Question

If $|\overrightarrow{a}|$ and $|\overrightarrow{b}|$ are equal then $\overrightarrow{a}+\overrightarrow{b}$ is equally inclined to both vectors.
If true enter 1, else enter 0.

Answer 1

Given $|\overrightarrow{a}|=|\overrightarrow{b}|$,

Let angle between $(\overrightarrow{a}+\overrightarrow{b})$ and $\overrightarrow{a}$ is ${\theta }_1$ and $(\overrightarrow{a}+\overrightarrow{b})$ and $\overrightarrow{b}$ is ${\theta }_2.$

$\Rightarrow (\overrightarrow{a}+\overrightarrow{b}).\overrightarrow{a}=|\overrightarrow{a}{|}^2+\overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}+\overrightarrow{b}||\overrightarrow{a}|cos\left({\theta }_1\right)$ ....1

$\Rightarrow (\overrightarrow{a}+\overrightarrow{b}).\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{b}+|\overrightarrow{b}{|}^2=|\overrightarrow{a}+\overrightarrow{b}||\overrightarrow{b}|cos\left({\theta }_2\right)$

$\Rightarrow$ Using $|\overrightarrow{a}|=|\overrightarrow{b}|$, $\Rightarrow |\overrightarrow{a}{|}^2+\overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}+\overrightarrow{b}||\overrightarrow{a}|cos\left({\theta }_2\right)$ ....2

$\Rightarrow$ Equating equations 1 and 2, $\Rightarrow Cos\left({\theta }_1\right)=cos\left({\theta }_2\right)$ $\Rightarrow {\theta }_1={\theta }_2$