## nickymarden Group Title does anyone want to teach me Multivariable Calculus? :P 2 years ago 2 years ago

1. amistre64 Group Title

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

2. nickymarden Group Title

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 Group Title

|dw:1334382554391:dw|

4. amistre64 Group Title

vector just point in a direction

5. nickymarden Group Title

wait, ill show u

6. amistre64 Group Title

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

7. amistre64 Group Title

... or typing lol

8. nickymarden Group Title

haha

9. ald0024 Group Title

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

10. nickymarden Group Title

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 Group Title

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

12. amistre64 Group Title

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

13. nickymarden Group Title

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

14. amistre64 Group Title

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. nickymarden Group Title

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

16. amistre64 Group Title

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

17. nickymarden Group Title

i got this:

18. amistre64 Group Title

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

19. nickymarden Group Title

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

20. amistre64 Group Title

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

21. nickymarden Group Title

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

22. amistre64 Group Title

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

23. nickymarden Group Title

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

24. amistre64 Group Title

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

25. amistre64 Group Title

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. nickymarden Group Title

there's a third part :X

27. amistre64 Group Title

and the negative is just -1/3 right?

28. nickymarden Group Title

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

29. amistre64 Group Title

now your just making up words :P

30. nickymarden Group Title

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

31. nickymarden Group Title

or trying anyway

32. amistre64 Group Title

standard basis perhaps?

33. nickymarden Group Title

yeees

34. amistre64 Group Title

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

35. amistre64 Group Title

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

36. amistre64 Group Title

or does it matter?

37. nickymarden Group Title

i did it like this

38. amistre64 Group Title

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

39. nickymarden Group Title

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

40. amistre64 Group Title

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 Group Title

your vectors need to be put in columns, not rows

42. amistre64 Group Title

multiply both sides by 3 to get rid of the scalar

43. nickymarden Group Title

ooh yeah

44. amistre64 Group Title

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

45. amistre64 Group Title

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

46. nickymarden Group Title

got it.

47. nickymarden Group Title

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 Group Title

youre welcome :)

49. Rogue Group Title

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 :(