Home
Class 11
MATHS
ABCDEF is a regular hexagon with point O...

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

Promotional Banner

Topper's Solved these Questions

  • MARCH - 2016 (ANDHRA PRADESH)

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise Section- C(Long answer type questions)|7 Videos
  • MARCH - 2016 (ANDHRA PRADESH)

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise Section- C(Long answer type questions)|7 Videos
  • LOCUS

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise TEXTUAL EXERCISES (EXERCISE -1(A))|17 Videos
  • MARCH - 2016 (TELANGANA)

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise Section -C (Long answer type quesitons)|7 Videos

Similar Questions

Explore conceptually related problems

Find the value of (vec(a) + vec(b)) xx (vec(a) - vec(b)) =

Find the angle between the vectors vec(i) + 2vec(j) + 3vec(k) and 3vec(i) - vec(j) + 2vec(k) .

ABCDEF is a regualar hexagon whose centre is O. Then bar(AB)+bar(AC)+bar(AD)+bar(AE)+bar(AF) is

If ABCDEF is a regular hexagon with centre O , then P.T bar(AB)+bar(AC)+bar(AD)+bar(AE)+bar(AF)=3bar(AD)=6bar(AO)

For any vector vec(a) , the value of (vec(a) xx vec(i))^(2) + (vec(a) xx vec(j))^(2) + (vec(a) xx vec(k))^(2)=

Let vec(a)= vec(i) + vec(j), vec(b)= 2vec(i) -vec(k) . Then the point of intersection of the lines vec(r ) xx vec(a) = vec(b) xx vec(a) and vec(r ) xx vec(b)= vec(a) xx vec(b) is

If ABCD is a quadrillateral such that vec(AB) = vec(i) + 2vec(j), vec(AD)= vec(j) + 2vec(k) and vec(AC)= 2 (vec(i) + 2vec(j)) + 3(vec(j) + 2vec(k)) , then area of the quadrilateral ABCD is