Q 23. - 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.

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

Step-by-Step Solution

Answer

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

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

Part (c) $x=\frac{y}{2}=\frac{z}{-4}$

Part (a) Step 1: Given information

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

Part (a) Step 2: Explanation

The goal is to determine the vector parametrization for the given point and vector of direction. 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 $P(0,0,0), d=(1,2,-4)$ the equation is,

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

The equation is written as follows,

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

The equation of vector parametrization of a line $\mathcal{L}$ is $r\left(t\right)=(t, 2 t,-4 t)$ Therefore, the required equation is $r\left(t\right)=(t, 2 t,-4 t)$

Part (b) Step 1: Explanation

The goal is to write all of the parametric equations.

The vector parametrization equation is $r\left(t\right)=(t, 2 t,-4 t)$

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 $r\left(t\right)=\left(x\right(t), y(t), z(t\left)\right)=(t, 2 t,-4 t)$

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

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

Part (c) Step 1: Explanation

The goal is to write the equation in its symmetric form.

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

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

Take the parametric equation $y=2 t$

Divide both sides of the equation by two now.

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

$\frac{y}{2}=t \dots \dots \left(2\right)$

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

On both sides of the equation, divide by $-4$

$\frac{z}{-4}=\frac{-4t}{-4}$

$\frac{z}{-4}=t$

They can be represented in the following fashion by equating the equations $\left(1\right),\left(2\right),\left(3\right)$ which are $x=t\frac{y}{2}=t\frac{z}{-4}=t$

Then,

$x=\frac{y}{2}=\frac{z}{-4}=t$

Thus, the symmetric equations are $x=\frac{y}{2}=\frac{z}{-4}$ Therefore, the required answer is $x=\frac{y}{2}=\frac{z}{-4}$