ABC is a triangle and P any point on BC. if ` vec P Q` is the sum of ` vec A P` + ` vec P B` +` vec P C` , show that ABPQ is a parallelogram and Q , therefore , is a fixed point.

ABC is a triangle and P any point on BC .If bar(P)Q is the sum of bar(A)P+bar(P)B+bar(P)C; show that ABQC is a parallelogram and Q therefore; is a fixed point.

IN any /_\ABC, a point p is on the side BC. If vec(PQ) is the resultant of the vectors vec(AP) , vec(PB) and vec(PC) the prove that ABQC is a parallelogram and hence Q is a fixed point.

If P is a point and ABCD is a quadrilateral and vec AP+vec PB+vec PD=vec PC, show that ABCD is a parallelogram.

Two vectors vec P and vec Q

If vec P and vec Q are two vectors, then the value of (vec P + vec Q) xx (vec P - vec Q) is

If vec P xx vec Q= vec R, vec Q xx vec R= vec P and vec R xx vec P = vec Q then

If vec PO+vec OQ=vec QO+vec OR, show that the point,P,Q,R are collinear.

What is the angle between vec P and the resultant of (vec P+vec Q) and (vec P-vec Q)

What is the anlge between vec P xx vec Q and vec Q xx vec P is

If vec p xxvec q = vec m xxvec n and vec p xxvec m = vec q xxvec n show that vec p-vec n is parallel to vec q-vec m where vec p! = Vec n and vec q! = Vec m