If a rigid body is rotating about an axis passing through the point 2i-j-k and parallel to i-2j+2k with an angular velocity 3 radians/sec,then find the velocity of the point of the rigid body whose position vector is 2i+3j-4k(a)-2i+3j+4k(b)2i-3j+4k(c)-2i+3j-4k(d)-2i-3j-4k

Obviously,the unit vector (^n) in the direction of

                              i-2j+2k = (i-2j+2k)/l i-2j+2k l

                                                         =(i-2j+2k)/3

angular velocity of a rigid body= 3*[(i-2j+2k)/3]= i-2j+2k

Let the point whose velocity is desired to be determined be P.Then its +ve w.r.t. the point 2i-j-k on the axis given by

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

Hence,if v be the velocity of pt. P,we have v=w(omegha)cross r

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

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