Does anyone know Unit vectors and length stuff really well? I'm boggled..

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Does anyone know Unit vectors and length stuff really well? I'm boggled..

Mathematics
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somewhat really well, but kinda sorta maybe like.... sure :)
i and j are unit vectors.... i = <1,0>
Hahaha, okay. I think I understand how to find length of a vector- that's pretty straight forward. How can I use that kind of math to find a vector parallel to another?

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since vectors arent tied to any specific spot, any vector with the same direction is parallel; if they are in the same direction and of the same magnitude they are equal vectors
if the "slope" of the vectors are equal, they are parallel right?
Yeah, okay. So if I am given a vector in R^4, I solve for it's length, and then what?
Yeah, totally got that.
<4,7> is parallel to <12,21>
Agreed.
r^4?.... 4d stuff eh..
Yeah. That doesn't really matter though, does it? length is length.
distance follows the same rules into other dimensions; you just keep tacking on the extra stuff under the radical.
I'm given a solution here, and they just took 1/(length)*(my original vector)
\[distance of R^n = \sqrt{x^2 + y^2 + z^2 + ......n^2}\]
Yep.
So them multiplying by the reciprocal of my length... Is that just them essentially 'rescaling' my vector, giving me one that is parallel (since it is basically underneath my original? It's on the same spot, but just not as long)?
not to sure about the answer to that.... its been awhile since I got down and dirty with vectors...
Does that make sense, though? If I took the vector <1, 0, 0, 0> and multiplied it by two, I would get <2, 0, 0, 0>, which is parallel.
that makes sense yes, its magnitude has increased but its direction is the same. it is straight in the air if i see it correctly
Right. But then why would they choose to multiply it by 1/(length)? there must be some significance..
must be, but I would have to review the material that you are getting it from to understand it better :/
Darn. Thanks though!
youre welcome..wish i could have been more helpful :)
It's strange too, since they are telling me to check whether the vector I got had the length of 1...
all unit vectors are of length "1". I recall that much :)
Maybe that's why.. Haha, thanks. Stay tuned. I'm sure I'll be back soon! ;)
and all vectors can be expressed as a sum of their unit vectors and scalars right?
Sheeeeeeeeesh. They wanted it as a unit vector parallel to the original....
Yeah, as a linear combination. Gottttt it!
Thanks!
yay!!
How do I become your fan? Lol
you might have to hit f5 to refresh your browser; but if you want there is a "become a fan" to the side of my name.
:D
lol...thanx ;)
Are you savvy with dot products?
sorta savvy; . = xa xb + ya yb right?
Yeah
For this question, I have (xv) . v And the answer is 2x.
They give me the length of v to be √2
I recall multiplication not being associative with vectors for some odd reason.
Is that just since SQRT(2)*SQRT(2) = 2?
Maybe..
The associative property is meaningless for the dot product because is not defined since is a scalar and therefore cannot itself be dotted. However, it does satisfy the property (rX).Y = r(X.Y) for a scalar.
The associative property is meaningless for the dot product because (a.b).c is not defined since a.b is a scalar and therefore cannot itself be dotted. However, it does satisfy the property
i know this means something lol
Okay... so. That gives me X(V.V)=2X, when ||V||=√2
What's the significance of a vector dotted with itself?
WAIT
lets see :) <1,2>.<1,2> 1(1) + 2(2) = 2+4 = 2(1+2)
1+2 = 1i + 2j right?
If ||V|| = √2, and ||V||= √(V.V), then (V.V)= (||V||)^2!
Where ||V|| is the length of V.
making sense for you :)
But not for you? :P
little bit, but it aint my question lol.
hahahaha, fair nuff!
if the length is sqrt(2) then yes, twice the length = 2
Nooooo. Twice the length is 2√2. Lol. √2 squared is 2.
thats better :) been staring at math too long
Lol. Bedtime!
I think you need to be my fan now..
We all just want to be loved, don't we? Hahahahaha
what? these aint free ya know ;)
Pfffffft
Just trying to get some street cred, yo.
yeah, pupil and 0 is soooo unbecoming these days :)
;)
I do actually have to head out, library is closing up soon... Ciao :)
Adios amigo!

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