Question

find a vector equation and parametric equations for the line. (use the parameter t.) the line through the point (8, 0, -4) and parallel to the line x = 4 - 2t, y = -1 + 3t, z = 6 + 8t r(t) = (x(t), y(t), z(t)) = ()

Explanation:

Step1: Identify the direction vector

The direction vector of the given line $x = 4 - 2t,y=-1 + 3t,z = 6+8t$ is $\vec{v}=\langle - 2,3,8
angle$. Since the required line is parallel to this line, it has the same direction vector.

Step2: Write the vector - equation of the line

The vector - equation of a line passing through the point $P(x_0,y_0,z_0)$ and having direction vector $\vec{v}=\langle a,b,c
angle$ is $\vec{r}(t)=\vec{r}_0+t\vec{v}$, where $\vec{r}_0=\langle x_0,y_0,z_0
angle$. Here, $x_0 = 8,y_0 = 0,z_0=-4$ and $\vec{v}=\langle - 2,3,8
angle$. So, $\vec{r}(t)=\langle8,0, - 4
angle+t\langle - 2,3,8
angle=\langle8 - 2t,3t,-4 + 8t
angle$.

Step3: Write the parametric equations of the line

If $\vec{r}(t)=\langle x(t),y(t),z(t)
angle=\langle8 - 2t,3t,-4 + 8t
angle$, then the parametric equations are $x(t)=8 - 2t$, $y(t)=3t$, $z(t)=-4 + 8t$.

Answer:

$\vec{r}(t)=\langle8 - 2t,3t,-4 + 8t
angle$
$(x(t),y(t),z(t))=(8 - 2t,3t,-4 + 8t)$