Question

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

Answer 1

Answer:k=2

Step-by-step explanation:

Answer 2

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.