NotTim
  • NotTim
Consider u=[2,3] and v=[-4,-6]. Find 2 unit vectors parallel to u.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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NotTim
  • NotTim
I THINKING OF treating it as a parallel thing, but that's not correct, is it?
TuringTest
  • TuringTest
\[{\vec u\over||\vec u||}\]makes it a unit vector, did you know that?
NotTim
  • NotTim
what's that.

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TuringTest
  • TuringTest
dividing by the norm/magnitude of the vector
anonymous
  • anonymous
Unit vector parallel to \(\vec{u}\) is : \( \frac{1}{\sqrt{13}} <2,3> \)
NotTim
  • NotTim
ok... I don't understand, but I don't think I really need further explaination @TuringTest
NotTim
  • NotTim
Also, what did you type FFM? That looks...strange
anonymous
  • anonymous
But there is one one unique vector (parallel) isn't ?
NotTim
  • NotTim
uhhh...i dunno. Was just gunna treat this as collinear. NEver saw how to deal with parallels in my txtbook
TuringTest
  • 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
  • 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
  • TuringTest
what is the magnitude of u ?
NotTim
  • NotTim
[2,3]
TuringTest
  • 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
  • TuringTest
to amke a vector into a unit vector, divide by its magnitude
TuringTest
  • TuringTest
*make
NotTim
  • NotTim
aww...what exactly is a uni vector. I though the point [2,3] was a vector itself...
TuringTest
  • TuringTest
A unit vector is a vector that has a length (i.e. "magnitude) of 1
NotTim
  • NotTim
ok. um. gimme a moment to do the math here
TuringTest
  • 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
  • NotTim
wait, how do i divide [2,3] by sqrt 13?
TuringTest
  • TuringTest
do each component individually: it is a scalar multiplication
NotTim
  • NotTim
im really fuzzy on this. I divide 2 by sqrt 13, and then 3 by sqrt 13 right?
TuringTest
  • TuringTest
yep, and those will be you new components
NotTim
  • NotTim
umm.srry. pencil is breaking on me here...
TuringTest
  • TuringTest
try a calculator ;)
NotTim
  • NotTim
im going to get a new pencil. bak in a flash
TuringTest
  • TuringTest
\[{2\over\sqrt{13}}\]just requires a calculator!!!!
NotTim
  • NotTim
argh. srry. wat nxt now
NotTim
  • NotTim
got [0.55,0.83]
TuringTest
  • TuringTest
hooray
TuringTest
  • 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
  • NotTim
I guess you don't wanna explain both methods?
TuringTest
  • TuringTest
well they are really based on the same idea: multiplying a vector be a scalar will leave it parallel to the original...
NotTim
  • NotTim
wat do yu mean "exact same trick"
TuringTest
  • 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
  • TuringTest
let \(c=2\) and we double the length of \(\vec u\):|dw:1333606018155:dw|notice that they remain parallel
TuringTest
  • 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
  • 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
  • NotTim
oohhh...i see. scalar multitplier of -2.
TuringTest
  • 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
  • NotTim
ok.
NotTim
  • NotTim
so, v is parallel to u. actually. wait. we already got the answer. thats magical.
TuringTest
  • 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
  • TuringTest
yeah you got it :)
NotTim
  • NotTim
wait...what's the "first" way.
TuringTest
  • TuringTest
that \(\vec v\) is a (negative) scalar multiple of \(\vec u\), and therefor has the same unit vector in the opposite direction
NotTim
  • NotTim
does that mean its parallel to u?
TuringTest
  • TuringTest
though sorry, now that I reread the question v isn't even important
TuringTest
  • 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
  • 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
  • TuringTest
scalar multiples only change the length of the vector
TuringTest
  • TuringTest
hence they do not change whether or not they are parallel
TuringTest
  • 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
  • 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
  • TuringTest
unit vectors parallel to*
TuringTest
  • 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
  • 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
  • TuringTest
no, unless there is some huge exception that is slipping my mind...
TuringTest
  • TuringTest
scalar multiplication of a vector changes only \(direction\) and/or \(magnitude\) (i.e. length)
NotTim
  • NotTim
so i pick, say [3,2,6]. Then it is parallel to [6,4,12]?
TuringTest
  • 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
  • TuringTest
look at it like slope
NotTim
  • 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
  • TuringTest
ok then, glad I could help rest well, a weary mind is not terribly useful in mathematics ;)
NotTim
  • NotTim
should i sleep now, and risk not studying later, or study, but not sleep ever?

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