Let `O bet the origin and let PQR be an arbitrary triangle. The point S is such that `vec(OQ).vec(OQ)+vec(OR).vec(OP)+vec(OQ).vec(OR)+vec(OP).vec(OS)` (A) centroid (B) circucentre (C) orthocenter (D) incentre

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

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that bar(OP)*bar(OQ)+bar(OR)*bar(OS)=bar(OR)*bar(OP)+bar(OQ)*bar(OS)=bar(OQ)*bar(OR)+bar(OP)*bar(OS) Then the triangle PQR has S as its

Let vec(C )= vec(A)+vec(B) then

If O is the circumcentre and P the orthocentre of Delta ABC , prove that vec(OA)+ vec(OB) + vec(OC) =vec(OP) .

Let vec(A)D be the angle bisector of angle A" of " Delta ABC such that vec(A)D=alpha vec(A)B+beta vec(A)C, then

Orthocenter of an equilateral triangle ABC is the origin O. If vec(OA)=veca, vec(OB)=vecb, vec(OC)=vecc , then vec(AB)+2vec(BC)+3vec(CA)=

Let O be the origin and vec(OX) , vec(OY) , vec(OZ) be three unit vector in the directions of the sides vec(QR) , vec(RP),vec(PQ) respectively , of a triangle PQR. |vec(OX)xxvec(OY)|=

Let vec(A), vec(B) and vec(C) , be unit vectors. Suppose that vec(A).vec(B)=vec(A).vec(C)=0 and the angle between vec(B) and vec(C) is pi/6 then

Consider the cube in the first octant with sides OP,OQ and OR of length 1, along the x-axis, y-axis and z-axis, respectively, where O(0,0,0) is the origin. Let S(1/2,1/2,1/2) be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If vec(p)=vec(SP), vec(q)=vec(SQ), vec(r)=vec(SR) and vec(t)=vec(ST) then the value of |(vec(p)xxvec(q))xx(vec(r)xx(vect))| is

If [ 2 vec (a) - 3 vec(b) vec( c ) vec(d)] =lambda [vec(a) vec(c ) vec(d) ] + mu [ vec(b) vec( c ) vec( d) ] , then 2 lambda + 3 mu=