चित्र में वैक्टर `vec(OA), vec(OB) और vec (OC)` का परिमाण बराबर है। दिशा खोजें...

The magnitude of vectors vec(OA), vec(OB) and vec (OC) in figure are equal. Find the direction of vec(OA)+vec(OB)-vec(OC) . .

Find the resultant of the three vectors vec(OA), vec(OB) and vec(OC) shown in figure. Radius of the circle is R.

The resultant of the three vectors vec(OA), vec(OB) and vec(OC) shown in figure is

If E is the intersection point of diagonals of parallelogram ABCD and vec( OA) + vec (OB) + vec (OC) + vec (OD) = xvec (OE) where O is origin then x=

ABCD is a quadrilateral and E is the point of intersection of the lines joining the middle points of opposite side. Show that the resultant of vec (OA) , vec(OB) , vec(OC) and vec(OD) = 4 vec(OE) , where O is any point.

Given a parallelogram OACB. The lengths of the vectors vec (OA), vec (OB) & vec (AB) are a, b & c respectively. The scalar product of the vectors vec (OC) & vec (OB) is (A) (a^2-3b^2+c^2)/2 (B) (3a^2+b^2-c^2)/2 (C) (3a^2-b^2+c^2)/2 (D) (a^2+3b^2-c^2)/2

If | vec AO + vec OB | = | vec BO + vec OC |, then A, B, C form

Let O, O' and G be the circumcentre, orthocentre and centroid of a Delta ABC and S be any point in the plane of the triangle. Statement -1: vec(O'A) + vec(O'B) + vec(O'C)=2vec(O'O) Statement -2: vec(SA) + vec(SB) + vec(SC) = 3 vec(SG)

Let vec(OA)=vecas, vec(OB)=10veca+2vecb and vec(OC)=vecb where O A and C are non collinear points. Let p denote the area of the quadrilaterial OABCand q denote the area of the parallelogram with OA and OC as adjacent sides. Then p/q= (A) 2 (B) 6 (C) 1 (D) 1/2|veca+vecb+vecc]

Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin. What is vec(OA) + vec(OB) + vec(OC ) + vec(OD) equal to