Question

Vector $\overrightarrow{R}$ having magnitude $25\text{N}$ makes ${37}^{\circ }$ with $x$-axis counterclockwise and vector $\overrightarrow{Q}$ having magnitude of $10\text{N}$ makes ${53}^{\circ }$ with negative $x$-axis clockwise as shown in the figure,
if $\overrightarrow{R}=\overrightarrow{P}+\overrightarrow{Q}$, then find direction of vector $\overrightarrow{P}$

• A. ${\tan }^{-1}\left(\frac{13}{14}\right)$ from positive x-axis counterclockwise
• B. ${\tan }^{-1}\left(\frac{7}{13}\right)$ from positive x-axis counterclockwise
• C. ${\tan }^{-1}\left(\frac{7}{26}\right)$ from positive x-axis counterclockwise
• D. ${\tan }^{-1}\frac{17}{26}$ from positive x-axis counterclockwise

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{7}{26}\right)$ from positive x-axis counterclockwise

According to the question,
$|\overrightarrow{R}|=25\text{N}$, $|\overrightarrow{Q}|=10\text{N}$ and
$\overrightarrow{R}=\overrightarrow{P}+\overrightarrow{Q}$
So, we have
$\overrightarrow{R}=\left(25\cos {37}^{\circ }\right)\hat{i}+\left(25\sin {37}^{\circ }\right)\hat{j}$
$\overrightarrow{R}=\left(20\right)\hat{i}+\left(15\right)\hat{j}$
$\overrightarrow{Q}=(-10\cos {53}^{\circ })\hat{i}+\left(10\sin {53}^{\circ }\right)\hat{j}$
$\overrightarrow{Q}=(-6)\hat{i}+\left(8\right)\hat{j}$
Now,

$\overrightarrow{P}=\overrightarrow{R}-\overrightarrow{Q}$

$\overrightarrow{P}=\left(26\right)\hat{i}+\left(7\right)\hat{j}$
Thus, direction of the $\overrightarrow{P}$ is given by
$\tan \alpha =\frac{7}{26}$
$\alpha ={\tan }^{-1}\left(\frac{7}{26}\right)$ from positive $x$-axis counterclockwise.