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Home  >>  TN XII Math  >>  Vector Algebra
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Show that the vector $\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$ is equally inclined with the coordinate axes.

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  • 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$
answered May 31, 2013 by thanvigandhi_1
 

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