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

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0somewhat really well, but kinda sorta maybe like.... sure :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i and j are unit vectors.... i = <1,0>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hahaha, 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since 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
 5 years ago
Best ResponseYou've already chosen the best response.0if the "slope" of the vectors are equal, they are parallel right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, okay. So if I am given a vector in R^4, I solve for it's length, and then what?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, totally got that.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0<4,7> is parallel to <12,21>

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0r^4?.... 4d stuff eh..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. That doesn't really matter though, does it? length is length.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0distance follows the same rules into other dimensions; you just keep tacking on the extra stuff under the radical.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm given a solution here, and they just took 1/(length)*(my original vector)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[distance of R^n = \sqrt{x^2 + y^2 + z^2 + ......n^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0not to sure about the answer to that.... its been awhile since I got down and dirty with vectors...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does 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
 5 years ago
Best ResponseYou've already chosen the best response.0that 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
 5 years ago
Best ResponseYou've already chosen the best response.0Right. But then why would they choose to multiply it by 1/(length)? there must be some significance..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0must be, but I would have to review the material that you are getting it from to understand it better :/

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0youre welcome..wish i could have been more helpful :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's strange too, since they are telling me to check whether the vector I got had the length of 1...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0all unit vectors are of length "1". I recall that much :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Maybe that's why.. Haha, thanks. Stay tuned. I'm sure I'll be back soon! ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and all vectors can be expressed as a sum of their unit vectors and scalars right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sheeeeeeeeesh. They wanted it as a unit vector parallel to the original....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, as a linear combination. Gottttt it!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How do I become your fan? Lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0Are you savvy with dot products?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sorta savvy; <xa,ya>.<xb,yb> = xa xb + ya yb right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For this question, I have (xv) . v And the answer is 2x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They give me the length of v to be √2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I recall multiplication not being associative with vectors for some odd reason.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that just since SQRT(2)*SQRT(2) = 2?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0i know this means something lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay... so. That gives me X(V.V)=2X, when V=√2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What's the significance of a vector dotted with itself?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets see :) <1,2>.<1,2> 1(1) + 2(2) = 2+4 = 2(1+2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If V = √2, and V= √(V.V), then (V.V)= (V)^2!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where V is the length of V.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0making sense for you :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0little bit, but it aint my question lol.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if the length is sqrt(2) then yes, twice the length = 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nooooo. Twice the length is 2√2. Lol. √2 squared is 2.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats better :) been staring at math too long

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you need to be my fan now..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We all just want to be loved, don't we? Hahahahaha

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what? these aint free ya know ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just trying to get some street cred, yo.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, pupil and 0 is soooo unbecoming these days :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I do actually have to head out, library is closing up soon... Ciao :)
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