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.

The position vectors of the vertices A ,Ba n dC of a triangle are three unit vectors vec a , vec b ,a n d vec c , respectively. A vector vec d is such that vec d ⋅ vec a = vec d ⋅ vec b = vec d ⋅ vec c a n d vec d=lambda( vec b+ vec c)dot Then triangle A B C is a. acute angled b. obtuse angled c. right angled d. none of these

The position vectors of four points A, B, C and D in the plane are vec(a),vec(b),vec( c ) and vec(d) . If (vec(a)-vec(d)).(vec(b)-vec( c ))=(vec(b)-vec(d)).(vec( c )-vec(a))=0 then D is a ………………is DeltaABC .

vec a , vec b and vec c are the position vectors of points A ,B and C respectively, prove that : vec a× vec b+ vec b× vec c+ vec c× vec a is vector perpendicular to the plane of triangle A B Cdot

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]] .

If vec a,vec b,vec c and vec d are distinct vectors such that vec a xxvec c=vec b xxvec d and vec a xxvec b=vec c xxvec d prove that (vec a-vec d)vec b-vec c!=0