For any two vectors ` vec ua n d vec v` prove that `( vec u. vec v)^2+| vec uxx vec v|^2=| vec u|^2| vec v|^2`

For any two vectors vec(a) and vec(b) , show that | vec (a).vec(b)| le |vec(a)||vec(b)|

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] .

Let vec a , vec b ,a n d vec c be non-coplanar vectors and let the equation vec a^' , vec b^' , vec c ' are reciprocal system of vector vec a , vec b , vec c , then prove that vec axx vec a^'+ vec bxx vec b^'+ vec cxx vec c ' is a null vector.

Let vec a , vec b ,and vec c be any three vectors, then prove that [ vec axx vec b vec bxx vec c vec cxx vec a ]= [vec a vec b vec c]^2

If vec aa n d vec b are two vectors and angle between them is theta, then | vec axx vec b|^2+( vec adot vec b)^2=| vec a|^2| vec b|^2 | vec axx vec b|=( vec adot vec b),iftheta=pi//4 vec axx vec b=( vec adot vec b) hat n ,(w h e r e hat n is unit vector, ) if theta=pi//4 ( vec axx vec b)dot( vec a+ vec b)=0

If vec u , vec v and vec w are three non-coplanar vectors, then prove that ( vec u+ vec v- vec w) . [ [( vec u- vec v)xx( vec v- vec w)]]= vec u . (vec v xx vec w)

If vec a_|_ vec b , then vector vec v in terms of vec aa n d vec b satisfying the equation s vec vdot vec a=0a n d vec vdot vec b=1a n d[ vec v vec a vec b]=1 is a. vec b/(| vec b|^2)+( vec axx vec b)/(| vec axx vec b|^2) b. vec b/(| vec b|^)+( vec axx vec b)/(| vec axx vec b|^2) c. vec b/(| vec b|^2)+( vec axx vec b)/(| vec axx vec b|^) d. none of these

If vec a , vec ba n d vec c are unit vectors satisfying | vec a- vec b|^2+| vec b- vec c|^2+| vec c- vec a|^2=9, then |2 vec a+5 vec b+5 vec c| is.

Let vec aa n d vec b be two non-collinear unit vector. If vec u= vec a-( vec adot vec b) vec ba n d vec v= vec axx vec b ,t h e n| vec v| is a. | vec u| b. | vec u|+| vec udot vec a| c. | vec u|+| vec udot vec b| d. | vec u|+ hat udot| vec a+ vec b|

If vec a and vec b are two vectors, then prove that ( vec axx vec b)^2=|[vec a.vec a,vec a.vec b],[vec b.vec a,vec b.vec b]| .