Answer
Vector equation: $\quad \mathrm{r}(t)=\langle 2+4t,\ 2t, -2t\rangle,\quad 0\leq t\leq 1$.
Parametric equations: $\left\{\begin{array}{l}
x=2+4t\\
y=2t\\
z=-2t,
\end{array}\right.,\quad 0\leq t\leq 1$.
Work Step by Step
$\mathrm{r}(t)=(1-t)\mathrm{r}_{0}+t\mathrm{r}_{1},\quad 0\leq t\leq 1\quad$ (eq.4 in 12-5)
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Given $\quad\mathrm{r}_{0}=\langle 2,0,0\rangle$ and $\mathrm{r}_{1}=\langle 6,2, -2\rangle$,
$\mathrm{r}(t)=(1-t)\mathrm{r}_{0}+t\mathrm{r}_{1},\quad 0\leq t\leq 1$
$=(1-t)\langle 2,0,0\rangle+t\langle 6,2, -2\rangle,\quad 0\leq t\leq 1$
$=\langle 2-2t+6t,0+2t, 0-2t\rangle,\quad 0\leq t\leq 1$.
$=\langle 2+4t,\ 2t, -2t\rangle,\quad 0\leq t\leq 1$.
Parametric equations: $\left\{\begin{array}{l}
x=2+4t\\
y=2t\\
z=-2t,
\end{array}\right.,\quad 0\leq t\leq 1$.
