Answers

2015-05-17T21:00:26+05:30
R = √P²+Q²+2PQcosΘ   ( Θ is the angle between the vectors)
⇒ R² = P²+Q²+2PQcosΘ
if P is doubled & Q remains unaltered then resultant 
R'=√4P²+Q²+4PQcosΘ
   =√(2P²+2P²)+(2Q²-Q²)+4PQcosΘ
   =√2P²-Q²+2(P²+Q²+2PQcosΘ)
   =√2P²-Q²+2R²  (proved)
0
2015-05-18T03:49:24+05:30

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Let the angle between vectors P and Q be = Ф.

magnitude of the resultant = R,   Hence,   R² = P² + Q² + 2 P Q Cos Ф

Let  P' = 2 P and Q  as well as  the angle between them Ф  remains the same.  Let the resultant vector be R'.

   R' ²  =  (2 P)² + Q² + 2 (2 P) Q Cos Ф
          =  4 P² + Q² + 4 P Q Cos Ф
          =  2 (P² + Q² + 2 P Q Cos Ф) + 2 P² - Q²
           = 2 R² + 2 P² - Q²

  Hence, the resultant now is
     R' = √ [ 2 R² + 2 P² - Q² ]


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