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If the magnitude of difference of two unit vectors is $\sqrt{3}$, then the magnitude of sum of the two vectors is
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$|\hat{a}-\hat{b}|=\sqrt{3}$
$\Rightarrow|\hat{a}-\hat{b}|^{2}=(\sqrt{3})^{2}$
$\Rightarrow \hat{a} . \hat{a}+\hat{b} \cdot \hat{b}-2 \hat{a} \cdot \hat{b}=3$
$\Rightarrow 2 \hat{a} \cdot \hat{b}=-1$
Now; $|\hat{a}+\hat{b}|^{2}=\hat{a} \cdot \hat{a}+\hat{b} \cdot \hat{b}+2 \hat{a} \cdot \hat{b}=1+1-1$
$\Rightarrow|\hat{a}+\hat{b}|^{2}=1$
$\Rightarrow|\hat{a}+\hat{b}|=1$
$\Rightarrow|\hat{a}-\hat{b}|^{2}=(\sqrt{3})^{2}$
$\Rightarrow \hat{a} . \hat{a}+\hat{b} \cdot \hat{b}-2 \hat{a} \cdot \hat{b}=3$
$\Rightarrow 2 \hat{a} \cdot \hat{b}=-1$
Now; $|\hat{a}+\hat{b}|^{2}=\hat{a} \cdot \hat{a}+\hat{b} \cdot \hat{b}+2 \hat{a} \cdot \hat{b}=1+1-1$
$\Rightarrow|\hat{a}+\hat{b}|^{2}=1$
$\Rightarrow|\hat{a}+\hat{b}|=1$
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