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```        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).

```
6 years ago

## Answers : (1)

30 Points
```										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 questionDifficult 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 ExpertsPriyansh Bajaj
```
6 years ago
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