A) 20 unit
B) 30 unit
C) 40 unit
D) 50 unit
Correct Answer: C
Solution :
Let \[{{\vec{F}}_{1}}=4\hat{i}+\hat{j}-3\hat{k},\,\,\,{{\vec{F}}_{2}}=3\hat{i}+\hat{j}-\hat{k},\] \[\vec{A}=\hat{i}+2\hat{j}+3\hat{k}\] and \[\vec{B}=5\hat{i}+4\hat{j}+\hat{k}\] \[\therefore \] \[\vec{F}={{\vec{F}}_{1}}+{{\vec{F}}_{2}}=7\hat{i}+2\hat{j}-4\hat{k}\] and \[\overrightarrow{AB}=\vec{d}=4\hat{i}+2\hat{j}-2\hat{k}\] \[\therefore \] Work done \[=\vec{d}.\vec{F}\] \[=(4\hat{i}+2\hat{j}-2\hat{k}).(7\hat{i}+2\hat{j}-4\hat{k})\] \[=28+4+8=40\,\,unit\]
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