Given the following simultaneous equation for vectors ` vec xa n d vec y , vec x+ vec y= vec a , vec xxx vec y= vec b , vec xdot vec a=1,t h e n vec x= dot` `( vec a+( vec axx vec b))/( vec a^2)` (B) `( vec a+( vec axx vec b))/(| vec a|)` (C) `(( vec a+( vec axx vec b)))/( vec b^2)` (D) none of these

Given the following simultaneous equation for vectors vec x and vec y , vec x+ vec y= vec a , vec x xx vec y= vec b , vec x . vec a=1 , then vec x= (A) ( vec a+( vec axx vec b))/( vec a^2) (B) ( vec a+( vec axx vec b))/(| vec a|) (C) (( vec a+( vec axx vec b)))/( vec b^2) (D) none of these

If vec adot vec b=betaa n d vec axx vec b= vec c ,t h e n vec b is ((beta vec a- vec axx vec c))/(| vec a|^2) b. ((beta vec a+ vec axx vec c))/(| vec a|^2) c. ((beta vec c- vec axx vec c))/(| vec a|^2) d. ((beta vec a+ vec axx vec c))/(| vec a|^2)

If vec adot vec b= vec adot vec c\ a n d\ vec axx vec b= vec axx vec c ,\ vec a!=0, then

Show that ( vec a- vec b)xx( vec a+ vec b)=2( vec axx vec b)dot

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] .

For any two vectors vec a\ a n d\ vec b , fin d\ ( vec axx vec b). vecbdot

The vector component of vec b perpendicular to vec a is a. ( vec bdot vec c) vec a b. ( vec axx( vec bxx vec a))/(| vec a|^2) c. vec axx( vec bxx vec a) d. none of these

Find the angle between two vectors vec a\ a n d\ vec b , if | vec axx vec b|= vec adot vec bdot

If | vec a|=2,| vec b|=5a n d vec adot vec b=0, then vec a x ( vec ax( vec ax( vec ax( vec a x( vec ax vec b))))) is equal to 64 vec a (b) 64 vec b (c) -64 vec b (d) -64 vec a

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 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