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If $\overrightarrow{a}\;,\overrightarrow{b}\;,\overrightarrow{c}$ are three mutually perpendicular unit vectors,than prove that $|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{3}$

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  • For any two vectors $ \hat a \: and \: \hat b$ $(\hat a + \hat b)^2=(\hat a)^2+2\hat a.\hat b+(\hat b)^2=a^2+2\hat a.\hat b+b^2$ $(\hat a-\hat b)^2=a^2-2\hat a.\hat b+b^2$ $(\hat a+\hat b).(\hat a-\hat b)=a^2-b^2$
$ | \overrightarrow a + \overrightarrow b+\overrightarrow c|^2 = (\overrightarrow a+\overrightarrow b+\overrightarrow c)^2=(\overrightarrow a+\overrightarrow b+\overrightarrow c).(\overrightarrow a+\overrightarrow b+\overrightarrow c)$
$= \overrightarrow a^2+\overrightarrow b^2+\overrightarrow c^2+2\overrightarrow a.\overrightarrow b+2\overrightarrow b.\overrightarrow c+2\overrightarrow c.\overrightarrow a$
Now $ \overrightarrow a, \overrightarrow b, \overrightarrow c$ are unit vectors $ \therefore \overrightarrow a^2=\overrightarrow b^2=\overrightarrow c^2=1 $  and they are mutually perpendicular.
$ \therefore \overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow c=\overrightarrow c.\overrightarrow a=0$
$ \therefore |\overrightarrow a+\overrightarrow b+\overrightarrow c|^2=1+1+1=3$
$ |\overrightarrow a+\overrightarrow b+\overrightarrow c|= \sqrt 3$


answered Jun 1, 2013 by thanvigandhi_1
edited Jun 3, 2013 by thanvigandhi_1

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