Physics - Vectors Question with Solution | TestHub

Given vector $\vec{A}=2\hat{i}+3\hat{j}$

The magnitude of the vector is $|\vec{A}|=\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13}$

Let $\alpha$ be the angle made by the vector with the x-axis. Then,

$\cos \alpha =\frac{A_x}{|\vec{A}|}=\frac{2}{\sqrt{13}}\Rightarrow \alpha ={\cos }^{-1}\left(\frac{2}{\sqrt{13}}\right)$

Let $\beta$ be the angle made by the vector with the y-axis. Then,

$\cos \beta =\frac{A_y}{|\vec{A}|}=\frac{3}{\sqrt{13}}\Rightarrow \beta ={\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)$

Answer: ${\cos }^{-1}\frac{2}{\sqrt{13}}$ , ${\cos }^{-1}\frac{3}{\sqrt{13}}$