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A force $\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ acts on a particle to displace it from
the point $\mathrm{A}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$ to the point $\mathrm{B}(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}})$. The
work done by the force will be
Options:
the point $\mathrm{A}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$ to the point $\mathrm{B}(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}})$. The
work done by the force will be
Solution:
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Verified Answer
The correct answer is:
9 units
$\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{AB}}=(3-1) \hat{\mathrm{i}}+(-1-2) \hat{\mathrm{j}}+(5-(-3)) \hat{\mathrm{k}}$
$=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}$
Work done $=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{AB}}=(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+8 \hat{\mathrm{k}})$
$=(1 \times 2)+(3 \times(-3))+(2 \times 8)$
$=2-9+16=9$ units.
$\overrightarrow{\mathrm{AB}}=(3-1) \hat{\mathrm{i}}+(-1-2) \hat{\mathrm{j}}+(5-(-3)) \hat{\mathrm{k}}$
$=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}$
Work done $=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{AB}}=(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+8 \hat{\mathrm{k}})$
$=(1 \times 2)+(3 \times(-3))+(2 \times 8)$
$=2-9+16=9$ units.
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