Home
Class 12
MATHS
In Fig. ABCDEF is a regular hexagon. Pro...

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

Promotional Banner

Topper's Solved these Questions

  • VECTOR ALGEBRA

    MODERN PUBLICATION|Exercise EXERCISE 10 (c ) Short Answer Type Questions|39 Videos
  • VECTOR ALGEBRA

    MODERN PUBLICATION|Exercise EXERCISE 10 (c ) Long Answer Type Questions (I)|12 Videos
  • VECTOR ALGEBRA

    MODERN PUBLICATION|Exercise EXERCISE 10 (a) Short Answer Type Questions|21 Videos
  • THREE DIMENSIONAL GEOMETRY

    MODERN PUBLICATION|Exercise CHAPTER TEST 11|11 Videos

Similar Questions

Explore conceptually related problems

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, prove that vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)=3vec(AD)

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)

ABCDEF is a regular hexagon with point O as centre. The value of vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF) is

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

If ABCDEF is a regular hexagon , then A vec D + E vec B + F vec C equals

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

ABCDEF is a regular hexagon, Fig. 2 (c ) .65. What is the value of (vec (AB) + vec (AC) + vec (AD) + vec (AE) + vec (AF) ? .

In a regular hexagon ABCDEF,vec AE