`vec(i).(vec(j)xxvec(k))=`

The value of (vec(A)+vec(B)).(vec(A)xxvec(B)) is :-

If veca=-vec(2i)-vec(2j)+vec(4k),vecb=-vec(2i)+vec(4j)-vec(2k) and vecc=vec(4i)-vec(2j)-vec(2k) Calculate the value of [veca vecb vecc\] and interpret the result.

If vec(A)xxvec(B)=vec(B)xxvec(A) , then the angle between vec(A) and vec(B) is-

Given vec(A)xxvec(B)=vec(0) and vec(B)xxvec(C )=vec(0) Prove that vec(A)xxvec(C )=vec(0)

If vec(a)= hat(i) + hat(j) + hat(k) and vec(b)= hat(j)-hat(k) , then find vec(c ) such that vec(a ) xx vec(c )= vec(b) and vec(a).vec(c )=3 .

Find the projection of vec(b)+ vec(c ) on vec(a) where vec(a)= 2 hat(i) -2hat(j) + hat(k), vec(b)= hat(i) + 2hat(j)- 2hat(k), vec(c ) = 2hat(i) - hat(j) + 4hat(k)

If vec(a) = hat(i) + hat(j) + hat(k), vec(a).vec(b) =1 and vec(a) xx vec(b) = hat(j)-hat(k) , then the vector vec(b) is

If there are two vectors vec(A) and vec(B) such that vec(A)+vec(B)=hat(i)+2hat(j)+hat(k) and vec(A)-vec(B)=(hat(i)-vec(k)) , then choose the correct options.

If vec(a) = hat(i) + hat(j) + 2 hat(k) , vec(b) = 2 hat(j) + hat(k) and vec (c) = hat(i) + hat(j) + hat(k) . Find a unit vector perpendicular to the plane containing vec(a), vec(b), vec(c)

The angle between Vectors (vec(A)xxvec(B)) and (vec(B)xxvec(A)) is