Prove that the line joining the points `vec(6a)-vec(4b)+vec(4c) and -vec(4c)` and the line joining the points `-veca-vec(2b)-vec(3c), veca+vec(2b)-vec(5c)` intersect at `-vec(4c)`

Prove that [vec a -vec b, vec b - vec c, vec c- vec a] =0

Show that perpendicular distance of the point vec c from the line joining vec a and vec b is (| vec bxx vec c+ vec cxx vec a+ vec axx vec b|)/(| vec b- vec a|)

Prove that if vec a,vec b,vec c and vec d are any four vectors, then (vec a xx vec b)*(vec c xx vec d)= [[vec a* vec c vec b* vec c], [vec a* vec d vec b* vec d]]

If |vec(a)|=5,|vec(b)|=4 and vec(a).vec(b)=16 , then |vec(a)xxvec(b)| is

If vec a, vec b, vec c are unit vectors such that vec a+ vec b+ vec c = vec 0 , then find the value of vec a* vecb + vec b* vec c +vec c* veca .

If vec a is perpendicular to vec b and vec r is non-zero vector such that p vec r+( vec rdot vec a) vec b= vec c , then vec r= vec c/p-(( vec adot vec c) vec b)/(p^2) (b) vec a/p-(( vec cdot vec b) vec a)/(p^2) vec a/p-(( vec adot vec b) vec c)/(p^2) (d) vec c/(p^2)-(( vec adot vec c) vec b)/p

Show that the points with position vectors vec a- 2 vec b+ 3 vec c , -2 vec a + 3 vec b+ 2 vec c and -8 vec a + 13 vec b are collinear, whatever vec a , vec b and vec c may be.

The vectors vec a and vec b are not perpendicular and vec c and vec d are two vectors satisfying : vec b""X vec c"" =vec b""X vec d""=""a n d"" vec adot vec d=0 . Then the vector vec d is equal to : (1) vec b-(( vec bdot vec c)/( vec adot vec d)) vec c (2) vec c+(( vec adot vec c)/( vec adot vec b)) vec b (3) vec b+(( vec bdot vec c)/( vec adot vec b)) vec c (4) vec c-(( vec adot vec c)/( vec adot vec b)) vec b