[2,3]

Answer 14

no, read what I wrote above
magnitude is the \(length\) of the vector, which we can find with pythagorus\[||\vec u||=\sqrt{2^2+3^2}\] and the magnitude is

Answer 15

to amke a vector into a unit vector, divide by its magnitude

Answer 16

*make

Answer 17

aww...what exactly is a uni vector. I though the point [2,3] was a vector itself...

Answer 18

A unit vector is a vector that has a length (i.e. "magnitude) of 1

Answer 19

ok. um. gimme a moment to do the math here

Answer 20

so a vector \(\vec u=<x,y>\) is a unit vector iff\[||\vec u||=\sqrt{x^2+y^2}=1\]after dividing a vector by its magnitude, it should make sense that it will become a unit vector
(you are essentially dividing a vector by it's length, which will logically give 1)

Answer 21

wait, how do i divide [2,3] by sqrt 13?

Answer 22

do each component individually: it is a scalar multiplication

Answer 23

im really fuzzy on this. I divide 2 by sqrt 13, and then 3 by sqrt 13 right?

Answer 24

yep, and those will be you new components

Answer 25

umm.srry. pencil is breaking on me here...

Answer 26

try a calculator ;)

Answer 27

im going to get a new pencil. bak in a flash

Answer 28

\[{2\over\sqrt{13}}\]just requires a calculator!!!!

Answer 29

argh. srry. wat nxt now

Answer 30

got [0.55,0.83]

Answer 31

hooray

Answer 32

now for the next one, do the exact same trick
...or use a shortcut (which takes a little explanation) and get the answer in 2 seconds
whichever you prefer

Answer 33

I guess you don't wanna explain both methods?

Answer 34

well they are really based on the same idea:
multiplying a vector be a scalar will leave it parallel to the original...

Answer 35

wat do yu mean "exact same trick"

Answer 36

actually that was a poor choice of phrasing; upi need to regognize what I'm about to say either way:
let's look at \(\vec u\) and \(\vec v\)let's say that \[\vec v=c\vec u\]then what does \(\vec v\) look like?

Answer 37

let \(c=2\) and we double the length of \(\vec u\):notice that they remain parallel