1.) A rigid body is rotating at 2.5 rad/sec about an axis AB,where A and B are the point (1,-2,1) and (3,-4,2). Find the velocity of the particle P of the body at the point(5,-1,-1).

Dear Pradeep,

Solution:-  AB = B -A ; where A and B are vector A and  vector B respectively.

AB = (3i -4j + 2k) - (i -2j +k) = 2i -2j + k

|AB| = 3

velocity of particle P = ωr; where ω = 2.5 and 'r' is the radius, which is the distance of P from AB axis.

i.e., r = perpendicular distance of point P from line AB

      r = | {AP - [(AP .AB)/ |AB|] AB} |

   AP = P - A = (5i-j-k) - (i -2j +k) = 4i +j -2k

AP. AB = 8- 2 -2 = 4

  r = | {(4i +j -2k) - [4/3] (2i -2j + k)} |

  r = | {(4i +j -2k)

[ANS]



Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.

All the best!!!

Regards,
Askiitians Experts
Priyansh Bajaj