Question

How to find the angle between the vector and $x-$axis?

Answer 1

Calculation of angle between $x-$axis and vector:

Angle between vectors $\alpha ={\cos }^{-1}\left(\frac{\overrightarrow{\mathrm{a}.}\overrightarrow{\mathrm{b}}}{\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|}\right)$.

x-axis is represented by $\overrightarrow{\mathrm{i}}$.

Let vector be $\mathrm{x}\overrightarrow{\mathrm{i}}+\mathrm{y}\overrightarrow{\mathrm{j}}+\mathrm{z}\overrightarrow{\mathrm{k}}$

$$\begin{gathered} \overrightarrow{a}.\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{i}}.(\mathrm{x}\overrightarrow{\mathrm{i}}+\mathrm{y}\overrightarrow{\mathrm{j}}+\mathrm{z}\overrightarrow{\mathrm{k}}) \\ \overrightarrow{a}.\overrightarrow{\mathrm{b}}=\mathrm{x}\left[\because (\overrightarrow{\mathrm{i}}.\overrightarrow{\mathrm{i}}=1),(\overrightarrow{\mathrm{i}}.\overrightarrow{\mathrm{j}}=0),(\overrightarrow{\mathrm{i}}.\overrightarrow{\mathrm{k}}=0)\right] \\ \left|\overrightarrow{\mathrm{a}}\right|=\sqrt{1^2}[\because \left|\overrightarrow{\mathrm{v}}\right|=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}] \\ \left|\overrightarrow{\mathrm{a}}\right|=1 \\ \left|\overrightarrow{\mathrm{b}}\right|=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2} \\ \mathrm{\alpha }={\cos }^{-1}\left(\frac{\mathrm{x}}{1.\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}}\right) \end{gathered}$$

Therefore, the angle between the vector and x-axis is ${\cos }^{-1}\left(\frac{\mathrm{x}}{1.\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2}}\right)$ .