Question

If vector $\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}$ and $|\overrightarrow{a}|+|\overrightarrow{b}|$ $=|\overrightarrow{c}|$ what is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$?

• A. ${90}^o$
• B. ${45}^o$
• C. $0^o$
• D. ${60}^o$

Answer 1

The correct option is C $0^o$

$\text{Step 1: Squaring the given equation and simplifying it}$

Let $\theta$ be the angle between $\overrightarrow{a}$ and $\overrightarrow{b}.$

Given: $\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}$

Squaring on both sides :

$\therefore (\overrightarrow{a}+\overrightarrow{b})\cdot (\overrightarrow{a}+\overrightarrow{b})=\overrightarrow{c}\cdot \overrightarrow{c}$

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

$\Rightarrow$ $|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2|\overrightarrow{a}||\overrightarrow{b}|\cos \theta =|\overrightarrow{c}{|}^2$ $....\left(1\right)$ $[\because$ $\overrightarrow{a}\cdot \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos \theta ]$

We are also given;

$|\overrightarrow{a}+|\overrightarrow{b}|=|\overrightarrow{c}|$

Squaring above equation

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

$\text{Step 2: Comparing the equations}$

Comparing $e{q}^n\left(1\right)$ and $\left(2\right)$

We get, $\cos \theta =1$

$\Rightarrow$ $\theta =0^{\circ }$

Hence option $C$ is correct.