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Home  >>  TN XII Math  >>  Vector Algebra
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If$|\overrightarrow{a}+\overrightarrow{b}|=60, |\overrightarrow{a}-\overrightarrow{b}|=40$ and $|\overrightarrow{b}|=46$ find$|\overrightarrow{a}|.$

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  • For any two vectors $ \hat a \: and \: \hat b$ $(\hat a + \hat b)^2=(\hat a)^2+2\hat a.\hat b+(\hat b)^2=a^2+2\hat a.\hat b+b^2$ $(\hat a-\hat b)^2=a^2-2\hat a.\hat b+b^2$ $(\hat a+\hat b).(\hat a-\hat b)=a^2-b^2$
$(\overrightarrow a+\overrightarrow b)^2=\overrightarrow a^2+\overrightarrow b^2+2\overrightarrow a.\overrightarrow b \cos\theta$
$ 60^2 = \overrightarrow a^2+46^2+2\overrightarrow a.\overrightarrow b \cos\theta$..............(i)
$ (\overrightarrow a-\overrightarrow b)^2=\overrightarrow a^2+\overrightarrow b^2-2\overrightarrow a.\overrightarrow b \cos\theta$
$ 40^2=\overrightarrow a^2+46^2-2\overrightarrow a.\overrightarrow b \cos\theta $...............(ii)
Adding (i) and (ii)
$ 60^2+40^2 = 2(\overrightarrow a^2+46^2)$
$ \large\frac{3600+1600}{2} = \overrightarrow a^2+46^2$
$2600=\overrightarrow a^2+46^2$
$ \overrightarrow a^2=2600-2116=484$
$ |\overrightarrow a| = \sqrt{484} = 22$

 

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

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