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  4. A force of magnitude 5 unit acting along the vector 2 hati-2 hatj+ hatk displaces the point of application from (1, 2, 3) to (5, 3, 7). Then the work done is:

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Q. A force of magnitude 5 unit acting along the vector $ 2\hat{i}-2\hat{j}+\hat{k} $ displaces the point of application from (1, 2, 3) to (5, 3, 7). Then the work done is:

KEAMKEAM 2002

Solution:

$ \therefore $ $ \overrightarrow{F}=\frac{5(2\hat{i}-2\hat{j}+\hat{k})}{\sqrt{4+4+1}}=\frac{5}{3}(2\hat{i}-2\hat{j}+\hat{k}) $ and $ \overrightarrow{d}=(5\hat{i}+3\hat{j}+7\hat{k})-(\hat{i}+2j+3\hat{k}) $ $ =4\hat{i}+\hat{j}+4\hat{k} $ $ \therefore $ $ W=\overrightarrow{F}.\overrightarrow{d}=\frac{5}{3}(2\hat{i}-2\hat{j}+\hat{k}) $ $ (4\hat{i}+\hat{j}+4\hat{k}) $ $ =\frac{5}{3}(8-2+4)=\frac{50}{3}unit $