# The resultant of two vectors P and Q acting at a point is R. Show that if P is doubled, Q remains unaltered, then the resultant will be of the magnitude √2P²-Q²+2R²

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by mrohit724

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by mrohit724

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⇒ 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)

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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 ]

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 ]