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 O be the origin , and vec(OX),vec(OY),vec(OZ) be three unit vector in the directions of the sides vec(OR) , vec(RP) , vec(PQ) respectively, of a triangle PQR, Then , |vec(OX) xx vec(OY)| =
The unit vector bisecting vec(OY) and vec(OZ) is
Let vec a , vec b , vec c be the three unit vectors such that vec a+5 vec b+3 vec c= vec0 , then vec a. ( vec bxx vec c) is equal to
Let vec a, vec b, vec c be three vectors from vec a xx (vec b xxvec c) = (vec a xxvec b) xxvec c, if
If vec a, vec b, vec c be three vectors such that [vec with bvec c] = 4 then [vec a xxvec bvec b xxvec cvec c xxvec a] =
vec a, vec b, vec c are three non-zero vectors. If o + be defined as vec x o + vec y = vec x + vec y + vec x xxvec y and
Let vec(p),vec(q),vec(r) be three unit vectors such that vec(p)xxvec(q)=vec(r) . If vec(a) is any vector such that [vec(a)vec(q)vec(r )]=1,[vec(a)vec(r)vec(p )]=2 , and [vec(a)vec(p)vec(q )]=3 , then vec(a)=
If vec a,vec b,vec c are three vectors such that vec a+vec b+vec c=vec 0, then prove that vec a xxvec b=vec b xxvec c=vec c xxvec a
if vec a is a unit vector and projection of vec x along vec a is 2 units and (vec a xxvec x)+vec b=vec x then vec x is given by
