Question

does anyone want to teach me Multivariable Calculus? :P

Answer 1

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

Answer 2

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

Answer 3

vector just point in a direction

Answer 4

wait, ill show u

Answer 5

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

Answer 6

... or typing lol

Answer 7

haha

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

i got this:

Answer 17

" Moreover, they are all required to have length one"

http://mathworld.wolfram.com/OrthonormalBasis.html

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

there's a third part :X

Answer 26

and the negative is just -1/3 right?

Answer 27

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

Answer 28

now your just making up words :P

Answer 29

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

Answer 30

or trying anyway

Answer 31

standard basis perhaps?

Answer 32

yeees

Answer 33

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

Answer 34

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

Answer 35

or does it matter?

Answer 36

i did it like this

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

your vectors need to be put in columns, not rows

Answer 41

multiply both sides by 3 to get rid of the scalar

Answer 42

ooh yeah

Answer 43

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

Answer 44

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

Answer 45

got it.

Answer 46

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

Answer 47

youre welcome :)

Answer 48

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 :(