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Q.

Three forces having magnitudes 5,4 and 3 units act on a particle in the directions 2i-2j-k ,  i + 2j + 2k and -2i + j-2k respectively and the particle gets displaced from the point A whose position vector is 9 i + 7j + 5k then the work done by these forces is

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a

30 unit

b

45 unit

c

9 unit 
 

d

43 unit 
 

answer is A.

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Detailed Solution

F=5(2i-2j-k)3+4(i+2j+2k)3+3(-2i+j-2k)3    =13(8i+j-3j) AB=3i+9j+2k F.AB=13(24+9-6)=273=9

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