Question: Prove that $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = 0$ in the given triangle ABC.

Solution:

To prove: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = 0$



We know,

In ∆ABC,
Using triangle law of vector addition,
$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$ - (i)

Taking LHS,

$= \vec{AB} + \vec{BC} + \vec{CA}$

$= \vec{AB} + \vec{BC} - \vec{AC}$

Put value of $\vec{AC}$ from equation (i)

$= \vec{AB} + \vec{BC} - ( \vec{AB} + \vec{BC})$

$= \vec{AB} + \vec{BC} - \vec{AB} - \vec{BC}$

$= 0$

RHS #proved.