Home
Class 12
MATHS
If ABCDEF is a regular hexagon then vec ...

If ABCDEF is a regular hexagon then `vec AD + vec EB + vec FC = `

A

0

B

2AB

C

3AB

D

4AB

Text Solution

Verified by Experts

The correct Answer is:
D
A regular hexagon ABCDEF.

We know from the hexagon that AD is parallel to BC orr `AD=2BC` is parallel to FA or EB=2FA and FC is parallel to AB or FC=2AB.
Thus, `AD+EB+FC=2BC+2FA+2AB`
`=2(FA+AB+BC)=2(FC)=2(2AB)=4AB`.
Promotional Banner

Similar Questions

Explore conceptually related problems

In a regular hexagon A B C D E F ,\ A vec B=a ,\ B vec C= vec b\ a n d\ Cvec D=vecC.T h e n\ vec A E=

If ABCD is a quadrilateral with vec AB = vec a, vec AD = vec b and vec AC = 2 vec a + 3 vec b . If its areas is alpha times the area of the parallelogram with AB, AD as adjacent sides than alpha =

If vec u, vec nu, vec w be such that |vec u| =1, |vec nu| = 2, |vec w| = 3 . If the projection vec nu along vec u is equal to that of vec w along vec u , vec nu and vec w are perpendicular to each other, then |vec u - vec nu + vec w| =

If vec a, vec b, vec c are mutually perpendicular unit vectors then |vec a + vec b + vec c| =

If vec a and vec b are not perpendicular to each other and vec r xx vec a = vec b xx vec a, vec r. vec c = vec 0 then vec r =

[vec a + vec b vec b + vec c vec c + vec a] =

In a trapezium ABCD the vector B vec C = lambda vec(AD). If vec p = A vec C + vec(BD) is coillinear with vec(AD) such that vec p = mu vec (AD), then

The contrapositive of the statement " If vec a = vec 0 or vec b = vec 0 then vec a xx vec b = vec 0 " is