The work done by the forces `vecF=2hati-3hatj+2hatk` in moving a particle from (3,4,5) to (1,2,3) is (A) `0` (B) `3/2` (C) `-4` (D) `-2`

The work done by the forces vecF = 2hati - hatj -hatk in moving an object along the vectors 3hati + 2hatj - 5hatk is:

The resultant of the forces vecF_(1) = 4hati-3hatj and vecF_(2) = 6hati + 8hatj is

Work done when a force F=(hati + 2hat(j) + 3hatk) N acting on a particle takes it from the point r_(1) =(hati + hatk) the point r_(2) =(hati -hatj + 2hatk) is .

Three constant forces vecF_1=2hati-3hatj+2hatk , vecF_2=hati+hatj-hatk , and vecF_3=3hati+hatj-2hatk in newtons displace a particle from (1, -1, 2) to (-1, -1, 3) and then to (2, 2, 0) (displacement being measured in meters). Find the total work done by the forces.

Find the torque of a force vecF=2hati+hatj+4hatk acting at the point vecr=7hati+3hatj+hatk :

Work done by a force F=(hati+2hatj+3hatk) acting on a particle in displacing it from the point r_(1)=hati+hatj+hatk to the point r_(2)=hati-hatj+hat2k is

The length of perpendicular from the origin to the line vecr=(4hati+2hatj+4hatk)+lamda(3hati+4hatj-5hatk) is (A) 2 (B) 2sqrt(3) (C) 6 (D) 7

The angle between force vecF =(3hati+4hatj-5hatk) unit and displacement vecd = (5hati+4hatj+3hatk) unit is

The torpue of force vecF=-2hati+2hatj+3hatk acting on a point vecr=hati-2hatj+hatk about origin will be :

The work done by a force F=(hati+2hatj+3hatk) N ,to displace a body from position A to position B is [The position vector of A is r_(1)=(2hati+2hatj+3hatk)m The position vector of B is r_(2)=(3hati+hatj+5hatk)m ]