Show that `(veca-vecb)xx(veca+vecb)=2(vecaxxvecb)`

veca,vecb,vec(c) and vecd are the position vectors of four complanar points such that (veca-vecd).(vecb-vec(c))=(vecb-vecd).(vec(c)-veca)=0 . Show that the point vecd represents the orthocentre of the triangle with veca,vecb and vec(c) as its vertices.

Find |veca|and|vecb| , if (veca+vecb)*(veca-vecb)=8and|veca|=8|vecb| .

If veca+vecb+vec(c)=vec0 , then prove that vecaxxvecb=vecbxxvec(c)=vec(c)xxveca .

If veca,vecb,vec(c) are non-coplanar, then show that the vectors veca-vecb,vecb+vec(c),vec(c)+vec(a) are planar.

For any three vectors veca,vecb,vec(c) prove that [vecbxxvec(c)" "vec(c)xxveca" "vecaxxvecb]=[vec(a)" "vecb" "vec(c)]^(2)

Let veca and vecb be two unit vectors such that |veca+vecb|=sqrt(3) . If vecc=veca+2vecb+3(veca xx vecb) , then 2|vecc| is equal to :