Q 29. - 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.

$P(0, 0, 0), Q(4,-1, 6)$

Step-by-Step Solution

Answer

Part (a) The required equation is $r\left(t\right)=(4 t,-t, 6 t)$

Part (b) $x\left(t\right)=4 t, y\left(t\right)=-t, z\left(t\right)=6 t$

Part (c) $\frac{x}{4}=-y=\frac{z}{6}$

Part (a) Step 1: Given information

The points $P(0,0,0)$ and $Q(4,-1,6)$

Part (a) Step 2: Calculation

The goal is to figure out how to vector parametrize the line segment joining $P$ to $Q$

We'll start by determining the direction vector of the line $\overrightarrow{PQ}$

The points are $P(0,0,0)$ and $Q(4,-1,6)$

$$\begin{gathered} \overrightarrow{PQ}=(4-0,-1-0,6-0) \\ \overrightarrow{PQ}=(4,-1,6) \end{gathered}$$

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 $P(0,0,0)$ and $\overrightarrow{PQ}=d=(4,-1,6)$ then the equation is,

$$\begin{gathered} r\left(t\right)=(0,0,0)+t(4,-1,6) \\ r\left(t\right)=(0+4t,0-t,0+6t) \end{gathered}$$

The equation is written as follows,

$r\left(t\right)=(4 t,-t, 6 t)$

The equation of a line $\mathcal{L}$ in the form of vector parameterization is $r\left(t\right)=(4 t,-t, 6 t)$

Therefore, the required equation is $r\left(t\right)=(4 t,-t, 6 t)$

Part (b) Step 1: Calculation

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

In the form of vector parametrization, the equation of line $\mathcal{L}$ is $r\left(t\right)=(4 t,-t, 6 t)$

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

$r\left(t\right)=\left(x\right(t), y(t), z(t\left)\right)=(4 t,-t, 6 t)$

Thus, the parametric equations are $x\left(t\right)=4 t, y\left(t\right)=-t, z\left(t\right)=6 t$

Therefore, the answer is $x\left(t\right)=4 t, y\left(t\right)=-t, z\left(t\right)=6 t$

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 $x\left(t\right)=4 t, y\left(t\right)=-t, z\left(t\right)=6 t$

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

$x=4 t$

Divide by four on both sides of the equation.

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

$\frac{x}{4}=t\text{,.....(1) }$

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

$y=-t$

Negatively multiply both sides of the equation.

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

Take $z\left(t\right)=6 t$

$z=6 t$

Divide by six on both sides of the equation.

$\frac{z}{6}=\frac{6t}{6}$

$\frac{z}{6}=t \cdots \cdots \left(3\right)$

They can be represented in the following form by equating the equations $\left(1\right),\left(2\right),\left(3\right)$ i.e. $\frac{x}{4}=t,-y=t,\frac{z}{6}=t$

Thus,

$\frac{x}{4}=-y=\frac{z}{6}=t$

Thus, the symmetric equations are $\frac{x}{4}=-y=\frac{z}{6}$

Therefore, the required answer is $\frac{x}{4}=-y=\frac{z}{6}$