Home
Class 12
MATHS
Consider the regular hexagon ABCDEF with...

Consider the regular hexagon ABCDEF with centre at O (origin).
Q. Five forces AB,AC,AD,AE,AF act at the vertex A of a regular hexagon ABCDEF. Then, their resultant is (a)3AO (b)2AO (c)4AO (d)6AO

A

` 3 vec(AO)`

B

`2 vec(AO)`

C

`4 vec(AO)`

D

`6 vec(AO)`

Text Solution

Verified by Experts

The correct Answer is:
D
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

Five points given by A,B,C,D and E are in a plane. Three forces AC,AD and AE act at A annd three forces CB,DB and EB act B. then, their resultant is

Five forces vec A B , vec A C , vec A D , vec A E and vec A F act at the vertex of a regular hexagon A B C D E Fdot Prove that the resultant is 6 vec A O , where O is the centre of hexagon.

Five forces vec A B , vec A C , vec A D , vec A E and vec A F act at the vertex of a regular hexagon A B C D E Fdot Prove that the resultant is 6 vec A O , where O is the centre of heaagon.

ABCDEF si a regular hexagon with centre at the origin such that A vec D + E vec B + F vec C = lambda E vec D . " Then, " lambda equals

In a regular hexagon ABCDEF, prove that vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)=3vec(AD)

ABCDEF is a regular hexagon. Find the vector vec AB + vec AC + vec AD + vec AE + vec AF in terms of the vector vec AD

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 ?

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

A B C D E F is a regular hexagon with centre O ). If the area of triangle O A B is 9\ c m^2 , find the area of: (i) the hexagon and (ii) the circle in which the hexagon is inscribed.