Question

State the law of parallelogram of vector addition and find the magnitude and direction of the resultant of vectors P and Q inclined at angle $\theta$ with each other. What happens when $\theta =0^0$ and $\theta ={90}^0$.

Answer 1

Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude & direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure. Let $\theta$ be the angle between P and Q and R be the resultant vectors. According to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.

So, $R=P+Q$

From triangle OCB

$O{B}^2=O{C}^2+B{C}^2$

$O{B}^2=(OA+AC{)}^2+B{C}^2$ .........$\left(1\right)$

In $\Delta ABC$

$\cos \theta =\frac{AC}{AB}$

$AC=AB\cos \theta$

$AC=OD\cos \theta =Q\cos \theta$

as $[AB=OD=Q]$

Also

$\cos \theta =\frac{BC}{AB}$

$BC=AB\sin \theta$

$BC=OD\sin \theta =Q\sin \theta$

[as $AB=OD=Q]$

Magnitude of Resultant substituting AC & BC in equation $\left(1\right)$ we get

$R^2=(P+Q\cos \theta {)}^2+(Q\sin \theta {)}^2$

$\therefore R=\sqrt{P^2+2PQ\cos \theta +Q^2}$ $\to$ Magnitude

From $\Delta$ ABC

$\tan \varphi =\frac{BC}{OC}=\frac{BC}{OA+AC}$

$\tan \varphi =\frac{Q\sin \theta }{P+Q\cos \theta }$

$\therefore \varphi ={\tan }^{-1}\left(\frac{Q\sin \theta }{P+Q\cos \theta }\right)$ .......$\left(2\right)$

When $\theta =0^o$

$\varphi ={\tan }^{-1}\left(\frac{Q\sin 0}{P+\cos 0}\right)$

$\varphi ={\tan }^{-1}\left(\frac{0}{1}\right)$

$\varphi ={\tan }^{-1}0$

$\varphi =0$

When $\theta ={90}^o$

$\varphi ={\tan }^{-1}\left(\frac{Q\sin {90}^o}{P+Q\cos {90}^o}\right)$

$\varphi ={\tan }^{-1}\left(\frac{Q}{P+0}\right)$

$\varphi ={\tan }^{-1}\left(\frac{Q}{P}\right)$.