Question

Let $L$ be the line determined by the equation

$\frac{x}{4}=\frac{y+2}{5}=\frac{z-8}{3}$

(a) Provide a vector parametrization for $L$

(b) Give parametric equations for $L$

Short answer

Part (a)$r\left(t\right)=(4t,5t-2,-3t+8)$

Part (b)$x\left(t\right)=4t,y\left(t\right)=5t-2,z\left(t\right)=-3t+8$

Step-by-step solution

Part (a) Step 1: Given information and calculation

Consider the line $\mathcal{L}$determined by the equations $\frac{x}{4}=\frac{y+2}{5}=-\frac{z-8}{3}$

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

Given equations of line $\mathcal{L}$are $\frac{x}{4}=\frac{y+2}{5}=-\frac{z-8}{3}$

Introduce a parameter $t$and identify the values of $\mathcal{L}$to parametrize the equation $x,y,z$

$\frac{x}{4}=\frac{y+2}{5}=-\frac{z-8}{3}=t$

Then,

$\frac{x}{4}=t,\frac{y+2}{5}=t,-\frac{z-8}{3}=t$

$\frac{x}{4}=t$

On both sides of the equation, multiply by four.

$$\begin{gathered} 4·\frac{x}{4}=4·t \\ x=4t\dots \dots \left(1\right) \end{gathered}$$

Take,$\frac{y+2}{5}=t$

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

$$\begin{gathered} 5·\frac{y+2}{5}=5·t \\ y+2=5t \end{gathered}$$

On both sides of the equation, add $-2$

$$\begin{gathered} y+2-2=5t-2 \\ y+\backslash \mathrm{not}2-\backslash \mathrm{not}2=5t-2 \end{gathered}$$

Thus,

$y=5t-2\dots \dots \left(2\right)$

Take, $-\frac{z-8}{3}=t$

Multiply both sides of the equation by three.

$$\begin{gathered} -3·\frac{z-8}{3}=3·t \\ -(z-8)=3t \end{gathered}$$

Part (a) Step 2: calculation

On the subject of additional simplicity,

$$\begin{gathered} -(z-8)=3t \\ -z+8=3t \end{gathered}$$

On both sides of the equation, add$-8$

$$\begin{gathered} -z+8-8=3t-8 \\ -z=3t-8 \end{gathered}$$

On both sides of the equation, multiply by $-1$

$$\begin{gathered} -1·-z=-1(3t-8) \\ z=-3t+8\dots \dots \left(3\right) \end{gathered}$$

Now take the equations $\left(1\right)\left(2\right)\left(3\right)$that is $x=4t,y=5t-2,z=-3t+8$

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

Then, $r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)=(4t,5t-2,-3t+8)$

Thus, the vector parametrization is $r\left(t\right)=(4t,5t-2,-3t+8)$Therefore, the answer is $r\left(t\right)=(4t,5t-2,-3t+8)$

Part (b) Step 1: calculation

The goal is to write the equation $\mathcal{L}$as a set of parametric equations.

The equation of line $\mathcal{L}$in the form of vector parametrization is $r\left(t\right)=(4t,5t-2,-3t+8)$

In a three-dimensional plane, the vector function $r\left(t\right)$ is represented by $r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)$

Then, $r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)=(4t,5t-2,-3t+8)$

Thus, the parametric equations are $x\left(t\right)=4t,y\left(t\right)=5t-2,z\left(t\right)=-3t+8$

Therefore, the answer is $x\left(t\right)=4t,y\left(t\right)=5t-2,z\left(t\right)=-3t+8$