Comment
Share
Q)

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

$\begin{array}{1 1} |\overrightarrow a|=|\overrightarrow b| \\ |\overrightarrow a|=2|\overrightarrow b| \\ 2 |\overrightarrow a|=|\overrightarrow b| \\ If \;\overrightarrow a\; and \;\overrightarrow b\; are\; \perp\end{array}$

Comment
A)
• $\cos \theta=\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Let us consider two non-collinear vectors $\overrightarrow a$ and $\overrightarrow b$
Let $\overrightarrow a+\overrightarrow b$ be the vector which bisects the angle between the two vectors.
Hence $\theta_1=\theta_2$
Therefore $\cos \theta_1=\large\frac{\overrightarrow a.(\overrightarrow a+\overrightarrow b)}{|\overrightarrow a||\overrightarrow a+\overrightarrow b|}$
$\cos \theta_2=\large\frac{\overrightarrow b.(\overrightarrow a+\overrightarrow b)}{|\overrightarrow b||\overrightarrow a+\overrightarrow b|}$
Since $\theta_1=\theta_2$=> $\cos \theta_1=\cos \theta_2$
$\large\frac{\overrightarrow a.(\overrightarrow a+\overrightarrow b)}{|\overrightarrow a||\overrightarrow a+\overrightarrow b|}=\large\frac{\overrightarrow b.(\overrightarrow a+\overrightarrow b)}{|\overrightarrow b||\overrightarrow a+\overrightarrow b|}$
$=>\overrightarrow a=\overrightarrow b$
The vector $\overrightarrow{a}+\overrightarrow{b}$ bisects the angle between the non-collinear vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are equal vectors