The position vectors of the vertices of a quadrilateral with `A` as origin are `B( vec b),D( vec d)a n dC(l vec b+m vec d)dot` Prove that the area of the quadrialateral is `1/2(l+m)| vec bxx vec d|dot`

ABCD is quadrilateral such that vec AB=vec b,vec AD=vec d,vec AC=mvec b+pvec d Show that he area of the quadrilateral ABCDis(1)/(2)|m+p||vec b xxvec d|

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 vecd dot vec a= vecd dot vec b= vec d dot vec ca 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

If vec a,vec b,vec c and vec d are the position vectors of the vertices of a cyclic quadrilateral ABCD prove that (|vec a xxvec b+vec b xxvec d+vec d xxvec a|)/((vec b-vec a)*(vec d-vec a))+(|vec b xxvec c+vec c xxvec d+vec d xxvec b|)/((vec b-vec c)*(vec d-vec c))=

For any three vectors adotb\ a n d\ c write the value of vec axx( vec b+ vec c)+ vec bxx( vec c+ vec a)+ vec cxx( vec a+ vec b)dot

If vec a\ a n d\ vec b are unit vectors then write the value of | vec axx vec b|^2+( vec adot vec b)^2dot

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

If vec rdot vec a= vec rdot vec b= vec rdot vec c=1/2 or some nonzero vector vec r , then the area of the triangle whose vertices are A( vec a),B( vec b)a n dC( vec c)i s( vec a , vec b , vec c are non-coplanar) |[ vec a vec b vec c]| b. | vec r| c. |[ vec a vec b vec c] vec r| d. none of these

For any two vectors vec a\ a n d\ vec b , find veca.( vecbxx veca)dot

If vec a , vec b , vec c , vec d are the position vector of point A , B , C and D , respectively referred to the same origin O such that no three of these point are collinear and vec a + vec c = vec b + vec d , than prove that quadrilateral A B C D is a parallelogram.

If O A B C is a tetrahedron where O is the orogin anf A ,B ,a n dC are the other three vertices with position vectors, vec a , vec b ,a n d vec c respectively, then prove that the centre of the sphere circumscribing the tetrahedron is given by position vector (a^2( vec bxx vec c)+b^2( vec cxx vec a)+c^2( vec axx vec b))/(2[ vec a vec b vec c]) .