## anonymous 4 years ago does anyone want to teach me Multivariable Calculus? :P

1. amistre64

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

2. anonymous

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

3. amistre64

|dw:1334382554391:dw|

4. amistre64

vector just point in a direction

5. anonymous

wait, ill show u

6. amistre64

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

7. amistre64

... or typing lol

8. anonymous

haha

9. anonymous

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

10. 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

11. amistre64

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

12. amistre64

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

13. anonymous

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

14. 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

15. anonymous

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

16. amistre64

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

17. anonymous

i got this:

18. amistre64

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

19. anonymous

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

20. amistre64

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

21. anonymous

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

22. amistre64

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

23. anonymous

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

24. amistre64

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

25. 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

26. anonymous

there's a third part :X

27. amistre64

and the negative is just -1/3 right?

28. anonymous

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

29. amistre64

now your just making up words :P

30. anonymous

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

31. anonymous

or trying anyway

32. amistre64

standard basis perhaps?

33. anonymous

yeees

34. amistre64

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

35. amistre64

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

36. amistre64

or does it matter?

37. anonymous

i did it like this

38. amistre64

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

39. anonymous

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

40. amistre64

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

41. amistre64

your vectors need to be put in columns, not rows

42. amistre64

multiply both sides by 3 to get rid of the scalar

43. anonymous

ooh yeah

44. amistre64

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

45. amistre64

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

46. anonymous

got it.

47. 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 :))

48. amistre64

youre welcome :)

49. anonymous

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 :(