Q 26. - Calculus

Question

Find an equation of the line containing the given point and parallel to the given vector. Express your answer 

(a) as a vector parametrization 

(b) in terms of parametric equations

(c) in symmetric form.

$P(x_0, y_0, z_0), d = \left(a, b, c\right)$

Step-by-Step Solution

Answer

Part (a) The required equation is $r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)$

Part (b) $x\left(t\right)=x_0+at,y\left(t\right)=y_0+bt,z\left(t\right)=z_0+ct$

Part (c) $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$

Part (a) Step 1: Given information

The point $P\left(x_0,y_0,z_0\right)$ and the direction vector $d=(a, b, c)$

Part (a) Step 2: Calculation

The goal is to discover the vector parametrization form equation of a line $\mathcal{L}$ for a given point and direction vector.

The following is the formula for determining the line $\mathcal{L}$ equation:

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

For $P\left(x_0,y_0,z_0\right)d=(a,b,c)$ the equation is,

$r\left(t\right)=\left(x_0,y_0,z_0\right)+t(a,b,c)$

The equation is written as follows,

$$\begin{gathered} r\left(t\right)=\left(x_0+a·t,y_0+b·t,z_0+c·t\right) \\ r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right) \end{gathered}$$

The equation $\mathcal{L}$ in the form of vector parametrization is $r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)$

 Therefore, the required equation is $r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)$

Part (b) Step 1: Calculation

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

In the form of vector parameterization, the equation of line $\mathcal{L}$ is

$r\left(t\right)=\left(x_0+at,y_0+bt,z_0+ct\right)$

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

Then, $r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)=\left(x_0+at,y_0+bt,z_0+ct\right)$

Thus, the parametric equations are $x\left(t\right)=x_0+at,y\left(t\right)=y_0+bt,z\left(t\right)=z_0+ct$

Therefore, the answer is $x\left(t\right)=x_0+at,y\left(t\right)=y_0+bt,z\left(t\right)=z_0+ct$

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 $x\left(t\right)=x_0+at,y\left(t\right)=y_0+bt,z\left(t\right)=z_0+ct$

Take $x\left(t\right)=x_0+at$

$x=x_0+at$

$-x_0$ should be added to both sides of the equation.

$$\begin{gathered} x-x_0=x_0+at-x_0 \\ x-x_0=x_0+at-y_0 \\ x-x_0=at \end{gathered}$$

Divide by $a$ on both sides of the equation.

$\frac{x-x_0}{a}=\frac{at}{a}$

$\frac{x-x_0}{a}=t \dots \dots \left(1\right)$

Take $y\left(t\right)=y_0+bt$

$y=y_0+bt$

$-y_0$ should be added to both sides of the equation.

$$\begin{gathered} y-y_0=y_0+bt-y_0 \\ y-y_0=y_0+bt-y_0 \\ y-y_0=bt \end{gathered}$$

Divide by$b$ on both sides of the equation.

$$\begin{gathered} \frac{y-y_0}{b}=\frac{bt}{b} \\ \frac{y-y_0}{b}=t\dots \dots \left(2\right) \end{gathered}$$

Take $z\left(t\right)=z_0+ct$

$z=z_0+ct$

Part (c) Step 2: Calculation

$-z_0$ should be added to both sides of the equation.

$$\begin{gathered} z-z_0=z_0+ct-z_0 \\ z-z_0=70+ct-\backslash \mathrm{not}0 \\ z-z_0=ct \end{gathered}$$

Divide by $c$ on both sides of the equation.

$$\begin{gathered} \frac{z-z_0}{c}=\frac{ct}{c} \\ \frac{z-z_0}{c}=t \dots \dots \left(3\right) \end{gathered}$$

They can be represented in the following form by equating the equations $\left(1\right),\left(2\right),\left(3\right)$, which are $\frac{x-x_0}{a}=t,\frac{y-y_0}{b}=t,\frac{z-z_0}{c}=t$

Thus,

$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}=t$

Thus the symmetric equations are $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$

Therefore, the required answer is $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$