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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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## 1 Answer

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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

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