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Forces $2i-5j+6k$ and $-i-2j-k$ move a particle from $A(4,-3,-2)$ to $B(6,-1,-3)$. The work done is:
A66 units
B33 units
C17 units
D11 units
Answer & Solution
Correct answer: C. 17 units
1. Work done is $W=\vec{F}\cdot\vec{AB}$, so first find the resultant force.
2. $\vec{F}=(2i-5j+6k)+(-i-2j-k)=i-7j+5k$.
3. The displacement is $\vec{AB}=(6-4)i+(-1+3)j+(-3+2)k=2i+2j-k$.
4. Dot product: $(1)(2)+(-7)(2)+(5)(-1)=2-14-5=-17$.
5. Taking the magnitude, the work done is $17$ units. The value $66$ comes from doubling a mis-formed dot product rather than summing the forces first.
_Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 3: Vectors_
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