does anyone want to teach me Multivariable Calculus? :P

- anonymous

does anyone want to teach me Multivariable Calculus? :P

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- schrodinger

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- amistre64

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

- 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

|dw:1334382554391:dw|

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## More answers

- amistre64

vector just point in a direction

- anonymous

wait, ill show u

- amistre64

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

- amistre64

... or typing lol

- anonymous

haha

- anonymous

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

- 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

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

- amistre64

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

- anonymous

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

- 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

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

- amistre64

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

- anonymous

i got this:

- amistre64

" Moreover, they are all required to have length one"
http://mathworld.wolfram.com/OrthonormalBasis.html

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

so since ||a||=||b||=||c||=3x=1 ---> x=1/3

- amistre64

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

- 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

there's a third part :X

- amistre64

and the negative is just -1/3 right?

- anonymous

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

- amistre64

now your just making up words :P

- anonymous

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

- anonymous

or trying anyway

- amistre64

standard basis perhaps?

- anonymous

yeees

- amistre64

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

- amistre64

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

- amistre64

or does it matter?

- anonymous

i did it like this

- amistre64

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

- anonymous

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

##### 1 Attachment

- 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

your vectors need to be put in columns, not rows

- amistre64

multiply both sides by 3 to get rid of the scalar

- anonymous

ooh yeah

- amistre64

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

- amistre64

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

- anonymous

got it.

- 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

youre welcome :)

- 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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