# 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.

In triangle OBN,

OB^{2 }= ON^{2 }+ BN^{2}

or, R^{2 }= (OA + AN)^{2} + BN^{2}

or, R^{2} = OA^{2 }+ 2OA · AN + AN^{2} + BN^{2}……… (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:

R^{2} = OA^{2} + AB^{2} + 2OA . AB .ABCosθ

or, R^{2} = OA^{2} + AB^{2} + 2OA .OC.ABCosθ (Since, AB = OC)

or, R^{2} = P^{2} + 2PQCosθ + Q^{2}

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

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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