does anyone want to teach me Multivariable Calculus? :P

- anonymous

does anyone want to teach me Multivariable Calculus? :P

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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 :(

Looking for something else?

Not the answer you are looking for? Search for more explanations.