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If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is $\sqrt{3}$.

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  • For any two vectors $ \hat a \: and \: \hat b$ $(\hat a + \hat b)^2=(\hat a)^2+2\hat a.\hat b+(\hat b)^2=a^2+2\hat a.\hat b+b^2$ $(\hat a-\hat b)^2=a^2-2\hat a.\hat b+b^2$ $(\hat a+\hat b).(\hat a-\hat b)=a^2-b^2$
Step 1
$ \hat a \: and \: \hat b$ are unit vectors with $ \hat a +\hat b=\hat c$ also a unit vector.
Step 2
$ | \hat a+\hat b|^2=|\hat c|^2=\hat a^2+2\hat a.\hat b+\hat b^2$
$ \therefore 1= 1+(1)(1) \cos\theta+1$
or $ \cos\theta = \large\frac{-1}{2}$
Step 3
Now $ |\hat a-\hat b|^2 = \hat a^2-2\hat a.\hat b+\hat b^2$
$ = 1-2(1)(1) \cos\theta+1$
$ = 2-2 \bigg( -\large\frac{1}{2} \bigg) = 2+1=3$
$ \therefore | \hat a-\hat b|= \sqrt 3$
answered Jun 1, 2013 by thanvigandhi_1

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