[A,B,C&D" are four points in a plane with position vectors "vec a,vec b,vec c&vec d" respectively such that "],[(vec a-vec d)*(vec b-vec c)=(vec b-vec d)*(vec c-vec a)=0." Then for the triangle "ABC,D" is its : "]

A,B,C and D are four points in a plane with position vectors vec a,vec b,vec c and vec d, respectively,such that (vec a-vec d)*(vec b-vec c)=(vec b-vec d)*(vec c-vec a)=0 Then point D is the of triangle ABC

If vec a,vec b,vec c are the position vectors of points A,B,C and D respectively such that (vec a-vec d)*(vec b-vec c)=(vec b-vec d)*(vec c-vec a)=0 then D is the

Show that the found points A,B,C,D with position vectors vec a,vec b,vec c,vec d respectively such that 3vec a-2vec b+5vec c-6vec d=vec 0 ,are coplanar .Also,find the position vector of the point of intersection of the line segments AC and BD.

Prove that [vec a,vec b,vec c+vec d]=[vec a,vec b,vec c]+[vec a,vec b,vec d]

Prove that (vec a × vec b).(vec c × vec d) = [[vec a.vec c,vec a.vec d],[vec b.vec c,vec b.vec d]] .

For any four vectors, vec a , vec b , vec c and vec d prove that vec d.( vec axx( vec bxx( vec cxx vec d)))=( vec b. vec d)[ vec a \ vec c \ vec d] .

For any four vectors, vec a , vec b , vec c and vec d prove that vec d.( vec axx( vec bxx( vec cxx vec d)))=( vec b. vec d)[ vec a \ vec c \ vec d] .