Home
Class 12
MATHS
If ABCDEF is regular hexagon, then AD+EB...

If ABCDEF is regular hexagon, then AD+EB+FC is

A

2 `vec(AB)`

B

3 `vec(AB)`

C

4` A vec(AB)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
Consider the regular hexagon ABCDEF with centre at O (origin)

`vec(AD) + vec(EB) + vec(FC) = 2 vec(AO) + 2 vec(OB) + 2 vec(OC)`
`" " = 2 ( vec(AO) + vec(OB) ) + 2 vec(OC)`
`" " = 2 vec(AB) + 2 vec(AB) `
`" " ( because vec(OC) = vec(AB))`
`" " = 4 vec(AB)`
`vecR = vec(AB) + vec(AC) + vec(AD) + vec(AE) + vec(AF)`
`= vec(ED) + vec(AC) + vec(AD) + vec(AE) +vec(CD)`
`( because vec(AB) = vec(ED) and vec(AF)= vec(CD))`
`= ( vec(AC) + vec(CD)) + ( vec(AE) + vec(ED)) + vec(AD)`
`= vec(AD) + vec(AD) + vec(AD) = 3 vec(AD) = 6 vec(AO)`
Promotional Banner

Topper's Solved these Questions

  • INTRODUCTION TO VECTORS

    CENGAGE ENGLISH|Exercise INTEGER TYPE|8 Videos

  • INTRODUCTION TO VECTORS

    CENGAGE ENGLISH|Exercise ARCHIVES SUBJECTIVE TYPE|9 Videos

  • INTRODUCTION TO VECTORS

    CENGAGE ENGLISH|Exercise REASONING TYPE|11 Videos

  • INTEGRALS

    CENGAGE ENGLISH|Exercise All Questions|764 Videos

  • INVERSE TRIGONOMETRIC FUNCTIONS

    CENGAGE ENGLISH|Exercise Archives (Numerical Value type)|2 Videos

Similar Questions

Explore conceptually related problems

If ABCDEF is a regualr hexagon, then vec(AC) + vec(AD) + vec(EA) + vec(FA)=

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

If ABCDEF is a regular hexagon with vec(AB) = veca and vec(BC)= vecb, then vec(CE) equals

If ABCDEF is a regular hexagon, prove that AD+EB+FC=4AB .

In the figure shown, ABCDEF is a regular hexagon and AOF is an equilateral triangle. The perimeter of DeltaAOF is 2a feet. What is the perimeter of the hexagon in feet?

ABCDEF is a regular hexagon with centre a the origin such that vec(AB)+vec(EB)+vec(FC)= lamda vec(ED) then lamda = (A) 2 (B) 4 (C) 6 (D) 3

If ABCDEF is a regular hexagon, prove that vec(AC)+vec(AD)+vec(EA)+vec(FA)=3vec(AB)

ABCDEF is a regular hexagon where centre O is the origin, if the position vector of A is hati-hatj+2hatk, then bar(BC) is equal to

ABCDEF is a regular hexagon , with the sides AB, AC, AD, AE and AF representing vectors 1 unit , 2 units , 3 units , 4 units and 5 units respetively . Find their resultant ?