Question

The angle that the vector $\overrightarrow{A}=2\hat{i}+3\hat{j}$ makes with y-axis is :

• A. $ta{n}^{-1}\left(3/2\right)$
• B. $ta{n}^{-1}\left(2/3\right)$
• C. $si{n}^{-1}\left(2/3\right)$
• D. $co{s}^{-1}\left(3/2\right)$

Answer 1

Answer: (B)

$ta{n}^{-1}\left(2/3\right)$

$\overrightarrow{A}=2\hat{i}+3\hat{j}$
$\overrightarrow{A}.\hat{j}=\sqrt{13}\cos \theta$
$3=\sqrt{13}\cos \theta$
$\theta ={\cos }^{-1}(\frac{3}{\sqrt{13}})$
Using formula, $\tan \left({\cos }^{-1}x\right)=\frac{\sqrt{1-x^2}}{x}$ we will get:
$\theta ={\tan }^{-1}\frac{\sqrt{1-\frac{9}{13}}}{\frac{3}{\sqrt{13}}}$

$\theta ={\tan }^{-1}(\frac{2}{3})$