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 thepoint(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]



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