If ABCDEF is a regular hexagon with AB-a and BC-b-where -a-b- then CE

In Δ ABC, AB+BC=AC ⇒AC=a+b .....(i) AD is parallel to BC and AD = 2BC, therefore nbsp;AD=2b. In Δ ACD, AC+CD=AD ⇒CD=2 b (a+b)=b a ∴ In Δ CDE, CE=CD+DE=b a+( ...

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Byju's Answer

Standard XIII

Mathematics

Addition of Vectors

If ABCDEF is ...

Question

If $A B C D E F$ is a regular hexagon with $- - \to A B = \to a$ and $- - \to B C = \to b$ ,
(where $| \to a | = | \to b |$ ) then $- - \to C E$

\to v - \to b

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- \to b

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\to b - 2 \to a

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None of these

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Solution

The correct option is C $\to b - 2 \to a$
In $Δ A B C, - - \to A B + - - \to B C = - - \to A C$
$\Rightarrow - - \to A C = \to a + \to b . . . . . (i)$
$A D$ is parallel to $B C$ and $A D = 2 B C,$ therefore $- - \to A D = 2 \to b$ .

In $Δ A C D, - - \to A C + - - \to C D = - - \to A D$
$\Rightarrow - - \to C D = 2 \to b - (\to a + \to b) = \to b - \to a$
$∴$ In
$Δ C D E, - - \to C E = - - \to C D + - - \to D E = \to b - \to a + (- \to a) = \to b - 2 \to a$

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If ABCDEF is a regular hexagon with −−→AB=→a and −−→BC=→b, (where |→a|=|→b|) then −−→CE

If $A B C D E F$ is a regular hexagon with $- - \to A B = \to a$ and $- - \to B C = \to b$ ,
(where $| \to a | = | \to b |$ ) then $- - \to C E$