1agree totally with swaraj.
OA has component OAsina in a direction perpendicular to OB.
That OAsina has a component OAsina.sin(90-a)=OAsina.cosa opposite to OC,thereby making the resultant in horizontal direction 0.
i wonder y i never asked it before.....its my age old doubt.......but now i am going to prove sumthing.........
WHAT??? that neone who says that any vector has zero magnitude component in a direction perpendicular to it is horribly wrong....
see OA vector has component OB=OAcos a
OB has component OC perp to OA=OBsin a=OA(sin a)(cos a)=1/2OAsin 2a
thus OA has a component along perpendicular.....
bt that depends on value of angle a......it is zero if reference vector OB is taken 45° from OA..
but as all i also know this is wrong.....so where is the fallacy????
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23 Answers
1i strongly believe that yu cant call da component of b(ie C) as da component of a just because b is a component of a
thats a wrong analogy
jagarans explanation is pretty appropriate:
consider B as X and line perpendicular to B as Y
you have 1st found components of A along X n Y
now you shift axes to C as X° and A as Y° and find components of B
obviously you get diff components....
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49the soln of swaraj is exactly in resonance with my solution.................and my q wasn't wrong....LOLzzz
1"""see OA vector has component OB=OAcos a
OB has component OC perp to OA=OBsin a=OA(sin a)(cos a)=(1/2)OAsin 2a
thus OA has a component along perpendicular"""
can any body tell me how u concluded tat OA has a component along OC
see its (1/2)OAsin2a.....so issey ye kahan prove ho raha hai ki OA has a component along OC
i mean teh doubt is itself rong ki u r saying tat u hav proved OA has a component along OC
1swaraj your explanation is pretty good
subhomoy pl put what solution you found...i guess most of us have completely deviated from what yu hav asked...
there are many ways of percepting the doubt n all da ppl hu hav given solns are right in their own way. your solution probably of a diff wavelength so tell us your thot process
11hi subhomoy,first when u take the component of OA along OB u left the another component of OA which is along the axis perpendicular to OB . this component has also component along OC,which cancels ur above found component and the net component along OC becomes 0.
49i give a hint here........itz something to do with something that we have missed out.....
LOL....all my hints are as usual graver and vaguer than my answer.....but anyways i start getting answers from the forum after such pokes only....
49let me tell you all something you are arguing that the question is vague...but believe me neither my question nor my solution is vague......it is wrong.....no not wrong also but still there is some flaw.....tho it is not from books but i think this needs perception....
so lets find out the single change that needed to be made in the solution....as for the matter of fact i got the answer as soon as i wrote the question on my copy.....
try that process out...hope you all succeed like i did.....
21hey tats actually quiet vague....the thing which u found out (OAsin2a)/2 tats actually teh component of OB NOT OA........ how r u saying it to be a component of OA ...i cant understand.......seriously....and it doesnt matter ki if u take out OB in terms of OA and then take component aLONG OC....then also it will remain component of OB not OA!
1da simplest explanation i can think of is tht when yu take component of b,it has one along a as well as along c
yu are calling da component of b along c as the component ofa...but in reality its da component along a tht shud be considered not c
49Devika read my question correctly....i ask as we all know vector OC must be zero while we re getting different answers.....how do u explain.............
lol.....as i said the secret lies in something superficial yet deep:P:D
1agts u have changed the axes of reference.thats where the mistake is
49@debmalya: then ans comes -OA/2........
@nishant bhaiya: LOL....but still i'm not leaving it though......
62lol..
trust me even i had this idea once [4]
But this one is a good doubt to have :)
subhomoy, you will figure out as i keep doing, no matter how much you think you are innovating, it will be tough to come out with something no one else has done before ;)
49no it is not vague....i have found out my mistake.....it is not as pointed out by others....the problem is more superficial yet deep
think india think!!![1][1][1][1][1][1][1]
1this one is vague
yu cant do tht bcoz once yu get da projection of a on b n then find da projection of b on c, yu are changing da vector of reference itself!
49i nw realise this could have been classified as one of <<<GOOGLY FROM ATGS>>>
49but still everyone try this..........i swear it is really easy.....
49yeah i got the answer..............really it is easy bt y did i not get earlier i wonder