Question

Let $\overrightarrow{\mathrm{A}}\mathrm{and}\overrightarrow{\mathrm{B}}$ be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angle 30° and 60° respectively, find the resultant.

Answer 1

Angle between $\overrightarrow{\mathrm{A}}\mathrm{and}\overrightarrow{\mathrm{B}}$, θ = 60° − 30° = 30°
$$\begin{gathered} \left|\overrightarrow{\mathrm{A}}\right|=\left|\overrightarrow{\mathrm{B}}\right|=10\mathrm{units} \\ \mathrm{The}\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{resultant}\mathrm{vector}\mathrm{is}\mathrm{given}\mathrm{by} \\ \mathrm{R}=\sqrt{A^2+A^2+2AA\cos \theta } \\ =\sqrt{{10}^2+{10}^2+2\times 10\times 10\times \cos 30°} \\ =\sqrt{200+200\cos 30°} \\ =\sqrt{200+200\times \frac{\sqrt{3}}{2}} \\ =19.3\mathrm{units} \end{gathered}$$



Let β be the angle between $\overrightarrow{\mathrm{R}}\mathrm{and}\overrightarrow{\mathrm{A}}$.
$$\begin{gathered} \therefore \mathrm{\beta }={\tan }^{-1}\left(\frac{A\sin 30°}{A+A\cos 30°}\right) \\ \Rightarrow \mathrm{\beta }={\tan }^{-1}\left(\frac{10\sin 30°}{10+10\cos 30°}\right) \\ \Rightarrow \mathrm{\beta }={\tan }^{-1}\left(\frac{1}{2+\sqrt{3}}\right)={\tan }^{-1}\left(\frac{1}{3.372}\right) \\ \Rightarrow \mathrm{\beta }={\tan }^{-1}\left(0.26795\right)=15° \end{gathered}$$

Angle made by the resultant vector with the X-axis = 15° + 30° = 45°

∴ The magnitude of the resultant vector is 17.3 and it makes angle of 45° with the X-axis.