The point of application of a forceF = (6j+8k) is displaced from A(1,-

1.	The point of application of a forceF = (6j+8k) is displaced from A(1,-1,2) to B(-1,1,2).Then the work done is
Answer» Explanation: the DISPLACEMENT VECTOR dr=( -i+j+2k)-(i-j+2k) dr= -i+j+2k-i+j-2k dr= -2i+2j the work done w =F.dr w= (6j+8k).(-2i+2j) w=12 unit. where we have used displacement vector = final POSITION vector- initial position vector and i.j = j.k =k.i =0

The point of application of a forceF = (6j+8k) is displaced from A(1,-1,2) to B(-1,1,2).Then the work done is

Answer»

Explanation:

the DISPLACEMENT VECTOR dr=( -i+j+2k)-(i-j+2k)

dr= -i+j+2k-i+j-2k

dr= -2i+2j

the work done w =F.dr

w= (6j+8k).(-2i+2j)

w=12 unit.

where we have used displacement vector = final POSITION vector- initial position vector

and i.j = j.k =k.i =0

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The point of application of a forceF = (6j+8k) is displaced from A(1,-1,2) to B(-1,1,2).Then the work done is​

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The point of application of a forceF = (6j+8k) is displaced from A(1,-1,2) to B(-1,1,2).Then the work done is