Show that if P,A,B are any three points, then `lambda vec(PA)+mu vec(PB)=(lambda+mu)vec(PC)`, where C divides [AB] in the ratio `mu : lambda`.

if A,B,C,D and E are five coplanar points, then vec(DA) + vec( DB) + vec(DC ) + vec( AE) + vec( BE) + vec( CE) is equal to :

If a, b and c are non-coplanar vectors and d=lambda vec a+mu vec b+nu vec c , then lambda is equal to

Find the position vector of the point R which divides the line (segment) joining the two points P and Q with position vectors vec(OP) = 2 vec a+ vec b and vec (OQ) = vec a-2 vec b respectively, in the ratio 1: 2. (i) internally (ii) externally. ‘

Statement 1: if three points P ,Qa n dR have position vectors vec a , vec b ,a n d vec c , respectively, and 2 vec a+3 vec b-5 vec c=0, then the points P ,Q ,a n dR must be collinear. Statement 2: If for three points A ,B ,a n dC , vec A B=lambda vec A C , then points A ,B ,a n dC must be collinear.

If A,B,C and D are any four points in space prove that |quad vec (AB) xx vec (CD)+ vec (BC) xx vec (AD) + vec (CA) xx vec (BD)|=4(area of Delta ABC) .

If |vec(a)| = 3, |vec(b)| = 4 , then the value 'lambda' for which vec(a) + lambda vec(b) is perpendicular to vec(a) - lambda vec(b) , is

If the position vector of a point A is vec a + 2 vec b and vec a divides AB in the ratio 2:3 , then the position vector of B, is

The position vector of the point which divides the join of points 2 vec a -3 vec b and vec a+vec b in the ratio 3: 1 is

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.

ABC is any triangle and D,E,F are the mid-points of sides vec(BC),vec(CA) and vec(AB) respectively. Express the vec (BE) and vec (CF) as linear combination of vectors vec (AB) and vec (AC) .