In the given hexagon ABCDEF, prove that: vec.AB + vec.BC + vec.CD + vec.DE + vec.ED + vec.FA = 0. | Vector Geometry | SciPiPupil

Question: In the given hexagon ABCDEF, prove that $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{FA} = 0$

Solution:

To prove: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{FA} = 0$

In a regular hexagon, opposite and parallel vectors in same direction are equal.

So,

$\vec{AB} = \vec{ED}$

$\vec{FA} = \vec{DC}$

$\vec{BC} = \vec{FE}$

Taking LHS,

$= \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{FA}$

[Put $\vec{AB} = \vec{ED}$, $\vec{FA} = \vec{DC}$, and $\vec{BC} = \vec{FE}$ from above]

$= \vec{ED} + \vec{FE} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{DC}$

$= \vec{ED} + \vec{DE} + \vec{FE} + \vec{EF} +  \vec{CD} + \vec{DC}$

$= - \vec{DE} + \vec{DE} + \vec{FE} - \vec{FE} + \vec{CD} - \vec{CD}$

$= 0 + 0 + 0$

$= 0$

= RHS

#proved