Find the magnitude and direction of the resultant of two vectors A and B in terms of theirmagnitudes and angle theta between them .

Find the magnitude and direction of the resultant of two vectors A and

1.	Find the magnitude and direction of the resultant of two vectors A and B in terms of theirmagnitudes and angle theta between them .
Answer» Solution :LET OP and OQ REPRESENT two vectors A and B making an angle `theta` (Fig . 4.10) . Then , using the parallelogram method of vector ADDITION , OS represents the resultant vector R: R = A+B SN is normal to OP and PM is normal to OS . From the geometry of the figure, `OS^(2)=ON^(2)+SN^(2)` but`ON=OP+PN=A+Bcostheta` `SN=Bsintheta` `OS^(2)=(A+Bcostheta)^(2)+(Bsintheta)^(2)` or , `R^(2)=A^(2)+B^(2)+2ABcostheta` `R=sqrt(A^(2)+B^(2)+2ABcostheta)` In `DeltaOSN SN =OS `sinalpha=Rsinalpha,and` in `DeltaPSN,SN=PSsintheta=Bsintheta` Therefore , `R sinalpha=Bsintheta` or`(R)/(sintheta)=(B)/(sinalpha)` similarly `PM=Asinalpha=Bsinbeta` or `(A)/(sinbeta)=(B)/(sinalpha)` or Combining Eqs . (4.24b) and (4.24c),we get `(R)/(sintheta)=(A)/(sinbeta)=(B)/(sinalpha)` Using Eq . (424a). or , `tanalpha=(SN)/(OP+PN)=(Bsintheta)/(A+Bcostheta)` Equation (4.24a) gives the magnitude of the resultant and Eqs . (4.24e) and (4.24f) its direction.Equantion (4.24a) is known as the Law of cosines and Eq . (4.24d) as the Law of sines.