Skip to content
Question of 68

Q.State Parallelogram law of vector. Derive an expression for the magnitude and direction of the resultant vector.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2020Subjective· 4mImportance★★★★★
0% · 0/68 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The parallelogram law says two vectors acting at a point are represented by two adjacent sides of a parallelogram; their resultant is the diagonal from that point. Its magnitude is R=P2+Q2+2PQcos⁡θR=\sqrt{P^2+Q^2+2PQ\cos\theta}.

Statement (Parallelogram law of vector addition): If two vectors P⃗\vec{P} and Q⃗\vec{Q}, acting simultaneously at a point, are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from that point, then their resultant R⃗\vec{R} is represented completely (in both magnitude and direction) by the diagonal of the parallelogram drawn from the same point.

Derivation:

Let P⃗=OA→\vec P = \overrightarrow{OA} and Q⃗=OB→\vec Q = \overrightarrow{OB}, with the angle between them θ\theta. Complete the parallelogram OACBOACB, so AC→=OB→=Q⃗\overrightarrow{AC} = \overrightarrow{OB} = \vec Q. The resultant is the diagonal R⃗=OC→\vec R = \overrightarrow{OC}.

Drop a perpendicular from CC to the extension of OAOA, meeting it at NN. In right triangle ONCONC:

  • CN=Qsin⁡θCN = Q\sin\theta
  • AN=Qcos⁡θAN = Q\cos\theta, so ON=OA+AN=P+Qcos⁡θON = OA + AN = P + Q\cos\theta

Applying Pythagoras in △ONC\triangle ONC:

R2=ON2+CN2=(P+Qcos⁡θ)2+(Qsin⁡θ)2R^2 = ON^2 + CN^2 = (P+Q\cos\theta)^2 + (Q\sin\theta)^2 …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.