A particle moves with the velocity `vec v = (5 vec i + 2 vec j - vec k) ms^(-1)` under the influence of a constant force, `vec F = (2 vec i + 5 vec j - 10 vec k)N`. The instantaneous power applied is

A particle moves with a velocity vec(v)=5hat(i)-3hat(j)+6hat(k) " "ms^(-1) under the influence of a constant force vec(f)= 10hat(i)+10hat(j)+20hat(k)N . The instantanteous power applied to the particle is

A force F = (2 vec i + vec j + vec k)N is acting on a particle moving with constant velocity vec v = (vec i + 2 vec j + vec k)m/s . Power delivered by force is

A particle is displaced from a position (2 vec i - vec j + vec k) metre to another position (3 vec i + 2 vec j -2 vec k) metre under the action of force (2 vec i + vec j -vec k) N. Work done by the force is

Find the angle between the line vec r=( vec i+2 vec j- vec k)+lambda( vec i- vec j+ vec k) and the normal to the plane vec rdot((2 vec i- vec j+ vec k))=4.

Find the unit vector in the direction of the vector vec a + vec b if vec a =vec i +2 vec j + 3 vec k and vec b = 2 vec i + 3 vec j + 5 vec k .

A point on the line r=2vec i+3vec j+4vec k+t(vec i+vec j+vec k) is

If vec a = 2vec i + 2vec j + vec kvec b = 5vec i + vec j + 2vec k then | vec a xxvec b | ^ (2) + (vec a * vec b) ^ (2) =

If vec a xx i+2vec a-4vec i-2vec j+vec k=0 then vec a=

Find the angle between the line vec r=(vec i+2vec j-vec k)+lambda(vec i-vec j+vec k) and the normal to the plane vec r(2vec i-vec j+vec k)=4

Prove that the vectors 3 vec I + 5 vec j + 2 vec k, 2 vec I - 3 vec j - 5 vec k and 5 vec I + 2 vec j -3 vec k form the sides of an equilatera, triangle.