Home
Class 11
PHYSICS
In a regular hexagon ABCDEF, prove that ...

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

A

`16hat(i)+ 24hat(j) + 32hat(k)`

B

`-16hat(i)- 24hat(j) + 32hat(k)`

C

`16hat(i)- 24hat(j)+ 32hat(k)`

D

`16hat(i) + 24hat(j)- 32hat(k)`

Text Solution

Verified by Experts

The correct Answer is:
D
We know, `therefore vec(OA)+ vec(OB)+ vec(OC) + vec(OD) + vec(OE) + vec(OF) + vec(OG) + vec(OH)= vec(0)`
By triangle law of vector addition, we can write
`vec(AB)= vec(AO) + vec(OB), vec(AC) = vec(AO)+ vec(OC)`
`vec(AD)= vec(AO) + vec(OD), vec(AE) = vec(AO) + vec(OE)`
`vec(AF)+ vec(AO) + vec(OF), vec(AG) = vec(AO)+ vec(OG)`
`vec(AH) = vec(AO) + vec(OH)`
Now, `vec(AB) + vec(AC) + vec(AD) + vec(AE) + vec(AF) + vec(AG) + vec(AH)`
`= (7 vec(AO)) + vec(OB) + vec(OC) + vec(OD) + vec(OE) + vecOF) + vec(OG)+ vec(OH)`
`= (7vec(AO)) + vec(0) - vec(OA)`
`=7 vec(AO)+ vec(AO)`
`=(8vec(AO))= 8 (2hat(i)+ 3hat(j)- 4hat(k))`
`=16hat(i)+ 24hat(j)- 32hat(k)`
Promotional Banner

Similar Questions

Explore conceptually related problems

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

In a regular hexagon ABCDEF,vec AE

In Fig. ABCDEF is a ragular hexagon. Prove that vec(AB) +vec(AC) +vec(AD) +vec(AE) +vec(AF) = 6 vec(AO) .

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

ABCDEF is a regular hexagon. Show that : vec(OA)+vec(OB)+vec(OC)+vec(OD)+vec(OE)+vec(OF)=vec(0)

ABCDE is a pentagon prove that vec(AB)+vec(BC)+vec(CD)+vec(DE)+vec(EA)=vec0

Let O be the centre of the regular hexagon ABCDEF then find vec(OA)+vec(OB)+vec(OD)+vec(OC)+vec(OE)+vec(OF)

In the regular hexagon shown in figure vec(AB)+vec(BC)+vec(CD)+vec(DE)+vec(EF)+vec(AF) can be expressed as :

In a regular hexagon ABCDEF,vec AB=a,vec BC=b and vec CD=c. Then ,vec AE=