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2014-07-31T20:46:02+05:30

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Please draw a parallelogram AB CD.  Let CD be parallel to AB. Let BC be parallel to AD.  Let vector P be represented by AB. Let Q be represented by AD.  Now the resultant force of P & Q is given by parallelo gram law as
         R =  square root [ P²+ Q² + 2 P Q cos A ]  = This is diagonal AC
If the angle is reversed, That is, reverse vector AD or Q. Now draw parallelogram AEFB so that AE = - Q = - AD.   FB = EA.  Now the diagonal  AF will be parallel to diagonal BD.  The angle at A in AEFB is  180 - A.  SO its cosine is - Cos A.
     S = squareroot [ P² + Q² - 2 P Q cos A ]
 So R² + Q² = 2 [P² + Q² ]              as the other terms cancel.
This is also a relation between sides of a parallelogram and diagonals.

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2014-07-31T23:03:52+05:30

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So given P + Q = R
after reversing direction of R we get
 - R = - P - Q
S = - P - Q
so let angle between P  and Q be θ
so resultants
R² = P² +Q² +2PQCOSθ       ....................   i        
S² = P² + Q² - 2PQCOSθ      .....................  ii
SO ADDING i AND ii
 R² + S² = 2(P² +Q² )
2PQCOSθ    CANCELLES OUT THUS GIVING RESULT
     
  R² + S² = 2(P² +Q² )
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hope it helps
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