Q 25. - 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.

$P(-1, 3, 7), d = \left(2, 0, 4\right)$

Step-by-Step Solution

Answer

Part (a) The required equation is $r\left(t\right)=(-1+2 t, 3,7+4 t)$

Part (b) $x\left(t\right)=-1+2 t, y\left(t\right)=3, z\left(t\right)=7+4 t$

Part (c) $y=3,\frac{x+1}{2}=\frac{z-7}{4}$

Part (a) Step 1: Given information

The point $P(-1,3,7)$ and the direction vector $d=(2,0,4)$

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 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.

For $P(-1,3,7), d=(2,0,4)$ the equation is,

$r\left(t\right)=(-1,3,7)+t(2,0,4)$

The equation is written as follows,

$$\begin{gathered} r\left(t\right)=(-1+2t,3+0·t,7+4t) \\ r\left(t\right)=(-1+2t,3,7+4t) \end{gathered}$$

The equation $\mathcal{L}$ in the form of vector parametrization is $r\left(t\right)=(-1+2 t, 3,7+4 t)$ Therefore, the required equation is $r\left(t\right)=(-1+2 t, 3,7+4 t)$

Part (b) Step 1: Explanation

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 $r\left(t\right)=(-1+2 t, 3,7+4 t)$

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

Then, $r\left(t\right)=\left(x\right(t), y(t), z(t\left)\right)=(-1+2 t, 3,7+4 t)$

Thus, the parametric equations are $x\left(t\right)=-1+2 t, y\left(t\right)=3, z\left(t\right)=7+4 t$

Therefore, the answer is $x\left(t\right)=-1+2 t, y\left(t\right)=3, z\left(t\right)=7+4 t$

Part (c) Step 1: Explanation

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

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

The parametric equations are $x\left(t\right)=-1+2 t, y\left(t\right)=3, z\left(t\right)=7+4 t$ then,

Take $x\left(t\right)=-1+2 t$

$x=-1+2 t$

On both sides of the equation, add $1$

$x+1=-1+2 t+1$

Thus,

$$\begin{gathered} x+1=A+2t+\backslash \mathrm{not} \\ x+1=2t \end{gathered}$$

On both sides of the equation, divide by two.

$$\begin{gathered} \frac{x+1}{2}=\frac{2t}{2} \\ \frac{x+1}{2}=t\dots \dots \left(1\right) \end{gathered}$$

Take the parametric equation $y\left(t\right)=3$

$y=3 \dots \dots \left(2\right)$

Now take the parametric equation $z\left(t\right)=7+4 t$

$z=7+4 t$

On both sides of the equation, add $-7$

$$\begin{gathered} z-7=7+4t-7 \\ z-7=7+4t>7 \\ z-7=4t \end{gathered}$$

Part (c) Step 2: Explanation

On both sides of the equation, divide by four.

$$\begin{gathered} \frac{z-7}{4}=\frac{4t}{4} \\ \frac{z-7}{4}=t\dots \dots \left(3\right) \end{gathered}$$

By equating the equations $\left(1\right),\left(3\right)$ that is $\frac{x+1}{2}=t,\frac{z-7}{4}=t$ and equation $\left(2\right) y=3$, they can be written as following way.

Thus,

$\frac{x+1}{2}=\frac{z-7}{4}=t,y=3$

Thus the symmetric equations are $y=3,\frac{x+1}{2}=\frac{z-7}{4}$

Therefore, the required answer is $y=3,\frac{x+1}{2}=\frac{z-7}{4}$