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.

$P(2,3,5),d=\left(2,3,5\right)$

Short answer

Part (a) The required equation is$r\left(t\right)=(2+2t,3+3t,5+5t)$

Part (b)$x\left(t\right)=2+2t,y\left(t\right)=3+3t,z\left(t\right)=5+5t$

Part (c)$\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-5}{5}$

Step-by-step solution

Part (a) Step 1: Given information

The point $P(2,3,5)$ and the direction vector $d=(2,3,5)$

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 $P(2,3,5),d=(2,3,5)$the equation is,

$r\left(t\right)=(2,3,5)+t(2,3,5)$

The equation is written as follows,

r(t)=(2+2 t, 3+3 t, 5+5 t)

The equation $\mathcal{L}$in the form of vector parametrization is $r\left(t\right)=(2+2t,3+3t,5+5t)$ Therefore, the required equation is $r\left(t\right)=(2+2t,3+3t,5+5t)$

Part (b) Step 1: Explanation

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

The equation of a line in vector parameterization is $r\left(t\right)=(2+2t,3+3t,5+5t)$

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

Then, $r\left(t\right)=\left(x\right(t),y(t),z(t\left)\right)=(2+2t,3+3t,5+5t)$

Thus, the parametric equations are $x\left(t\right)=2+2t,y\left(t\right)=3+3t,z\left(t\right)=5+5t$

Therefore, the answer is $x\left(t\right)=2+2t,y\left(t\right)=3+3t,z\left(t\right)=5+5t$

Part (c) Step 1: Explanation

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 $x\left(t\right)=2+2t,y\left(t\right)=3+3t,z\left(t\right)=5+5t$. then,

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

On both sides of the equation, add $-2$

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

Thus,

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

On both sides, divide the equation by two.

$\frac{x-2}{2}=\frac{2t}{2}$

$\frac{x-2}{2}=t\dots \dots \left(1\right)$

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

$y=3+3t$

On both sides of the equation, add -3.

$$\begin{gathered} y-3=3+3t-3 \\ y-3=\backslash \mathrm{not}\beta +3t-\backslash \mathrm{not}\beta \\ y-3=3t \end{gathered}$$

Subtract $3$ from both sides of the equation.

$$\begin{gathered} \frac{y-3}{3}=\frac{3t}{3} \\ \frac{y-3}{3}=t\dots \dots \left(2\right) \end{gathered}$$

Now take the parametric equation $z\left(t\right)=5+5t$

$z=5+5t$

On both sides of the equation, add $-5$

$$\begin{gathered} z-5=5+5t-5 \\ z-5=\backslash \mathrm{not}+5t-5 \\ z-5=5t \end{gathered}$$

Part (c) Step 2: Explanation

On both sides of the equation, multiply by $5$

$$\begin{gathered} \frac{z-5}{5}=\frac{5t}{5} \\ \frac{z-5}{5}=t\dots \dots \left(3\right) \end{gathered}$$

By equating the equations $\left(1\right),\left(2\right),\left(3\right)$that is $\frac{x-2}{2}=t,\frac{y-3}{3}=t,\frac{z-5}{5}=t$they can be written as following way.

Then, $\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-5}{5}=t$

Thus the symmetric equations are $\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-5}{5}$

Therefore, the required answer is $\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-5}{5}$