Parallelogram Law of Vectors

Parallelogram law of vectors: It states that, “If two vector quantities acting simultaneously on a body are represented in both magnitude and direction by two adjacent sides of a parallelogram drawn from a point, their resultant is given by the diagonal passing through the same point.”

To find the magnitude and direction of the resultant of the vector P and Q, a line OA is produced.

Parallelogram law of vectors

In triangle OBN,


OB2 = ON2 + BN2

or, R2 = (OA + AN)2 + BN2

or, R2 = OA2 + 2OA · AN + AN2 + BN2……… (i)

From right angled triangle ABN,

$$cos \theta = \frac{AN}{AB}$$
$$\therefore AN = ABCos \theta$$

Substituting the value of AN and OC in equation (i), we get:

R2 = OA2 + AB2 + 2OA . AB .ABCosθ

or, R2 = OA2 + AB2 + 2OA .OC.ABCosθ (Since, AB = OC)

or, R2 = P2 + 2PQCosθ  + Q2

$$\therefore R=  \sqrt{P^2 + 2PQcos \theta + Q^2}$$

The above expression gives the magnitude of resultant of P and Q.

To find the direction of R, let β be the angle made by the resultant R with direction of P.

Now, In triangle OBN,$$Tan\beta = \frac{BN}{ON} = \frac{BN}{OA + AN}…….(ii)$$

In triangle ABN,
$$Sin \theta = \frac{BN}{OB}$$
$$or, BN = ABSin \theta $$

Putting the value of BN in equation (ii), we get:

$$Tan\beta = \frac{AB . sin \theta }{OA + ABCos \theta }$$
$$or, Tan\beta = \frac{OC .sin \theta }{OA + OC.Cos \theta } $$

$$or, Tan \beta = \frac{Q.Sin \theta}{P + Q.Cos \theta } $$
$$\therefore \beta = Tan^{-1}( \frac{Q.Sin \theta}{P + Q.Cos \theta })$$

The above expression gives the direction of resultant R.


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