ABCDEF is a regular hexagon with centre at the origin such that vector AD + vector EB+ vector FC= k vectorED . Then the value of k is
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Answer:k=2
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Value of k is 2a.
Regular hexagon = ABCDEF, (Given)
Vectors = AF = CD, FE = BC, ED = AB (Given)
Therefore,
AB+AC+AD+AE+AF
= AB+(AB+BC)+(AB+BC+CD)+(AF+FE)+AF
= AB+(AB+BC)+(AB+BC+CD)+(CD+BC)+CD
= 3AB+3BC+3CD
= 3(AB+BC+CD)
= 3AD
Magnitude of vector AC-
AC² = AB² + BC² − 2AB × BC cos(120°)
AC² = a²+a²−2a²×(−12)
AC² = 2a²+a² = 3a²
|AC| = a√3
∠ACD = ∠BCD - 30°
= 120° - 30°
= 90°
In right ΔACD
AD² = AC² + CD²
= 3a² + a² = 4a²
AD = 2a
Thus, the value of k is 2a.
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