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cmmon guys.......survey's not over yet.......

Given equations are..........

----------------------(1)

-----------------(2)

-----------------------(3)

--------------------(4)

solving in pairs,,,,,,,(1) & (3),,,,,,(1) & (4),,,,,,,(2)&(4).........

we get three vertices......... $(\frac{1}{m-n},\frac{m}{m-n})$ ....(0,1) and (0,0)........

finding the area of the triangle formed by them.....using this determinant

$Area = \frac{1}{2}\begin{vmatrix}{x}_{1} & {y}_{1} & 1\\ {x}_{2} & {y}_{2} & 1\\ {x}_{3} & {y}_{3} & 1\end{vmatrix}$

multiply it with two.........to get the area of parallelogram..........

my answer is.......... $\frac{1}{m-n}$ .........

posting my full explanation soon........

Well I also think it will be a straight line motion....as seen by another projectile..

angles???.......which angles.....please be more specific......

SOME EXPERTS PLS SEE INTO IT>>>>>>>>>>>>>>>

and more importantly.......u shud know its electronic configuration.....

what the right answer then?????

well this is getting more and more confusing for me..........I think risin is correct........

So far these are the demands...........

AEROSPACE------------------ 3 votes........

MECHANICAL---------------- 1 vote........

CSE----------------------------- 3 votes.......

IT-------------------------------- 1 vote........

Electronics & Comm------- 1 vote........

Civil------------------------------ 1 vote.....

Electricals----------------------1 vote.........

more response awaited..........

@risin.....i think i didnt get u exactly what ur point was....if u cud make it more clear pls.........

well it may be........i think we need an expert opinion.....

I have noted one bug sir.........

after posting....when we go back to edit the post.......

SOOOO MANNNYYY spaces come.....automatically.....

hope this will be corrected soon.........

Rest is gr8....

@risin......... have numbered them already.....

neways.......i wud prefer,,,,,,,

CSE or CIVIL.....

But hey........bro......u also have to consider the refractive index of glass.........

I mean to say.....while calculating the focal length of the concave lens.......the value of refractive index shud be

$\frac{1.7}{1.5}$ .......i.e

$\frac{1}{f} = (\frac{{\mu}_{g}}{{\mu}_{l}}-1)(\frac{-1}{12.5}+\frac{-1}{12.5})$ .....

I think we have to do it this way/////////