nickymarden
does anyone want to teach me Multivariable Calculus? :P



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amistre64
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you know what you did to x in calc 1 and 2? now do it to y and z

nickymarden
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im good with matrices and all that. it's when it comes a part of vectors that i'm not understanding really well

amistre64
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dw:1334382554391:dw

amistre64
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vector just point in a direction

nickymarden
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wait, ill show u

amistre64
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good, casue i aint none to goog at mindreading :)

amistre64
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... or typing lol

nickymarden
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haha

ald0024
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I'm in the same position! Some teacher failed me along the way in the vector department!

nickymarden
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given vectors a=2xi+2xj+xk, b=xi2xj+2xk and c=2xixj2xk show that {a,b,c} is a negative orthogonal base for x<0

amistre64
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an orthogonal base is such that each vector is orthogonal to each other

amistre64
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since x is a common factor of everyting; we can ignore it

nickymarden
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for wich value(s) of x, will {a,b,c} be an orthornomal base?

amistre64
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a.b =2xi+2xj+xk,
xi2xj+2xk

2  4 + 2 = 0
b.c=2xixj 2xk
xi2xj+2xk

2 + 2  4 = 0
a.c =2xi+2xj+xk,
xi2xj+2xk

2  4 + 2 = 0
so it is orthonormal

nickymarden
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i did that, but when it comes to the second part i get lost.

amistre64
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and since x is just a common "scalar" id assume x can be anything; but maybe zero since people hate zeros

nickymarden
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i got this:


nickymarden
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\[a=\sqrt{4x^2+4x^2+x^2}=\sqrt{9x^2}=3x\]

amistre64
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looks like x needs to be the 1/... yeah, that

nickymarden
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so i did it for all 3 vectors and they all were 3x

amistre64
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x is the scalar that makes them go unit i believe; which means the x = 1 over magnitude

nickymarden
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so since a=b=c=3x=1 > x=1/3

amistre64
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x is not part of the length; its factored out and eqauted to the reciprocal of the magnitude

amistre64
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each vector has the same basic components so they are going to have the same lengths
<2,2,1> ; sqrt(4+4+1) = sqrt(9) = 3
x = 1/3 yep

nickymarden
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there's a third part :X

amistre64
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and the negative is just 1/3 right?

nickymarden
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Find the coordinates to v in the orthornomal base obtained, in wich v in the canonical base has coordinates (1,2,3)

amistre64
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now your just making up words :P

nickymarden
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im sorry, its in portuguese, i'm translating :P

nickymarden
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or trying anyway

amistre64
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standard basis perhaps?

nickymarden
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yeees

amistre64
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row reduce your vector basis next to this new vector from the standard bases

amistre64
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how should we define our vectors in the basis? what should our x y and z parts correlate to?

amistre64
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or does it matter?

nickymarden
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i did it like this

amistre64
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it matters; switching columns of a matrix alters things; switching rows doent

nickymarden
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for the first vector and i did the same to the other 2

amistre64
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v1=2,2,1
v2=1,2,2
v3=2,1,2
\[\frac{1}{3}\begin{vmatrix}v_1&v_2&v_3\\2&1&2\\2&2&1\\1&2&2\end{vmatrix}\ \begin{vmatrix}c_1\\c_2\\c_3\end{vmatrix}=\begin{vmatrix}1\\2\\3\end{vmatrix}\]

amistre64
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your vectors need to be put in columns, not rows

amistre64
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multiply both sides by 3 to get rid of the scalar

nickymarden
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ooh yeah

amistre64
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\[RREF\begin{vmatrix}2&1&2&&3\\2&2&1&&6\\1&2&2&&9\end{vmatrix}\]

amistre64
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the left goes to identity and the right goes to the coord vector in the basis

nickymarden
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got it.

nickymarden
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guess im not as bad as i thought? at least i got the first 2 parts correct :p and thanks for explaining the rest :))

amistre64
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youre welcome :)

Rogue
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I'd like to learn multivariate as well to get a head start on next year, but I'm not a fan of matrices and vectors :(