Find the magnitude and direction of the resultant of two vectors $A$ and $B$ in terms of their magnitudes and the angle $\theta$ between them.
(N/A) Let $OP$ and $OQ$ represent the two vectors $A$ and $B$ making an angle $\theta$. Using the parallelogram method of vector addition,$OS$ represents the resultant vector $R = A + B$. $SN$ is drawn normal to $OP$ extended. From the geometry of the figure:
$ON = OP + PN = A + B \cos \theta$
$SN = B \sin \theta$
In right-angled triangle $\Delta OSN$,by Pythagoras theorem:
$OS^2 = ON^2 + SN^2$
$R^2 = (A + B \cos \theta)^2 + (B \sin \theta)^2$
$R^2 = A^2 + B^2 \cos^2 \theta + 2AB \cos \theta + B^2 \sin^2 \theta$
$R^2 = A^2 + B^2(\cos^2 \theta + \sin^2 \theta) + 2AB \cos \theta$
$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
For the direction,let $\alpha$ be the angle that the resultant $R$ makes with vector $A$. In $\Delta OSN$:
$\tan \alpha = \frac{SN}{ON} = \frac{B \sin \theta}{A + B \cos \theta}$
$\alpha = \tan^{-1} \left( \frac{B \sin \theta}{A + B \cos \theta} \right)$
$ON = OP + PN = A + B \cos \theta$
$SN = B \sin \theta$
In right-angled triangle $\Delta OSN$,by Pythagoras theorem:
$OS^2 = ON^2 + SN^2$
$R^2 = (A + B \cos \theta)^2 + (B \sin \theta)^2$
$R^2 = A^2 + B^2 \cos^2 \theta + 2AB \cos \theta + B^2 \sin^2 \theta$
$R^2 = A^2 + B^2(\cos^2 \theta + \sin^2 \theta) + 2AB \cos \theta$
$R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$
For the direction,let $\alpha$ be the angle that the resultant $R$ makes with vector $A$. In $\Delta OSN$:
$\tan \alpha = \frac{SN}{ON} = \frac{B \sin \theta}{A + B \cos \theta}$
$\alpha = \tan^{-1} \left( \frac{B \sin \theta}{A + B \cos \theta} \right)$
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