Question

Two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ have magnitudes $3$ units and $4$ units respectively. $\overrightarrow{R}$ is the resultant $\overrightarrow{P}$ and $\overrightarrow{Q}$ . Find the magnitude and direction of $\overrightarrow{R}$ when the angle $\theta$ between $\overrightarrow{P}$ and $\overrightarrow{Q}$ is ${30}^o$.

Answer 1

The magnitude of the resultant is given as,

$R^2=P^2+Q^2+2PQ\cos \theta$

$R^2={\left(3\right)}^2+{\left(4\right)}^2+2\times 3\times 4\cos {30}^{\circ }$

$R=6.766units$

Let the angle between the resultant and $P$ is $\alpha$.

The direction of resultant is given as,

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

$\tan \alpha =\frac{4\times \sin {30}^{\circ }}{4+3\cos {30}^{\circ }}$

$\alpha ={16.86}^{\circ }$

Thus, the magnitude and direction of resultant is $6.766units$ and ${16.86}^{\circ }$ respectively.