A point of application of a force F = 5 i -3 j+2 k is moved from ri -2i+7 j+4k to r2 =-5i+ 2j+3k. The work done is

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- 22 units

22 units

11 units

0 unit

answer is A.

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Detailed Solution

Here displacement is given byr2 -r1 = (-5i+2j+ 3k) -(2i+7j+4k)= - 7 i - 5 j - kWork done =F⋅r2−r1∴W=(5i−3j+2k)⋅(−7i−5j−k)or W = -35+15-2=-22 units