Answer
$r(t) =\lt \dfrac{t}{2},-1+\dfrac{4t}{3},1-\dfrac{3t}{4} \gt$
$x=\dfrac{t}{2}$ and $ y= -1+\dfrac{4t}{3}$ and $z=1-\dfrac{3t}{4} $ ; $0 \leq t \leq 1$
Work Step by Step
The General vector line equation for the given two points is defined as:
$r(t)=(1-t) r_0+t \times r_1$
Now, we have
$r(t)=(1-t) \times \lt 0,-1, 1 \gt +t \times \lt \dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4} \gt$
$\implies \lt 0,-1+t,1-t \gt + \lt (\dfrac{1}{2})t,(\dfrac{1}{3}) t,(\dfrac{1}{4})t \gt$
$\implies \lt \dfrac{t}{2},-1+\dfrac{4t}{3},1-\dfrac{3t}{4} \gt$
Answer: $r(t) =\lt \dfrac{t}{2},-1+\dfrac{4t}{3},1-\dfrac{3t}{4} \gt$
Hence, the parametric equations are:
$x=\dfrac{t}{2}$ and $ y= -1+\dfrac{4t}{3}$ and $z=1-\dfrac{3t}{4} $ ; $0 \leq t \leq 1$
