anonymous
  • anonymous
Does anyone know Unit vectors and length stuff really well? I'm boggled..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
somewhat really well, but kinda sorta maybe like.... sure :)
amistre64
  • amistre64
i and j are unit vectors.... i = <1,0>
anonymous
  • anonymous
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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amistre64
  • amistre64
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
amistre64
  • amistre64
if the "slope" of the vectors are equal, they are parallel right?
anonymous
  • anonymous
Yeah, okay. So if I am given a vector in R^4, I solve for it's length, and then what?
anonymous
  • anonymous
Yeah, totally got that.
amistre64
  • amistre64
<4,7> is parallel to <12,21>
anonymous
  • anonymous
Agreed.
amistre64
  • amistre64
r^4?.... 4d stuff eh..
anonymous
  • anonymous
Yeah. That doesn't really matter though, does it? length is length.
amistre64
  • amistre64
distance follows the same rules into other dimensions; you just keep tacking on the extra stuff under the radical.
anonymous
  • anonymous
I'm given a solution here, and they just took 1/(length)*(my original vector)
amistre64
  • amistre64
\[distance of R^n = \sqrt{x^2 + y^2 + z^2 + ......n^2}\]
anonymous
  • anonymous
Yep.
anonymous
  • anonymous
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)?
amistre64
  • amistre64
not to sure about the answer to that.... its been awhile since I got down and dirty with vectors...
anonymous
  • anonymous
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.
amistre64
  • amistre64
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
anonymous
  • anonymous
Right. But then why would they choose to multiply it by 1/(length)? there must be some significance..
amistre64
  • amistre64
must be, but I would have to review the material that you are getting it from to understand it better :/
anonymous
  • anonymous
Darn. Thanks though!
amistre64
  • amistre64
youre welcome..wish i could have been more helpful :)
anonymous
  • anonymous
It's strange too, since they are telling me to check whether the vector I got had the length of 1...
amistre64
  • amistre64
all unit vectors are of length "1". I recall that much :)
anonymous
  • anonymous
Maybe that's why.. Haha, thanks. Stay tuned. I'm sure I'll be back soon! ;)
amistre64
  • amistre64
and all vectors can be expressed as a sum of their unit vectors and scalars right?
anonymous
  • anonymous
Sheeeeeeeeesh. They wanted it as a unit vector parallel to the original....
anonymous
  • anonymous
Yeah, as a linear combination. Gottttt it!
anonymous
  • anonymous
Thanks!
amistre64
  • amistre64
yay!!
anonymous
  • anonymous
How do I become your fan? Lol
amistre64
  • amistre64
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.
anonymous
  • anonymous
:D
amistre64
  • amistre64
lol...thanx ;)
anonymous
  • anonymous
Are you savvy with dot products?
amistre64
  • amistre64
sorta savvy; . = xa xb + ya yb right?
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
For this question, I have (xv) . v And the answer is 2x.
anonymous
  • anonymous
They give me the length of v to be √2
amistre64
  • amistre64
I recall multiplication not being associative with vectors for some odd reason.
anonymous
  • anonymous
Is that just since SQRT(2)*SQRT(2) = 2?
anonymous
  • anonymous
Maybe..
amistre64
  • amistre64
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.
amistre64
  • amistre64
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
amistre64
  • amistre64
i know this means something lol
anonymous
  • anonymous
Okay... so. That gives me X(V.V)=2X, when ||V||=√2
anonymous
  • anonymous
What's the significance of a vector dotted with itself?
anonymous
  • anonymous
WAIT
amistre64
  • amistre64
lets see :) <1,2>.<1,2> 1(1) + 2(2) = 2+4 = 2(1+2)
amistre64
  • amistre64
1+2 = 1i + 2j right?
anonymous
  • anonymous
If ||V|| = √2, and ||V||= √(V.V), then (V.V)= (||V||)^2!
anonymous
  • anonymous
Where ||V|| is the length of V.
amistre64
  • amistre64
making sense for you :)
anonymous
  • anonymous
But not for you? :P
amistre64
  • amistre64
little bit, but it aint my question lol.
anonymous
  • anonymous
hahahaha, fair nuff!
amistre64
  • amistre64
if the length is sqrt(2) then yes, twice the length = 2
anonymous
  • anonymous
Nooooo. Twice the length is 2√2. Lol. √2 squared is 2.
amistre64
  • amistre64
thats better :) been staring at math too long
anonymous
  • anonymous
Lol. Bedtime!
anonymous
  • anonymous
I think you need to be my fan now..
anonymous
  • anonymous
We all just want to be loved, don't we? Hahahahaha
amistre64
  • amistre64
what? these aint free ya know ;)
anonymous
  • anonymous
Pfffffft
anonymous
  • anonymous
Just trying to get some street cred, yo.
amistre64
  • amistre64
yeah, pupil and 0 is soooo unbecoming these days :)
anonymous
  • anonymous
;)
amistre64
  • amistre64
I do actually have to head out, library is closing up soon... Ciao :)
anonymous
  • anonymous
Adios amigo!

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