Question

If vectors AB, AC,AD,AE and AF act at the regular hexagon ABCDEF . prove that resultant is 6 AO where O is the centre of the hexagon

Answer 1

given: ABCDEF is a regular hexagon with center O.

AB+AC+AD+AE+AF=AB+(AC+AF)+AD+AE=AB+(AC+CD)+AD+AE [since AF=CD]=AB+AD+AD+AE=2AD+(AB+AE)=2AD+(ED+AE) [since AB=ED=2AD+AD=3AD=3*(2AO) [since O is the center and AO=OD=6AO

Answer 2

Vector AB + AC + AD + AE + AF = 6AO

Given,

Five vectors,

AB, AC, AD, AE, and AF

To Find,

Prove that the resultant of the given vectors are 6AO.

Solution,

Given that, the five vectors are acting at a hexagon ABCDEF.

So in a hexagon,

$\vec AO$ = $\vec AB$ + $\vec AF$

$\vec AC$ = $\vec AB$+ $\vec BC$ which is equal to $\vec AB$ + $\vec AO$.

$\vec AE$ = $\vec AF$ +$\vec FE$ which is also equal to $\vec AF$ +$\vec AO$A$\vec AO$

$\vec AD$  = 2* $\vec AO$

$\vec AB + \vec AC + \vec AD+ \vec AE + \vec AF$= $\vec AB +\vec AF + \vec AC + \vec AF +\vec AD$

2($\vec AB + \vec AF$) + 2$\vec AD$ + 2$\vec AD$  ,

$\vec AB$ + $\vec AF$ = $\vec AO$

$\vec AD$ = 2$\vec AO$

$2\vec AO + 2 \vec AO +2 \vec AO$ = 6$\vec AO$

Hence, 6AO is the resultant of five vectors.

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