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NotTimBest ResponseYou've already chosen the best response.0
I THINKING OF treating it as a parallel thing, but that's not correct, is it?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
\[{\vec u\over\vec u}\]makes it a unit vector, did you know that?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
dividing by the norm/magnitude of the vector
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
Unit vector parallel to \(\vec{u}\) is : \( \frac{1}{\sqrt{13}} <2,3> \)
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
ok... I don't understand, but I don't think I really need further explaination @TuringTest
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
Also, what did you type FFM? That looks...strange
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
But there is one one unique vector (parallel) isn't ?
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
uhhh...i dunno. Was just gunna treat this as collinear. NEver saw how to deal with parallels in my txtbook
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
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]?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
what is the magnitude of u ?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
to amke a vector into a unit vector, divide by its magnitude
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
aww...what exactly is a uni vector. I though the point [2,3] was a vector itself...
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
A unit vector is a vector that has a length (i.e. "magnitude) of 1
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
ok. um. gimme a moment to do the math here
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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)
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
wait, how do i divide [2,3] by sqrt 13?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
do each component individually: it is a scalar multiplication
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
im really fuzzy on this. I divide 2 by sqrt 13, and then 3 by sqrt 13 right?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
yep, and those will be you new components
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
umm.srry. pencil is breaking on me here...
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
im going to get a new pencil. bak in a flash
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
\[{2\over\sqrt{13}}\]just requires a calculator!!!!
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
I guess you don't wanna explain both methods?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
well they are really based on the same idea: multiplying a vector be a scalar will leave it parallel to the original...
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
wat do yu mean "exact same trick"
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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:dwlet's say that \[\vec v=c\vec u\]then what does \(\vec v\) look like?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
let \(c=2\) and we double the length of \(\vec u\):dw:1333606018155:dwnotice that they remain parallel
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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 :antiparallel")
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
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
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
oohhh...i see. scalar multitplier of 2.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
dw:1333606190135:dwyeah, 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
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
so, v is parallel to u. actually. wait. we already got the answer. thats magical.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
...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
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
wait...what's the "first" way.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
that \(\vec v\) is a (negative) scalar multiple of \(\vec u\), and therefor has the same unit vector in the opposite direction
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
does that mean its parallel to u?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
though sorry, now that I reread the question v isn't even important
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
yeah. i kidna just included every thing related to the question. given the chance, i would include the background info by the text too
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
scalar multiples only change the length of the vector
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
hence they do not change whether or not they are parallel
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
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.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
unit vectors parallel to*
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
is there a situation when the vectors are not parallel, but are only different due to a scalar multiplier. many question i give
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
no, unless there is some huge exception that is slipping my mind...
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
scalar multiplication of a vector changes only \(direction\) and/or \(magnitude\) (i.e. length)
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
so i pick, say [3,2,6]. Then it is parallel to [6,4,12]?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
look at it like slope
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
ok. thank you very much. all my questions have been cleared up, im not done my review, and i wish to take a nappy.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.2
ok then, glad I could help rest well, a weary mind is not terribly useful in mathematics ;)
 2 years ago

NotTimBest ResponseYou've already chosen the best response.0
should i sleep now, and risk not studying later, or study, but not sleep ever?
 2 years ago
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