Q 31. - Calculus

Question

Find an equation of the line containing the

given pair of points. Express your answer 

(a) as a vector parametrization.

(b) in terms of parametric equations.

(c) in symmetric form.

$P(-4, 11, 0), Q(4, 11, 2)$

Step-by-Step Solution

Answer

Part (a) The required equation is $r\left(t\right)=(-4+8 t, 11,2 t)$

Part (b) $x\left(t\right)=-4+8 t, y\left(t\right)=11, z\left(t\right)=2 t$

Part (c) $y=11,\frac{x+4}{8}=\frac{z}{2}$

Part (a) Step 1: Given information

The points $P(-4,11,0)$ and $Q(4,11,2)$

Part (a) Step 2: Calculation

The goal is to figure out how to vector parametrize the line segment that connects $P$ and $Q$

First we will find the direction vector for the line $\overrightarrow{PQ}$

The points are $P(-4,11,0)$ and $Q(4,11,2)$

$$\begin{gathered} \overrightarrow{PQ}=(4-(-4),11-11,2-0) \\ \overrightarrow{PQ}=(8,0,2) \end{gathered}$$

The formula to find the line $\mathcal{L}$ equation is as follows,

$r\left(t\right)=P_0+td$ Where, $P_0$ is the point and $d$ is the direction vector.

Here $P(-4,11,0)$ and $\overrightarrow{PQ}=d=(8,0,2)$ then the equation is,

$$\begin{gathered} r\left(t\right)=(-4,11,0)+t(8,0,2) \\ r\left(t\right)=(-4+8t,11+0·t,0+2t) \end{gathered}$$

The equation is written as follows,

$r\left(t\right)=(-4+8 t, 11,2 t)$

The equation of a line $\mathcal{L}$ in the form of vector parametrization is,

$r\left(t\right)=(-4+8 t, 11,2 t)$

Therefore, the required equation is $r\left(t\right)=(-4+8 t, 11,2 t)$

Part (b) Step 1: Calculation

The goal is to use parametric equations to represent the line equation $\mathcal{L}$

The equation of line $\mathcal{L}$ in the form of vector parametrization is,

$r\left(t\right)=(-4+8 t, 11,2 t)$

The vector function $r\left(t\right)$ in three -dimensional plane represents $r\left(t\right)=\left(x\right(t), y(t), z(t\left)\right). r\left(t\right)=\left(x\right(t), y(t), z(t\left)\right)=(-4+8 t, 11,2 t)$

Thus, the parametric equations are $x\left(t\right)=-4+8 t, y\left(t\right)=11, z\left(t\right)=2 t$

Therefore, the answer is $x\left(t\right)=-4+8 t, y\left(t\right)=11, z\left(t\right)=2 t$

Part (c) Step 1: Calculation

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

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

The parametric equations are $x\left(t\right)=-4+8 t, y\left(t\right)=11, z\left(t\right)=2 t$

Take $x\left(t\right)=-4+8 t$

$x=-4+8 t$

On all sides of the equation, add $4$

$$\begin{gathered} x+4=-4+8t+4 \\ x+4=4+8t+4 \\ x+4=8t \end{gathered}$$

Divide by eight on both sides of the equation.

$$\begin{gathered} \frac{x+4}{8}=\frac{8t}{8} \\ \frac{x+4}{8}=t\dots \dots \left(1\right) \end{gathered}$$

Take $y\left(t\right)=11$

$y=11\dots \dots \left(2\right)$

Take $z\left(t\right)=2 t$

$z=2 t$

Divide by two on both sides of the equation.

$\frac{z}{2}=\frac{2t}{2}$

$\frac{z}{2}=t\dots ..\left(3\right)$

By equating the equations $\left(1\right),\left(3\right)$ that is $\frac{x+4}{8}=t,\frac{z}{2}=t$ and the equation $\left(2\right)$ is $y=11$ they can be written in the following way.

Thus,

$y=11,\frac{x+4}{8}=\frac{z}{2}=t$

Thus the symmetric equations are $y=11,\frac{x+4}{8}=\frac{z}{2}$

Therefore, the required answer is $y=11,\frac{x+4}{8}=\frac{z}{2}$