anonymous
  • anonymous
does anyone want to teach me Multivariable Calculus? :P
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
you know what you did to x in calc 1 and 2? now do it to y and z
anonymous
  • anonymous
im good with matrices and all that. it's when it comes a part of vectors that i'm not understanding really well
amistre64
  • amistre64
|dw:1334382554391:dw|

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amistre64
  • amistre64
vector just point in a direction
anonymous
  • anonymous
wait, ill show u
amistre64
  • amistre64
good, casue i aint none to goog at mindreading :)
amistre64
  • amistre64
... or typing lol
anonymous
  • anonymous
haha
anonymous
  • anonymous
I'm in the same position! Some teacher failed me along the way in the vector department!
anonymous
  • anonymous
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
amistre64
  • amistre64
an orthogonal base is such that each vector is orthogonal to each other
amistre64
  • amistre64
since x is a common factor of everyting; we can ignore it
anonymous
  • anonymous
for wich value(s) of x, will {a,b,c} be an orthornomal base?
amistre64
  • amistre64
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
anonymous
  • anonymous
i did that, but when it comes to the second part i get lost.
amistre64
  • amistre64
and since x is just a common "scalar" id assume x can be anything; but maybe zero since people hate zeros
anonymous
  • anonymous
i got this:
amistre64
  • amistre64
" Moreover, they are all required to have length one" http://mathworld.wolfram.com/OrthonormalBasis.html
anonymous
  • anonymous
\[||a||=\sqrt{4x^2+4x^2+x^2}=\sqrt{9x^2}=3x\]
amistre64
  • amistre64
looks like x needs to be the 1/|...| yeah, that
anonymous
  • anonymous
so i did it for all 3 vectors and they all were 3x
amistre64
  • amistre64
x is the scalar that makes them go unit i believe; which means the x = 1 over magnitude
anonymous
  • anonymous
so since ||a||=||b||=||c||=3x=1 ---> x=1/3
amistre64
  • amistre64
x is not part of the length; its factored out and eqauted to the reciprocal of the magnitude
amistre64
  • amistre64
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
anonymous
  • anonymous
there's a third part :X
amistre64
  • amistre64
and the negative is just -1/3 right?
anonymous
  • anonymous
Find the coordinates to v in the orthornomal base obtained, in wich v in the canonical base has coordinates (1,-2,-3)
amistre64
  • amistre64
now your just making up words :P
anonymous
  • anonymous
im sorry, its in portuguese, i'm translating :P
anonymous
  • anonymous
or trying anyway
amistre64
  • amistre64
standard basis perhaps?
anonymous
  • anonymous
yeees
amistre64
  • amistre64
row reduce your vector basis next to this new vector from the standard bases
amistre64
  • amistre64
how should we define our vectors in the basis? what should our x y and z parts correlate to?
amistre64
  • amistre64
or does it matter?
anonymous
  • anonymous
i did it like this
amistre64
  • amistre64
it matters; switching columns of a matrix alters things; switching rows doent
anonymous
  • anonymous
for the first vector and i did the same to the other 2
amistre64
  • amistre64
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
  • amistre64
your vectors need to be put in columns, not rows
amistre64
  • amistre64
multiply both sides by 3 to get rid of the scalar
anonymous
  • anonymous
ooh yeah
amistre64
  • amistre64
\[RREF\begin{vmatrix}2&1&2&|&3\\2&-2&-1&|&-6\\1&2&-2&|&-9\end{vmatrix}\]
amistre64
  • amistre64
the left goes to identity and the right goes to the coord vector in the basis
anonymous
  • anonymous
got it.
anonymous
  • anonymous
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
  • amistre64
youre welcome :)
Rogue
  • Rogue
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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