Discussion
A particle is subjected to forces P = 2 i - 5 j + 6 k and Q = - i + 2 j - k. The positional vectors are \(\overrightarrow{OA}\) = 4 i - 3 j - 2 k and \(\overrightarrow{OB}\) = 6 i + j - 3 k. Work done by the resultant force (F = P + Q) to move the particle from A to B is
Answer: D
The work done is −15.
The resultant force and the displacement from A to B are
\[ \begin{aligned} \mathbf{F} &= (2 - 1)\mathbf{i} + (-5 + 2)\mathbf{j} + (6 - 1)\mathbf{k} \\ &= \mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \\ \overrightarrow{AB} &= 2\mathbf{i} + 4\mathbf{j} - \mathbf{k} \end{aligned} \]
Work is their dot product:
\[ W = (1)(2) + (-3)(4) + (5)(-1) = 2 - 12 - 5 = -15 \]
The negative sign means the force opposes the displacement on balance. 2 and −5 are single terms of the sum, and 12 drops the signs.
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