Home
Class 12
PHYSICS
In the regular hexagon shown in figure v...

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

Text Solution

Verified by Experts

The correct Answer is:
`2bar(AF)`
Promotional Banner

Topper's Solved these Questions

  • VECTOR & CALCULUS

    MOTION|Exercise EXERCISE -3 (LEVEL - II) SUBJECTIVE|8 Videos
  • VECTOR & CALCULUS

    MOTION|Exercise EXERCISE -4 (LEVEL - I) PREVIOUS YEAR|22 Videos
  • VECTOR & CALCULUS

    MOTION|Exercise EXERCISE -2 (LEVEL - II) MULTIPLE CORRECT|13 Videos
  • VECTOR

    MOTION|Exercise Exercise - 3|18 Videos
  • WAVE MOTION

    MOTION|Exercise Exercise - 3 Section - B (Previous Years Problems)|7 Videos

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)

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

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

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

In a regular hexagon ABCDEF,vec AE

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

If ABCDE is a pentagon, then vec(AB) + vec(AE) + vec(BC) + vec(DC) + vec(ED) + vec(AC) is equal to

Let O be the centre of a regular pentagon ABCDE and vec(OA) = veca , then vec(AB) +vec(2BC) + vec(3CD) + vec(4DE) + vec(5EA) is equals:

Let O be the centre of a regular pentagon ABCDE and vec(OA) = veca , then vec(AB) +vec(2BC) + vec(3CD) + vec(4DE) + vec(5EA) is equals: