Answer
$r(t) =\lt \dfrac{t}{2},-1+\dfrac{4t}{3},1-\dfrac{3t}{4} \gt$
$x=\dfrac{t}{2} \\ y= -1+\dfrac{4t}{3}; \\z=1-\dfrac{3t}{4} ; \\0 \leq t \leq 1$
Work Step by Step
The vector line equation of a line segment joining the points with position vectors $r_0$ and $r_1$ for the given two points is as follows:
$r(t)=(1-t) r_0+tr_1=(1-t) \lt 0,-1, 1 \gt +t \lt \dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4} \gt$
or, $=\lt 0,-1+t,1-t \gt + \lt \dfrac{t}{2},\dfrac{t}{3},\dfrac{t}{4} \gt$
or, $= \lt \dfrac{t}{2},-1+\dfrac{4t}{3},1-\dfrac{3t}{4} \gt$
Thus, we have $r(t) =\lt \dfrac{t}{2},-1+\dfrac{4t}{3},1-\dfrac{3t}{4} \gt$
Now, the parametric equations are:
$x=\dfrac{t}{2} \\ y= -1+\dfrac{4t}{3}; \\z=1-\dfrac{3t}{4} ; \\0 \leq t \leq 1$
