Question

Vector $\overrightarrow{A}$ having magnitude $12\text{N}$ makes ${37}^{\circ }$ North of East and another vector $\overrightarrow{B}$ having magnitude $12\text{N}$ makes ${53}^{\circ }$ North of West. The resultant of these two vectors is

• A. $|\overrightarrow{R}|=12\sqrt{2}\text{N}$ and angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is $\alpha ={37}^{\circ }$.
• B. $|\overrightarrow{R}|=12\text{N}$ and angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is $\alpha ={53}^{\circ }$.
• C. $|\overrightarrow{R}|=15\text{N}$ and angle between vector $\overrightarrow{B}$ and resultant $\overrightarrow{R}$ is $\alpha ={37}^{\circ }$.
• D. $|\overrightarrow{R}|=12\sqrt{2}\text{N}$ and angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is $\alpha ={45}^{\circ }$ .

Answer 1

The correct option is D $|\overrightarrow{R}|=12\sqrt{2}\text{N}$ and angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is $\alpha ={45}^{\circ }$ .


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, $\cos {90}^{\circ }=0$
$\Rightarrow R=\sqrt{{12}^2+{12}^2}$
$\Rightarrow R=12\sqrt{2}\text{N}$
Also, angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is
$\tan \alpha =\frac{12}{12}=\text{1}$
$\Rightarrow \alpha ={\tan }^{-1}\text{1}={45}^{\circ }$

Hence, magnitude of resultat is $12\sqrt{2}\text{N}$ and angle between vector $\overrightarrow{A}$ and resultant $\overrightarrow{R}$ is $\alpha ={45}^{\circ }$.