A) 11 unit
B) 33 unit
C) 10 unit
D) 30 unit
E) 44 unit
Correct Answer: B
Solution :
\[\because \] \[{{\overrightarrow{F}}_{1}}=\frac{5(6\hat{i}+2\hat{j}+3\hat{k})}{7}\] \[{{\overrightarrow{F}}_{2}}=\frac{3(3\hat{i}-2\hat{j}+6\hat{k})}{7}\] and \[{{\overrightarrow{F}}_{3}}=\frac{1(2\hat{i}-3\hat{j}-6\hat{k})}{7}\] \[\therefore \] \[\overrightarrow{F}={{\overrightarrow{F}}_{1}}+{{\overrightarrow{F}}_{2}}+{{\overrightarrow{F}}_{3}}\] \[=\frac{1}{7}(30\hat{i}+10\hat{j}+15\hat{k}+9\hat{i}-6\hat{j}+18\hat{k}\] \[+2\hat{i}-3\hat{j}-6\hat{k})\] \[=\frac{1}{7}[41\hat{i}+\hat{j}+27\hat{k}]\] and \[\overset{\to }{\mathop{AB}}\,=5\hat{i}-\hat{j}+\hat{k}-2\hat{i}+\hat{j}+3\hat{k}\] \[=3\hat{i}+4\hat{k}\] \[\therefore \] Work done\[=\frac{1}{7}[41\hat{i}+\hat{j}+27\hat{k}].[3\hat{i}+4\hat{k}]\] \[=\frac{1}{7}[123+108]=33\]unit
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