In a triangle OAC, if `B` is the mid point of side AC and ` vec O A= vec a ,\ vec O B= vec b ,\ ` then what is ` vec O C ?`

If D is the mid-point of the side BC of a triangle ABC, prove that vec AB+vec AC=2vec AD .

[vec a + vec b, vec b + vec c, vec c + vec a] = 2 [vec a, vec b, vec c]

If O is a point in space, A B C is a triangle and D , E , F are the mid-points of the sides B C ,C A and A B respectively of the triangle, prove that vec O A + vec O B+ vec O C= vec O D+ vec O E+ vec O Fdot

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that vec O Pdot vec O Q+ vec O Rdot vec O S= vec O Rdot vec O P+ vec O Qdot vec O S= vec O Q . vec O R+ vec O Pdot vec O S Then the triangle PQ has S as its: circumcentre (b) orthocentre (c) incentre (d) centroid

If vec a , vec b , vec c , vec d are the position vector of point A , B , C and D , respectively referred to the same origin O such that no three of these point are collinear and vec a + vec c = vec b + vec d , than prove that quadrilateral A B C D is a parallelogram.

Let G be the centroid of Delta ABC , If vec(AB) = vec a , vec(AC) = vec b, then the vec(AG), in terms of vec a and vec b, is

[[vec a + vec b-vec c, vec b + vec c-vec a, vec c + vec a-vec b is equal to

Three vectors vec a, vec b and vec c satisfy the condition vec a + vec b + vec b + vec c = vec 0 .Evaluate the quantity mu = vec a * vec b + vec b * vec c + vec c * vec a , if | vec a | = 1 | vec b | = 4 and | vec c | = 2

If vec a + vec b + vec c = o, prove that vec a xxvec b = vec b xxvec c = vec c xxvec a

The vertices A , B , C of triangle A B C have respectively position vectors vec a , vec b , vec c with respect to a given origin O . Show that the point D where the bisector of /_A meets B C has position vector vec d=(beta vec b+gamma vec c)/(beta+gamma), where beta=| vec c- vec a| and, gamma=| vec a- vec b|dot Hence, deduce that incentre I has position vector (alpha vec a+beta vec b+gamma vec c)/(alpha+beta+gamma) where alpha=| vec b- vec c|