Question

Can we add 3 vectors of equal magnitude and get a zero vector?

• A. True
• B. False

Answer 1

The correct option is A

True

Hmmmmm.....

Let's take 3 vectors $\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}$ with equal magnitude a

Now if I start by adding $\overrightarrow{A}$ & $\overrightarrow{B}$ and I get a resultant $\overrightarrow{R}+\overrightarrow{C}$. This should give me a null vector.

$\Rightarrow \overrightarrow{R}+\overrightarrow{C}=0$

$\Rightarrow \overrightarrow{R}$ should be negative of $\overrightarrow{C}$

$\overrightarrow{R}=-\overrightarrow{C}$

$\Rightarrow$R will have the same magnitude as $\overrightarrow{C}$ just opposite direction

$\Rightarrow |\overrightarrow{R}|=a$

Let me verify that mathematically if it is possible or not.

Okay,$\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{R}$

Let's assume they have angle $\theta$ between them

We know $|\overrightarrow{A}|=|\overrightarrow{B}|=a$

Also$|\overrightarrow{R}|=a$

$|\overrightarrow{R}|=\sqrt{|\overrightarrow{A}{|}^2+|\overrightarrow{B}{|}^2+2|\overrightarrow{A}||\overrightarrow{B}|cos\theta }$

$a=\sqrt{a^2+a^2+2a^{2}cos\theta }$

$a^2=2a^2+2a^{2}cos\theta$

$cos\theta =-\frac{1}{2}$

$\Rightarrow \theta ={120}^{\circ }$

$tan\varphi =\frac{|\overrightarrow{B}|sin\theta }{|\overrightarrow{A}|+|\overrightarrow{B}|cos\theta }=\frac{asin{120}^{\circ }}{a+acos{120}^{\circ }}=\frac{a\frac{\sqrt{3}}{2}}{2a+a\frac{1}{2}}=\sqrt{3}$

Now let me draw the image

Now

$\because \overrightarrow{R}=-\overrightarrow{C}$

$\Rightarrow$

Now we see that all the 3 vectors should make ${120}^{\circ }$ with each other