## ajprincess 3 years ago Plzzz can someone help me? Thanx.:)

1. ajprincess

2. timo86m

I understand the q and I know it is true I jsut dont know how to prove it. maybe something about f(x) and shifting it f(x-a)+b :)

3. dpaInc

sorry... i don' have open office or word... how 'bout a pdf?

4. timo86m

if you have hotmail it has a free viewer :) via skydrive

5. ajprincess

jst a moment let me convert it.

6. dpaInc

it says Advance.. i hope it's not too advanced otherwise you're doing this for nothing.. :)

7. ajprincess

8. ajprincess

@dpaInc hw abt a screenshot? I hav posted a screenshot.

9. dpaInc

hmmm.. yep.. too advanced for me... sorry.... :(

10. ajprincess

it's k. Thanxx a lot for tryng to help me. :)

11. dpaInc

when ur done with this i can help with the times table... :)

12. ajprincess

:)

13. ajprincess

Thanxx a lot @timo86m for the suggestion.

14. shubhamsrg

r1,r2,r3,r4 are vectors a1r1 + a2r2 + a3r3 + a4r4 =0 now r1 is position vector w.r.t O,,for any other point O' ,,vector is r1 - OO' and the others become r2 - OO' , r3 -OO', r4-OO' we need to find a1(r1 - OO') + a2(r2-OO') + a3(r3-OO') + a4(r4 -OO') => ( 0 ) - OO' ( a1 + a2 +a3 +a4) clearly is becomes 0 for a1 +a2 + a3 + a4 =0 hence proved..

15. ajprincess

if u dnt mind can u please tell me hw u got r1-OO'? That part is nt clear to me.

16. shubhamsrg

|dw:1338704382266:dw| we can see easily that r1 = OO' + (required vector) thus req. vector is r1 - OO'

17. ajprincess

Oh k. Thanxxxx a lotttt.

18. shubhamsrg