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ABCDEF is a regular hexagon with point ‘O’ as centre. The value of $\vec{A B} + \vec{A C} + \vec{A D} + \vec{A E} + \vec{A F}$ is

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\vec{O}

6 \vec{A O}

2 \vec{A O}

4 \vec{A O}

answer is C.

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Detailed Solution

Given that ABCDEF is a regular hexagon with centre O as shown

Question Image

We are asked to evaluate $\vec{A B} + \vec{A C} + \vec{A D} + \vec{A E} + \vec{A F}$ .

We can have using triangle law of vectors

Question Image

$\vec{O A} + \vec{A B} = \vec{O B} o r \vec{A B} = \vec{O B} - \vec{O A}$

Using this analogy we can have from regular hexagon.

$\vec{A B} = \vec{O B} - \vec{O A}$ , $| \vec{O A} | = | \vec{O B} | = | \vec{O C} | = | \vec{O D} | = | \vec{O E} | = | \vec{O F} |$ ,

$\vec{O A} + \vec{O D} = 0$

$\vec{A C} = \vec{O C} - \vec{O A}$ $\vec{O E} + \vec{O B} = 0$

$\vec{A D} = \vec{O D} - \vec{O A}$ $\vec{O F} + \vec{O C} = 0$

$\vec{A E} = \vec{O E} - \vec{O A}$

$\vec{A F} = \vec{O F} - \vec{O A}$

$\Rightarrow \vec{A B} + \vec{A C} + \vec{A D} + \vec{A E} + \vec{A F} = (\vec{O B} + \vec{O C} + \vec{O D} + \vec{O E} + \vec{O F}) - 5 \vec{O A}$

$\Rightarrow \vec{O D} - 5 \vec{O A} ∵ \vec{O D} = \vec{A O}$

$\Rightarrow \vec{A O} + 5 (\vec{A O})$

$∴ \vec{A B} + \vec{A C} + \vec{A D} + \vec{A E} + \vec{A F} = 6 \vec{A O}$

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