Line ` vec r= vec a+lambda vec b` will not meet the plane ` vec rdot vec n=q ,` if a. ` vec bdot vec n=0, vec adot vec n=q` b. ` vec bdot vec n!=0, vec adot vec n!=q` c. ` vec bdot vec n=0, vec adot vec n!=q` d. ` vec bdot vec n!=0, vec adot vec n=q`

The condition for equations vec rxx vec a= vec ba n d vec rxx vec c= vec d to be consistent is a. vec bdot vec c= vec adot vec d b. vec adot vec b= vec c dot vec d c. vec bdot vec c+ vec adot vec d=0 d. vec adot vec b+ vec c dot vec d=0

Given that vec adot vec b= vec adot vec c , vec axx vec b= vec axx vec c and vec a is not a zero vector. Show that vec b= vec c dot

For non-zero vectors vec a , vec b ,a n d vec c ,|( vec axx vec b)dot vec c|=| vec a|| vec b|| vec c| holds if and only if a. vec a* vec b=0, vec b* vec c=0 b. vec b* vec c=0, vec c* vec a=0 c. vec c* vec a=0, vec a* vec b=0 d. vec a* vec b=0, vec b* vec c=0, vec c* vec a=0

If vec a= vec p+ vec q , vec pxx vec b=0a n d vec qdot vec b=0, then prove that ( vec bxx( vec axx vec b))/( vec bdot vec b)= vec qdot

The reflection of the point vec a in the plane vec rdot vec n=q is a. vec a+(( vec q- vec adot vec n))/(| vec n|) b. vec a+2((( vec q- vec adot vec n))/(| vec n|^2)) vec n c. vec a+(2( vec q+ vec adot vec n))/(| vec n|^2) vec n d. none of these

The lines vec r= vec a+lambda( vec bxx vec c)a n d vec r= vec b+mu( vec cxx vec a) will intersect if a. vec axx vec c= vec bxx vec c b. vec adot vec c= vec bdot vec c c. bxx vec a= vec cxx vec a d. none of these

Let vec pa n d vec q be any two orthogonal vectors of equal magnitude 4 each. Let vec a , vec b ,a n d vec c be any three vectors of lengths 7sqrt(15)a n d2sqrt(33), mutually perpendicular to each other. Then find the distance of the vector ( vec adot vec p) vec p+( vec adot vec q) vec q+( vec adot( vec pxx vec q))( vec pxx vec q)+( vec bdot vec p) vec p( vec bdot vec q) vec q+( vec bdot( vec pxx vec q))( vec pxx vec q)+( vec cdot vec p) vec p+( vec cdot vec q) vec q+( vec cdot( vec pxx vec q))( vec pxx vec q) from the origin.

If vec axx( vec bxx vec c) is perpendicular to ( vec axx vec b)xx vec c , we may have a. ( vec a. vec c)| vec b|^2=( vec a. vec b)( vec b.vec c)( vec c.vec a) b. vec adot vec b=0 c. vec adot vec c=0 d. vec bdot vec c=0

Distance of point P( vec p) from the plane vec rdot vec n=0 is a. | vec pdot vec n| b. (| vec pxx vec n|)/(| vec n|) c. (| vec pdot vec n|)/(| vec n|) d. none of these

If vec a , vec b , vec ca n d vec d are distinct vectors such that vec axx vec c= vec bxx vec da n d vec axx vec b= vec cxx vec d , prove that ( vec a- vec d). (vec b- vec c)!=0,