State parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector.

Explanation

Statement : If two vectors acting at a point are represented by the adjacent sides of a parallelogram in magnitude and direction, then their resultant is represented by the diagonal of the parallelogram in magnitude and direction drawn from the same point.
Explanation : Let two forces `vec(P)` and `vec(Q)` act at a point O. Let `theta` be the angle between two forces. Let the side `OA = vec(P)` and `OB = vec(Q)`. The parallelogram OACB is completed. The points O and C are joined.
Now `OC = vec(R)`
Resultant magnitude :
In fig `vec(OA) = vec(P), vec(OB) = vec(Q), vec(OC) = vec(R)`
In the triangle `COD, OC^(2) = OD^(2) + CD^(2)`
`OC^(2) = (OA + AD)^(2) + CD^(2) (because OD = OA + AD)`
`OC^(2) = OA^(2) + AD^(2) + 2OA. D + CD^(2)`

`OC^(2) = OA^(2) + AC^(2) + 2OA.AD` ..........(1)
from `Delta^(l e) CAD, AD^(2) + CD^(2) = AC^(2)`
From `Delta^(l e) CAD, cos theta = (AD)/(AC)` ...........(2)
`AD = AC cos theta`
`therefore R^(2) = P^(2) + Q^(2) + 2PQ cos theta`
`R = sqrt(P^(2) + Q^(2) + 2PQ cos theta)` ...........(3)
Resultant direction :
Let `alpha` be the angle made by the resultant vector `vec(R)` with `vec(P)`
Then `tan alpha = (CD)/(OD)`
`tan alpha = (CD)/(OA + AD)` ..........(4)
In the triangle CAD, `sin theta = (CD)/(AC)`
`CD = AC sin theta`
`CD = Q sin theta` ...........(5)
`therefore tan alpha = (Q sin theta)/(P + Q cos theta) (because AD = Q cos theta)`
`alpha = tan^(-1)((Q sin theta)/(P + Q cos theta))` ..............(6)