Question

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

Answer 1

Let the two vectors be $\overrightarrow{a}$ and $\overrightarrow{b}$
$\text{Then,}|\overrightarrow{a}|=1$ and $|\overrightarrow{b}|=1\cdots \left(1\right)$
$\text{Also,}|\overrightarrow{a}+\overrightarrow{b}|=1$
We need to prove $|\overrightarrow{a}-\overrightarrow{b}|=\sqrt{3}$
$|\overrightarrow{a}+\overrightarrow{b}|=1$
$\Rightarrow |\overrightarrow{a}+\overrightarrow{b}{|}^2=1$
$\Rightarrow |\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}=1$
$\Rightarrow 1^2+1^2+2\overrightarrow{a}\cdot \overrightarrow{b}=1\left[\text{From (1)}\right]$
$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=\frac{-1}{2}\cdots \left(2\right)$

$\text{Now,}|\overrightarrow{a}-\overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2-2\overrightarrow{a}\cdot \overrightarrow{b}$
$=1+1-(-1)\left[\text{From}\right(1\left)\&\right(2\left)\right]$
$=3$
$\Rightarrow |\overrightarrow{a}-\overrightarrow{b}|=\sqrt{3}$

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