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

- anonymous

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

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- schrodinger

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- amistre64

somewhat really well, but kinda sorta maybe like.... sure :)

- amistre64

i and j are unit vectors.... i = <1,0>

- 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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## More answers

- 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

if the "slope" of the vectors are equal, they are parallel right?

- anonymous

Yeah, okay.
So if I am given a vector in R^4, I solve for it's length, and then what?

- anonymous

Yeah, totally got that.

- amistre64

<4,7> is parallel to <12,21>

- anonymous

Agreed.

- amistre64

r^4?.... 4d stuff eh..

- anonymous

Yeah. That doesn't really matter though, does it? length is length.

- amistre64

distance follows the same rules into other dimensions; you just keep tacking on the extra stuff under the radical.

- anonymous

I'm given a solution here, and they just took 1/(length)*(my original vector)

- amistre64

\[distance of R^n = \sqrt{x^2 + y^2 + z^2 + ......n^2}\]

- anonymous

Yep.

- 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

not to sure about the answer to that.... its been awhile since I got down and dirty with vectors...

- 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

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

Right. But then why would they choose to multiply it by 1/(length)? there must be some significance..

- amistre64

must be, but I would have to review the material that you are getting it from to understand it better :/

- anonymous

Darn. Thanks though!

- amistre64

youre welcome..wish i could have been more helpful :)

- anonymous

It's strange too, since they are telling me to check whether the vector I got had the length of 1...

- amistre64

all unit vectors are of length "1". I recall that much :)

- anonymous

Maybe that's why.. Haha, thanks. Stay tuned. I'm sure I'll be back soon! ;)

- amistre64

and all vectors can be expressed as a sum of their unit vectors and scalars right?

- anonymous

Sheeeeeeeeesh. They wanted it as a unit vector parallel to the original....

- anonymous

Yeah, as a linear combination. Gottttt it!

- anonymous

Thanks!

- amistre64

yay!!

- anonymous

How do I become your fan? Lol

- 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

:D

- amistre64

lol...thanx ;)

- anonymous

Are you savvy with dot products?

- amistre64

sorta savvy; . = xa xb + ya yb right?

- anonymous

Yeah

- anonymous

For this question, I have (xv) . v
And the answer is 2x.

- anonymous

They give me the length of v to be √2

- amistre64

I recall multiplication not being associative with vectors for some odd reason.

- anonymous

Is that just since SQRT(2)*SQRT(2) = 2?

- anonymous

Maybe..

- 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

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

i know this means something lol

- anonymous

Okay... so. That gives me X(V.V)=2X, when ||V||=√2

- anonymous

What's the significance of a vector dotted with itself?

- anonymous

WAIT

- amistre64

lets see :)
<1,2>.<1,2>
1(1) + 2(2) = 2+4 = 2(1+2)

- amistre64

1+2 = 1i + 2j right?

- anonymous

If ||V|| = √2, and ||V||= √(V.V), then (V.V)= (||V||)^2!

- anonymous

Where ||V|| is the length of V.

- amistre64

making sense for you :)

- anonymous

But not for you? :P

- amistre64

little bit, but it aint my question lol.

- anonymous

hahahaha, fair nuff!

- amistre64

if the length is sqrt(2) then yes, twice the length = 2

- anonymous

Nooooo. Twice the length is 2√2. Lol. √2 squared is 2.

- amistre64

thats better :) been staring at math too long

- anonymous

Lol. Bedtime!

- anonymous

I think you need to be my fan now..

- anonymous

We all just want to be loved, don't we? Hahahahaha

- amistre64

what? these aint free ya know ;)

- anonymous

Pfffffft

- anonymous

Just trying to get some street cred, yo.

- amistre64

yeah, pupil and 0 is soooo unbecoming these days :)

- anonymous

;)

- amistre64

I do actually have to head out, library is closing up soon... Ciao :)

- anonymous

Adios amigo!

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