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

- NotTim

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

- chestercat

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- NotTim

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

- TuringTest

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

- NotTim

what's that.

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## More answers

- TuringTest

dividing by the norm/magnitude of the vector

- anonymous

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

- NotTim

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

- NotTim

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

- anonymous

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

- NotTim

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

- TuringTest

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

- NotTim

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]?

- TuringTest

what is the magnitude of u ?

- NotTim

[2,3]

- TuringTest

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

- TuringTest

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

- TuringTest

*make

- NotTim

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

- TuringTest

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

- NotTim

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

- TuringTest

so a vector \(\vec u=\) 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)

- NotTim

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

- TuringTest

do each component individually: it is a scalar multiplication

- NotTim

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

- TuringTest

yep, and those will be you new components

- NotTim

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

- TuringTest

try a calculator ;)

- NotTim

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

- TuringTest

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

- NotTim

argh. srry. wat nxt now

- NotTim

got [0.55,0.83]

- TuringTest

hooray

- TuringTest

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

- NotTim

I guess you don't wanna explain both methods?

- TuringTest

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

- NotTim

wat do yu mean "exact same trick"

- TuringTest

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 \(\vec v\)|dw:1333605852213:dw|let's say that \[\vec v=c\vec u\]then what does \(\vec v\) look like?

- TuringTest

let \(c=2\) and we double the length of \(\vec u\):|dw:1333606018155:dw|notice that they remain parallel

- TuringTest

here we are doing the same thing: \(\vec v\) is a scalar multiple of \(\vec u\) with \(c=-1\)
this will change the direction of the vector
(hence the term :anti-parallel")

- NotTim

so wait, v and u are actually related? im suppose to use them both? i was just thinking of using u and random other vectrs

- NotTim

oohhh...i see. scalar multitplier of -2.

- TuringTest

|dw:1333606190135:dw|yeah, sorry my drawing is wrong, it should be -2
but note: that won't change the \(unit\) \(vector\) because all unit vectors have length 1
so all we need to do is notice that \(\vec v\) is a scalar multiple of \(\vec u\), so it has the same unit vector, but the - sign changes the direction
it is still parallel though, in the sense that they never intersect

- NotTim

ok.

- NotTim

so, v is parallel to u. actually. wait. we already got the answer. thats magical.

- TuringTest

...another way to say it is that a unit vector has only one parallel vector to it: it's negative
therefor no calculation is really needed.
it's better to understand it the first way though, in my opinion

- TuringTest

yeah you got it :)

- NotTim

wait...what's the "first" way.

- TuringTest

that \(\vec v\) is a (negative) scalar multiple of \(\vec u\), and therefor has the same unit vector in the opposite direction

- NotTim

does that mean its parallel to u?

- TuringTest

though sorry, now that I reread the question v isn't even important

- TuringTest

but yes, as I said v is a scalar multiple of u
when one vector is a scalar multiple of the other, they are parallel

- NotTim

yeah. i kidna just included every thing related to the question. given the chance, i would include the background info by the text too

- TuringTest

scalar multiples only change the length of the vector

- TuringTest

hence they do not change whether or not they are parallel

- TuringTest

but it only asks for two unit vectors perpendicular to \(\vec u\), so \(\vec v\) has nothing to do with it; it's coincidentally parallel but is not needed to find the answer

- NotTim

so vector, wit ha scalar multiple of the original vector, is not actually parallel to it?
man. we've taken up quite some time with this.

- TuringTest

unit vectors parallel to*

- TuringTest

I repeat:
scalar multiples only change the length of the vector, hence they do not change whether or not they are parallel
so in\[c\vec u=\vec v\] where \(c\) is a scalar, \(\vec u\) and \(\vec v\) are parallel

- NotTim

is there a situation when the vectors are not parallel, but are only different due to a scalar multiplier.
many question i give

- TuringTest

no, unless there is some huge exception that is slipping my mind...

- TuringTest

scalar multiplication of a vector changes only \(direction\) and/or \(magnitude\) (i.e. length)

- NotTim

so i pick, say [3,2,6]. Then it is parallel to [6,4,12]?

- TuringTest

yes. in higher math, like linear algebra, you will learn that vectors that can be written as scalar multiples of each other are called "linearly dependent"
that means they are parallel, which is very important in setting up coordinate systems and such, which is what you do in linear algebra

- TuringTest

look at it like slope

- NotTim

ok. thank you very much. all my questions have been cleared up, im not done my review, and i wish to take a nappy.

- TuringTest

ok then, glad I could help
rest well, a weary mind is not terribly useful in mathematics ;)

- NotTim

should i sleep now, and risk not studying later, or study, but not sleep ever?

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