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does anyone want to teach me Multivariable Calculus? :P

Mathematics
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you know what you did to x in calc 1 and 2? now do it to y and z
im good with matrices and all that. it's when it comes a part of vectors that i'm not understanding really well
|dw:1334382554391:dw|

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Other answers:

vector just point in a direction
wait, ill show u
good, casue i aint none to goog at mindreading :)
... or typing lol
haha
I'm in the same position! Some teacher failed me along the way in the vector department!
given vectors a=2xi+2xj+xk, b=xi-2xj+2xk and c=2xi-xj-2xk show that {a,b,c} is a negative orthogonal base for x<0
an orthogonal base is such that each vector is orthogonal to each other
since x is a common factor of everyting; we can ignore it
for wich value(s) of x, will {a,b,c} be an orthornomal base?
a.b =2xi+2xj+xk, xi-2xj+2xk ----------- 2 - 4 + 2 = 0 b.c=2xi-xj -2xk xi-2xj+2xk ----------- 2 + 2 - 4 = 0 a.c =2xi+2xj+xk, xi-2xj+2xk ----------- 2 - 4 + 2 = 0 so it is orthonormal
i did that, but when it comes to the second part i get lost.
and since x is just a common "scalar" id assume x can be anything; but maybe zero since people hate zeros
i got this:
" Moreover, they are all required to have length one" http://mathworld.wolfram.com/OrthonormalBasis.html
\[||a||=\sqrt{4x^2+4x^2+x^2}=\sqrt{9x^2}=3x\]
looks like x needs to be the 1/|...| yeah, that
so i did it for all 3 vectors and they all were 3x
x is the scalar that makes them go unit i believe; which means the x = 1 over magnitude
so since ||a||=||b||=||c||=3x=1 ---> x=1/3
x is not part of the length; its factored out and eqauted to the reciprocal of the magnitude
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
there's a third part :X
and the negative is just -1/3 right?
Find the coordinates to v in the orthornomal base obtained, in wich v in the canonical base has coordinates (1,-2,-3)
now your just making up words :P
im sorry, its in portuguese, i'm translating :P
or trying anyway
standard basis perhaps?
yeees
row reduce your vector basis next to this new vector from the standard bases
how should we define our vectors in the basis? what should our x y and z parts correlate to?
or does it matter?
i did it like this
it matters; switching columns of a matrix alters things; switching rows doent
for the first vector and i did the same to the other 2
v1=2,2,1 v2=1,-2,2 v3=2,-1,-2 \[\frac{1}{3}\begin{vmatrix}v_1&v_2&v_3\\2&1&2\\2&-2&-1\\1&2&-2\end{vmatrix}\ \begin{vmatrix}c_1\\c_2\\c_3\end{vmatrix}=\begin{vmatrix}1\\-2\\-3\end{vmatrix}\]
your vectors need to be put in columns, not rows
multiply both sides by 3 to get rid of the scalar
ooh yeah
\[RREF\begin{vmatrix}2&1&2&|&3\\2&-2&-1&|&-6\\1&2&-2&|&-9\end{vmatrix}\]
the left goes to identity and the right goes to the coord vector in the basis
got it.
guess im not as bad as i thought? at least i got the first 2 parts correct :p and thanks for explaining the rest :))
youre welcome :)
I'd like to learn multi-variate as well to get a head start on next year, but I'm not a fan of matrices and vectors :(

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