# Find the angles which the vector $\overrightarrow{i}-\overrightarrow{j}+\sqrt{2}\overrightarrow{k}$ makes 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
The direction cosines of $\overrightarrow a = \overrightarrow i-\overrightarrow j+\sqrt 2 \overrightarrow k$ are obtained.
$|\overrightarrow a|=\sqrt{1+1+2}=\sqrt 4=2$
$\therefore l = \large\frac{a_1}{|\overrightarrow a|} = \large\frac{1}{2},m = \large\frac{a_2}{|\overrightarrow a|} = \large\frac{-1}{2}. n = \large\frac{a_3}{|\overrightarrow a|}=\large\frac{\sqrt 2}{2}=\large\frac{1}{\sqrt 2}$
Step 2
Then the angles made by $\overrightarrow a$ with the coordinate axes are
$\alpha = \cos^{-1}l = \cos^{-1} \large\frac{1}{2} = \large\frac{\pi}{3}$
$\beta = \cos^{-1}m = \cos^{-1} -\large\frac{1}{2} = \large\frac{2\pi}{3}$
$\gamma = \cos^{-1}n = \cos^{-1} \large\frac{1}{\sqrt 2} = \large\frac{\pi}{4}$

answered May 31, 2013
edited Jun 20, 2013