# Show that the vector $\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$ is equally inclined with the coordinate axes.

Toolbox:
• If $\overrightarrow a = a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k$ then $|\overrightarrow a|=\sqrt{a_1^2+a_2^2+a_3^3}$
• The direction cosines of vector $\overrightarrow a=a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k$ are $l = \large\frac{a_1}{|\overrightarrow a|}, m = \large\frac{a_2}{|\overrightarrow a|}, n = \large\frac{a_3}{|\overrightarrow a|}$. The angles made by $\overrightarrow a$ with the coordinate axes are $\cos^{-1}l, \: \cos^{-1}m, \: \cos^{-1}n.$
Step 1
$\overrightarrow a = \overrightarrow i+\overrightarrow j+\overrightarrow k$
$|\overrightarrow a|=\sqrt{1^2+1^2+1^2}=\sqrt 3$
The DC's are $l=\large\frac{1}{\sqrt 3}, m = \large\frac{1}{\sqrt 3}, n = \large\frac{1}{\sqrt 3}$
Step 2
$\therefore$ the angles of inclination with the coordinate axes are $\cos^{-1}\large\frac{1}{\sqrt 3}= \alpha = \beta = \gamma$