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