# Question #b00d1

Nov 3, 2015

$\left\mid b + a \right\mid = \sqrt{3}$

#### Explanation:

Consider the diagram below.
$\vec{b} = \vec{P Q}$ and $\left\mid \vec{P Q} \right\mid = \left\mid \vec{b} \right\mid = 1$

$\vec{a} = \vec{P Q} = \vec{T S}$ and $\left\mid \vec{P Q} \right\mid = \left\mid \vec{T S} \right\mid = \left\mid \vec{a} \right\mid = 1$

$\vec{a + b} = \vec{P S}$

$\angle S T U = \angle Q P T = {60}^{\circ}$ (the angle between $\vec{a}$ and $\vec{b}$
$\Rightarrow \angle P T S = {120}^{\circ}$
$\Rightarrow \angle R P T = \angle R S T = {30}^{\circ}$

The bisector of $\angle P T S$
divides $\triangle P T S$ into 2 congruent triangles
$\left\mid P R \right\mid = \left\mid R S \right\mid$

Since $\angle P T R = {60}^{\circ}$ and $\left\mid P T \right\mid = 1$
$\Rightarrow \left\mid P R \right\mid = \frac{\sqrt{3}}{2}$

and
$\left\mid P S \right\mid = \sqrt{3}$

Therefore $\left\mid \vec{a + b} \right\mid = \sqrt{3}$