State parallelogram law of vector addition. Find the magnitude and direction of resultant of two vectors `vecA` and `vecB` inclined at an angle '`theta`' with each other by using this law.

Explanation

If two vectors are represented both in magnitude and direction by the adjacent side of a parallelogrm, then the resultant vector is represented both in magnetide and direction by the diagonal of parallelogram passing through same point.

Consider two vectors `vec P and vec Q` as shown in the such that tehy represent two adjacent side of parallelogrm OLMN. The angle betwen two vectors is `theta.` Fom `DeltaOLM`
`vec (OM) = vec(OL) + vec(LM) or vecR + vecP + vecQ`
Analytic method
Construction (i) Magnitude of resultant vector `(vec R)`
From right angled `Delta OMD`
`OM ^(2) = OD ^(2) + MD ^(2) = (O L + LD) ^(2) + MD ^(2) " "...(i)`
From light angle `Delta LDM`
`(LD)/( LM ) = cos theta or LD = LM cos theta = Q cos theta " "...(ii)`
Also `(MD)/( LM ) = sin theta or MD = LM sin theta = Q sin theta " " ...(iii)`
Also `OM = R and OL = P " "...(iv)`
Substituting the values fo (ii),(iii) and (iv) in (i) we get
`R ^(2) = ( P + Q cos theta ) ^(2) + Q ^(2) sin ^(2) theta = P^(2) + Q ^(2) cos theta + 2 PQcos theta + Q ^(2) sin ^(2) theta`
or `R^(2) = P ^(2) + Q ^(2) (cos ^(2) theta + sin ^(2) theta) + 2 PQ cos theta =P ^(2) + Q ^(2) + 2 P Q cos theta `
or `R = sqrt ( P ^(2) + Q ^(2) + 2 PQ cos theta) " "...(v)`
(ii) Direction of resultant vector , `vec R`
From right angled `Delta ODM`
`tan alpha = ( MD)/( OD) = ( MD)/( OL + LD) = ( Q sin theta)/( P + Qcos theta)`
`tan alpha = ( Q sin theta)/( P + Q cos theta) or alpha = tan ^(-1) ((Q sin theta)/( P + Q cos theta) ) " "...(iv)`
Equation (v) and (vi) represnt the magnitude and direction of rueulstant vector respectively.