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

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

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

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

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

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

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

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2FFM 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
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.2what is the magnitude of u ?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2no, 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
 3 years ago
Best ResponseYou've already chosen the best response.2to amke a vector into a unit vector, divide by its magnitude

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

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2so 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)

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

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

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

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

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

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

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

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

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2actually 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?

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2here 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")

NotTim
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.0oohhh...i see. scalar multitplier of 2.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2dw: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

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

TuringTest
 3 years ago
Best 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

NotTim
 3 years ago
Best ResponseYou've already chosen the best response.0wait...what's the "first" way.

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

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

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

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

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2but 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
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.2unit vectors parallel to*

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

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

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2yes. 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
 3 years ago
Best ResponseYou've already chosen the best response.2look at it like slope

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

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

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