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Q.If two vectors Ā and B̄ represent two adjacent sides of parallelogram, then prove that, magnitude of resultant R = √(A² + B² + 2AB cosθ).

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2022Subjective· 3mImportance★★★★★
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Dropping a perpendicular in the parallelogram of vectors and applying Pythagoras gives R = √(A² + B² + 2AB cosθ).

Setup. Let Aˉ=OP→\bar A = \overrightarrow{OP} and Bˉ=OQ→\bar B = \overrightarrow{OQ} be two vectors drawn from a common origin O, with angle θ\theta between them. Complete the parallelogram OPSQ (with PS→\overrightarrow{PS} parallel and equal to OQ→=Bˉ\overrightarrow{OQ} = \bar B). By the parallelogram law, the diagonal OS→=Rˉ=Aˉ+Bˉ\overrightarrow{OS} = \bar R = \bar A + \bar B is the resultant.

Derivation. Drop a perpendicular from S to the extension of OP, meeting it at N. In right triangle ONS:

  • ON=OP+PN=A+Bcos⁡θON = OP + PN = A + B\cos\theta (since PS=BPS = B, and the angle between PSPS and the extended line ONON is θ\theta, the angle between Aˉ\bar A and Bˉ\bar B)
  • SN=Bsin⁡θSN = B\sin\theta

By the Pythagorean theorem:

OS2=ON2+SN2OS^2 = ON^2 + SN^2 …

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