If two vectors acting at a point are represented by two sides of triangle taken in same order, then their resultant is represented by third side of triangle taken in opposite order:
Let us consider two vectors `vec A and vec B` as shown in figure such that they represent two sides of `Delta OMN,` then by triangle law of vector addition, we have `vec R = vec A + vec B`
Analytic method
Construction : From point N draw ND perpendicular on OM etended to meet at poind D.
(i) Magnitude of resultant vector `(vec R)`
In right angle `Delta OND,` we have
`ON ^(2) = OD ^(2) + ND ^(2)`
`or ON ^(2) = (OM + MD) ^(2) + ND ^(2) ...(i)`
From `Delta MND, cos theta = ( MD)/( MN)...(ii)`
`or MD = MN cos theta = |vec B| cos theta `
Also `sin theta = ( DN)/( MN)`
`or DN = MN sin theta =|vec B| cos theta`
Also `ON = |vec R| and OM = |vec A| ...(iv)`
Substituting the value of (ii), (iii) and (iv) in (i) we get
`|vec R | ^(2) = [ | vec A |+ |vec B|cos theta] ^(2) + |vec B|^(2) sin ^(2) theta `
`= |vec A | ^(2) + |vec B | ^(2) cos ^(2) theta + 2 |vec A| |vec B| cos theta + |vec B| ^(2) sin ^(2) theta `
`= |vec A | ^(2) + |vec B| ^(2) [cos ^(2) theta + sin ^(2) theta ] + 2 |vec A| |vec B|cos theta `
`R ^(2) = A ^(2) + B ^(2) + 2 AB cos theta (because sin ^(2) theta + cos ^(2) theta =1 )`
`or R = sqrt (A ^(2) + B ^(2) + 2 AB cos B )...(v)`
(ii) Directon of resultant vector `(vec R)`
It is clear from figure that `vec R` makes an angle `alpha ` with direction of `vec A.` From right angled `Delta ODN `
`tan alpha = (ND)/(OD) = (ND)/( OM + MD) = (B sin theta )/( A + B cos theta )`
`tan alpha = ( B sin theta )/( A + B cos theta ) implies alpha = tan ^(-1) ((B sin theta)/( A+ B cos theta)) ....(vi)`
Equations (v) and (vi) represent magnitude nd direction respectively.
