Question

Vector $\overrightarrow{A}$ having magnitude $12\text{N}$ makes ${30}^{\circ }$ North of East and another vector $\overrightarrow{B}$ having magnitude $9\text{N}$ makes ${60}^{\circ }$ North of West. The resultant of these two vectors is

• A. $|\overrightarrow{R}|=15\text{N}$ and angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is $\alpha ={\tan }^{-1}$ $\frac{3}{4}$.
• B. $|\overrightarrow{R}|=15\text{N}$ and angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is $\alpha ={\tan }^{-1}$ $\frac{5}{4}$.
• C. $|\overrightarrow{R}|=15\text{N}$ and angle between vector $\overrightarrow{B}$ and resultant $\overrightarrow{R}$ is $\alpha ={\tan }^{-1}$ $\frac{3}{4}$.
• D. $|\overrightarrow{R}|=15\text{N}$ and angle between vector $\overrightarrow{B}$ and resultant $\overrightarrow{R}$ is $\alpha ={\tan }^{-1}$ $\frac{5}{4}$.

Answer 1

The correct option is A $|\overrightarrow{R}|=15\text{N}$ and angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is $\alpha ={\tan }^{-1}$ $\frac{3}{4}$.


From the diagram it can be seen that angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is ${90}^{\circ }$. So, the magnitude of resultant can be given as
$R$=$\sqrt{|\overrightarrow{A}{|}^2+|\overrightarrow{B}{|}^2}$
since, $\cos {90}^{\circ }=0$
$\Rightarrow R=\sqrt{{12}^2+9^2}$
$\Rightarrow R=15\text{N}$
Also, angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is
$\tan \alpha =\frac{9}{12}=\frac{3}{4}$
$\Rightarrow \alpha ={\tan }^{-1}\frac{3}{4}$

Hence, magnitude of resultat is $15\text{N}$ and angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is $\alpha ={\tan }^{-1}$ $\frac{3}{4}$.