Answer
We verify the work performed:
(a) $W = {\bf{F}}\cdot\overrightarrow {PQ} $
(b) $W = {\bf{F}}\cdot\overrightarrow {PQ} $
Work Step by Step
We have the constant vector field ${\bf{F}} = \left( {2, - 1,4} \right)$.
(a) We have $P = \left( {0,0,0} \right)$, $Q = \left( {4,3,5} \right)$. The line segment $\overline {PQ} $ can be parametrized by
${\bf{r}}\left( t \right) = \left( {4t,3t,5t} \right)$ ${\ \ \ }$ for $0 \le t \le 1$
${\bf{r}}'\left( t \right) = \left( {4,3,5} \right)$
So, the work performed by the field is
$W = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^1 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$W = \mathop \smallint \limits_0^1 \left( {2, - 1,4} \right)\cdot\left( {4,3,5} \right){\rm{d}}t = 25\mathop \smallint \limits_0^1 {\rm{d}}t = 25$
Evaluate ${\bf{F}}\cdot\overrightarrow {PQ} $:
${\bf{F}}\cdot\overrightarrow {PQ} = \left( {2, - 1,4} \right)\cdot\left( {4,3,5} \right) = 25$
So, $W = {\bf{F}}\cdot\overrightarrow {PQ} $.
(b) We have $P = \left( {3,2,3} \right)$, $Q = \left( {4,8,12} \right)$. The line segment $\overline {PQ} $ can be parametrized by
${\bf{r}}\left( t \right) = \left( {3,2,3} \right) + t\left( {1,6,9} \right)$ ${\ \ \ }$ for $0 \le t \le 1$
${\bf{r}}\left( t \right) = \left( {t + 3,6t + 2,9t + 3} \right)$
${\bf{r}}'\left( t \right) = \left( {1,6,9} \right)$
So, the work performed by the field is
$W = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^1 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$W = \mathop \smallint \limits_0^1 \left( {2, - 1,4} \right)\cdot\left( {1,6,9} \right){\rm{d}}t = 32\mathop \smallint \limits_0^1 {\rm{d}}t = 32$
Evaluate ${\bf{F}}\cdot\overrightarrow {PQ} $:
${\bf{F}} \cdot \overrightarrow {PQ} = \left( {2, - 1,4} \right)\cdot\left( {1,6,9} \right) = 32$
So, $W = {\bf{F}}\cdot\overrightarrow {PQ} $.
