The vector $\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}}$ bisects the angle between the non-collinear vectors $\overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}} ,$ if.............

If vector $\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}}$ bisects the angle between the non-collinear vectors, then

$\overrightarrow {\rm{a}} \cdot (\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} ) = |\overrightarrow {\rm{a}} |\left| {\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} } \right|\cos \theta$

$\overrightarrow {\rm{a}} \cdot (\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} ) = a\sqrt {{a^2} + {b^2}} \cos \theta$

$\Rightarrow$ $\cos \theta = \frac{{\overrightarrow {\rm{a}} \cdot (\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} )}}{{a\sqrt {{a^2} + {b^2}} }}$

……(i)
and $\overrightarrow {\rm{b}} \cdot (\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} ) = |\overrightarrow {\rm{b}} | \cdot |\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} |\cos \theta$

$\begin{array}{llllllllllllllllllll}{\overrightarrow {\rm{b}} \cdot (\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} ) = b\sqrt {{a^2} + {b^2}} \cos \theta }&{}\end{array}$

[since, $\theta$ should be same]
$\Rightarrow$ $\cos \theta = \frac{{\overrightarrow {\rm{b}} \cdot (\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} )}}{{b\sqrt {{a^2} + {b^2}} }}$

…..(ii)
From Eqs. (i) and (ii),
$\frac{{\overrightarrow {\rm{a}} \cdot (\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} )}}{{a\sqrt {{a^2} + {b^2}} }}$

$= \frac{{\overrightarrow {\rm{b}} \cdot (\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} )}}{{b\sqrt {{a^2} + {b^2}} }} \Rightarrow \frac{{\overrightarrow {\rm{a}} }}{{|\overrightarrow {\rm{a}} |}}$

$= \frac{{\overrightarrow {\rm{b}} }}{{|\overrightarrow {\rm{b}} |}}$

$\therefore \widehat {\rm{a}} = \widehat {\rm{b}} \Rightarrow \overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}}$ are equal vectors.