Home
Class 12
MATHS
ABCDEF is a regular hexagon in the x-y p...

ABCDEF is a regular hexagon in the x-y plance with vertices in the anticlockwise direction. If `vecAB=2hati`, then `vecCD` is

A

`hati+3hatj`

B

`hati9+2hatj`

C

`-hati+sqrt(3)hatj`

D

none of these

Text Solution

Verified by Experts

AB is along the X-axis and BD is along the Y-axis.
`AB=2hatiimpliesAB=BC=CD=` . . . . .=2

From the figure, BC=BC`sin60^(@)=2sin60^(@)=sqrt(3)`
`thereforeBD=2sqrt(3)hatj`
`BC=BCcos60^(@)hati+BCsin60^(@)hatj=hati+sqrt(3)hatj`
`CD=BD-BC=2sqrt(3)hatj-(hati+sqrt(3)hatj)=-hati+sqrt(3)hatj`
Promotional Banner

Topper's Solved these Questions

  • VECTOR ALGEBRA

    ARIHANT MATHS|Exercise Exercise For Session 1|7 Videos
  • VECTOR ALGEBRA

    ARIHANT MATHS|Exercise Exercise For Session 2|17 Videos
  • TRIGONOMETRIC FUNCTIONS AND IDENTITIES

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|18 Videos

Similar Questions

Explore conceptually related problems

If ABCDEF is a regular hexagon, prove that AD+EB+FC=4AB .

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

OABCDE is a regular hexagon of side 2 units in the XY-plane in the first quadrant. O being the origin and OA taken along the x-axis. A point P is taken on a line parallel to the z-axis through the centre of the hexagon at a distance of 3 units from O in the positive Z direction. Then find vector vec(AP) .

If ABCDEF is a regular hexagon with vec(AB) = vec(a) ,vec(BC ) = vec(b) then vec(CE) equals :

Find the unit vector in the direction of the vector : vecb =2hati+hatj+2hatk .

Take a regular hexagon. Is x = y = z = p = q = r ?Why ?.