Question

Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the coordinate axes.

Answer 1

$$\begin{gathered} \text{Let}{\theta }_1\text{be the angle between}\overrightarrow{a}\text{and}x-\text{axis.} \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}=\sqrt{3} \\ \overrightarrow{b}=\stackrel{\wedge }{i}\text{(Because}\stackrel{\wedge }{i}\text{is the unit vector along}\mathit{\text{x}}\text{-axis)} \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(1\right)}^2}=\sqrt{1}=1 \\ \overrightarrow{a}.\overrightarrow{b}=1+0+0=1 \\ \cos {\theta }_1=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{1}{\left(\sqrt{3}\right)\left(1\right)}=\frac{1}{\sqrt{3}} \\ \Rightarrow {\theta }_1={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)...\left(1\right) \\ \text{Let}{\theta }_2\text{be the angle between}\overrightarrow{a}\text{and}y-\text{axis.} \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}=\sqrt{3} \\ \overrightarrow{b}=\stackrel{\wedge }{j}\text{(Because}\stackrel{\wedge }{j}\text{is the unit vector along}\mathit{\text{y}}\text{-axis)} \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(1\right)}^2}=\sqrt{1}=1 \\ \overrightarrow{a}.\overrightarrow{b}=0+1+0=1 \\ \cos {\theta }_2=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{1}{\left(\sqrt{3}\right)\left(1\right)}=\frac{1}{\sqrt{3}} \\ \Rightarrow {\theta }_2={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)...\left(2\right) \\ \text{Let}{\theta }_3\text{be the angle between}\overrightarrow{a}\text{and}z-\text{axis.} \\ \left|\overrightarrow{a}\right|=\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}=\sqrt{3} \\ \overrightarrow{b}=\stackrel{\wedge }{k}\text{(Because}\stackrel{\wedge }{k}\text{is the unit vector along}\mathit{\text{z}}\text{-axis)} \\ \left|\overrightarrow{b}\right|=\sqrt{{\left(1\right)}^2}=\sqrt{1}=1 \\ \overrightarrow{a}.\overrightarrow{b}=0+0+1=1 \\ \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{1}{\left(\sqrt{3}\right)\left(1\right)}=\frac{1}{\sqrt{3}} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)...\left(3\right) \\ \text{From (1), (2) and (3), the given vector is equally inclined to the coordinate axes.} \end{gathered}$$