Question
If $\hat{\text{a}},\hat{\text{b}}$ are unit vectors such that $\hat{\text{a}}+\hat{\text{b}}$ is a unit vector, write the value of $\big|\hat{\text{a}}-\hat{\text{b}}\big|.$
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Answer
Given that $\hat{\text{a}}$ and $\hat{\text{b}}$ are unit vectors such that $\hat{\text{a}}+\hat{\text{b}}$ is a unit vector.
$\Rightarrow|\hat{\text{a}}|=\big|\hat{\text{b}}\big|=\big|\hat{\text{a}}+\hat{\text{b}}\big|=1\dots(1)$
Now,
$\big|\hat{\text{a}}+\vec{\text{b}}\big|=1$
Squaring both sides, we get
$|\vec{\text{a}}|^2+\big|\hat{\text{b}}\big|^2+2\hat{\text{a}}.\hat{\text{b}}=1$
$\Rightarrow1+1+2\hat{\text{a}}.\hat{\text{b}}=1$ [Form (1)]
$\Rightarrow\hat{\text{a}}.\hat{\text{b}}=\frac{-1}{2}\dots(2)$
Now,
$\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=|\text{a}|^2+\big|\hat{\text{b}}\big|^2-2\hat{\text{a}}.\hat{\text{b}}$
$=1+1-2\big(\frac{-1}{2}\big)=3$ [From (1) and (2)]
$\therefore\big|\hat{\text{a}}-\hat{\text{b}}\big|=\sqrt{3}$
$\Rightarrow|\hat{\text{a}}|=\big|\hat{\text{b}}\big|=\big|\hat{\text{a}}+\hat{\text{b}}\big|=1\dots(1)$
Now,
$\big|\hat{\text{a}}+\vec{\text{b}}\big|=1$
Squaring both sides, we get
$|\vec{\text{a}}|^2+\big|\hat{\text{b}}\big|^2+2\hat{\text{a}}.\hat{\text{b}}=1$
$\Rightarrow1+1+2\hat{\text{a}}.\hat{\text{b}}=1$ [Form (1)]
$\Rightarrow\hat{\text{a}}.\hat{\text{b}}=\frac{-1}{2}\dots(2)$
Now,
$\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=|\text{a}|^2+\big|\hat{\text{b}}\big|^2-2\hat{\text{a}}.\hat{\text{b}}$
$=1+1-2\big(\frac{-1}{2}\big)=3$ [From (1) and (2)]
$\therefore\big|\hat{\text{a}}-\hat{\text{b}}\big|=\sqrt{3}$
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