A) 124/7
B) 120/7
C) 125/7
D) 121/7
E) 123/7
Correct Answer: A
Solution :
Let\[\overrightarrow{{{F}_{1}}}=6\hat{i}+2\hat{j}+3\hat{k}\] and \[\overrightarrow{{{F}_{2}}}=3\hat{i}-2\hat{j}+6\hat{k}\] \[\therefore \]\[\overrightarrow{{{F}_{1}}}=3\left( \frac{6\hat{i}+2\hat{j}+3\hat{k}}{7} \right)=\frac{18\hat{i}+6\hat{j}+9\hat{k}}{7}\] and \[\overrightarrow{{{F}_{2}}}=4\left( \frac{3\hat{i}-2\hat{j}+6\hat{k}}{7} \right)=\frac{12\hat{i}-8\hat{j}+24\hat{k}}{7}\] \[\overrightarrow{F}=\overrightarrow{{{F}_{1}}}+\overrightarrow{{{F}_{2}}}=\frac{1}{7}[18\hat{i}+6\hat{j}+9\hat{k}+12\hat{i}\] \[-8\hat{j}+24\hat{k}]\] \[=\frac{1}{2}[30\hat{i}-2\hat{j}+33\hat{k}]\] Let \[\overrightarrow{OA}=2\hat{i}+2\hat{j}-\hat{k}\] and \[\overrightarrow{OB}=4\hat{i}+3\hat{j}+\hat{k}\] \[\therefore \] \[\overrightarrow{d}=\overrightarrow{OB}-\overrightarrow{OA}=2\hat{i}+\hat{j}+2\hat{k}\] \[\therefore \]Work done \[=\overrightarrow{F}.\overrightarrow{d}=\frac{1}{7}(30\hat{i}-2\hat{j}+33\hat{k}).(2\hat{i}+\hat{j}+2\hat{k})\] \[=\frac{60-2+66}{7}=\frac{124}{7}\]
You need to login to perform this action.
You will be redirected in
3 sec
