Answer
Vector equation: $\quad \displaystyle \mathrm{r}(t)=\langle \frac{1}{2}t , -1+\displaystyle \frac{4}{3}t,\ 1-\displaystyle \frac{3}{4}t\rangle,\quad 0\leq t\leq 1$.
Parametric equations: $\left\{\begin{array}{l}
x= \frac{1}{2}t\\
y= -1+\frac{4}{3}t,\\
z=1-\frac{3}{4}t
\end{array}\right.,\quad 0\leq t\leq 1$.
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 $\quad\mathrm{r}_{0}=\langle 0, -1,1\rangle$ and $\displaystyle \mathrm{r}_{1}=\langle\frac{1}{2},\ \displaystyle \frac{1}{3},\ \displaystyle \frac{1}{4}\rangle$,
$\mathrm{r}(t)=(1-t)\mathrm{r}_{0}+t\mathrm{r}_{1},\quad 0\leq t\leq 1$
$=(1-t)\displaystyle \langle 0, -1,1\rangle+t\langle\frac{1}{2},\ \displaystyle \frac{1}{3},\ \displaystyle \frac{1}{4}\rangle,\quad 0\leq t\leq 1$
$=\displaystyle \langle 0+\frac{1}{2}t, -1+t+\displaystyle \frac{1}{3}t,\ 1-t+\displaystyle \frac{1}{4}t\rangle,\quad 0\leq t\leq 1$.
$=\displaystyle \langle\frac{1}{2}t, -1+\displaystyle \frac{4}{3}t,\ 1-\displaystyle \frac{3}{4}t\rangle,\quad 0\leq t\leq 1$.
Vector equation: $\quad \displaystyle \mathrm{r}(t)=\langle \frac{1}{2}t , -1+\displaystyle \frac{4}{3}t,\ 1-\displaystyle \frac{3}{4}t\rangle,\quad 0\leq t\leq 1$.
Parametric equations: $\left\{\begin{array}{l}
x= \frac{1}{2}t\\
y= -1+\frac{4}{3}t,\\
z=1-\frac{3}{4}t
\end{array}\right.,\quad 0\leq t\leq 1$.
