Let L be the line determined by the system of equations

$x\left(t\right)=4,y\left(t\right)=3-5t,z\left(t\right)=t,-\infty <t<\infty$

(a) Provide a vector parametrization for $L$

(b) Write an equation for $L$ in symmetric form.

Consider the line $\mathcal{L}$ determined by the equations$x\left(t\right)=4,y\left(t\right)=3-5t,z\left(t\right)=t$ $-\infty \le t\le \infty$

The goal is to use vector parametrization to express the equation $\mathcal{L}$

Given equations of the line $\mathcal{L}$are $x\left(t\right)=4,y\left(t\right)=3-5t,z\left(t\right)=t$

The vector parametrization of the line $\mathcal{L}$ is represented by $r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)$

Then, $r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)=(4,3-5t,t)$

Thus, the vector parameterization is $r\left(t\right)=(4,3-5t,t)$

Therefore, the answer is $r\left(t\right)=(4,3-5t,t)$

The goal is to write the symmetric form of the equation $\mathcal{L}$

Remove the parameter $t$from the parametric equations of the line $L$to write the symmetric form.

The parametric equations are $x\left(t\right)=4,y\left(t\right)=3-5t,z\left(t\right)=t$

Take $x\left(t\right)=4$

$x=4\dots \dots \left(1\right)$

Take $y\left(t\right)=3-5t$

$y=3-5t$

On both sides of the equation, add $-3$

$$\begin{gathered} y-3=3-5t-3 \\ y-3=\backslash \mathrm{not}\beta -5t-\backslash \mathrm{not}\beta \\ y-3=-5t \end{gathered}$$

On both sides of the equation, multiply by $-5$

$$\begin{gathered} \frac{y-3}{-5}=\frac{-5t}{-5} \\ \frac{y-3}{-5}=t\dots \dots \left(2\right) \end{gathered}$$

Take $z\left(t\right)=t$

$z=t\text{......(3)}$

By equating the equations $\left(2\right),\left(3\right)$ that is $\frac{y-3}{-5}=t,z=t$ and $\left(1\right)x=4$ they can be written in the following way.

Thus, the symmetric equations are $x=4,\frac{y-3}{-5}=z$

Therefore, the required answer is $x=4,\frac{y-3}{-5}=z$