Find `vec(a) . vec(b )` =
( A ) `-vecb . vec(a)`
( B ) `vecb . veca`
( C ) 1

Let A and B be points with position vectors veca and vecb with respect to origin O. If the point C on OA is such that 2vec(AC)=vec(CO), vec(CD) is parallel to vec(OB) and |vec(CD)|=3|vec(OB)| then vec(AD) is (A) vecb-veca/9 (B) 3vecb-veca/3 (C) vecb-veca/3 (D) vecb+veca/3

Let veca , vecb, vec c be three non coplanar vectors , and let vecp , vecq " and " vec r be the vectors defined by the relation vecp = (vecb xx vec c )/([veca vecb vec c ]), vec q = (vec c xx vec a)/([veca vecb vec c ]) " and " vec r = (vec a xx vec b)/([veca vecb vec c ]) Then the value of the expension (vec a + vec b) .vec p + (vecb + vec c) .q + (vec c + vec a) . vec r is equal to

If the vectors vec a, vec b, vec c are coplanar, then the value of | (vec a, vec b, vec c), (vec a * vec a, vec a * vec b, vec a * vec c), (vec b * vec a, vec b * vec b, vecb * vec c) | =

If veca, vecb, vec c are three non coplanar vectors , then the value of (vec a.(vec b xx vec c) )/((vec c xx vec a).vec b) + ( vecb.(vec a xx vec c ))/(vec c.(vec a xx vec b)) is

If the points P( veca + 2 vec b + vec c ), Q (2 veca + 3 vecb), R (vecb+ t vec c ) are collinear, where veca , vec b , vec c are non-coplanar vectors, the value of t is

Two vectors are presented as veca = 3.0 hati + 5.0 hatj and vecb = 2. 0 hati + 4.0 hatj. Find (a) vec a xx vecb , (b) veca. vecb ,(c ) (veca + vecb).vecb ,

Let vec a , vec b ,vec c are three non- coplanar vectors such that vecr_(1)=veca + vec c , vecr_(2)= vec b+vec c -veca , vec r_(3) = vec c + vec a + vecb, vec r = 2 vec a - 3 vec b + 4 vec c. If vec r = lambda_(1)vecr_(1)+lambda_(2)vecr_(2)+lambda_(3)vecr_(3) , then

Let veca vecb and vecc be three non-zero vectors such that veca + vecb + vecc = vec0 and lambda vecb xx veca xx veca + vecbxxvecc + vecc + veca = vec0 then find the value of lambda .

Statement -1 : If veca and vecb are non- collinear vectors, then points having position vectors x_(1) vec(a) + y_(1) vec(b) , x_(2)vec(a)+ y_(2) vec(b) and x_(3) veca + y_(3) vecb are collinear if |(x_(1),x_(2),x_(3)),(y_(1),y_(2),y_(3)),(1,1,1)|=0 Statement -2: Three points with position vectors veca, vecb , vec c are collinear iff there exist scalars x, y, z not all zero such that x vec a + y vec b + z vec c = vec 0, " where " x+y+z=0.