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.