Q 27. - 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.

$P(3, 1), d = \left(2, 5\right)$

Step-by-Step Solution

Answer

Part (a) The required equation is $r\left(t\right)=(3+2 t, 1+5 t)$

Part (b) $x\left(t\right)=3+2 t, y\left(t\right)=1+5 t$

Part (c) $\frac{x-3}{2}=\frac{y-1}{5}$

Part (a) Step 1: Given information

The point $P(3,1)$ and the direction vector $d=(2,5)$

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 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 $P(3,1) d=(2,5)$ the equation is,

$r\left(t\right)=(3,1)+t(2,5)$

The equation is written as follows,

$r\left(t\right)=(3+2 t, 1+5 t)$

The equation $\mathcal{L}$ in the form of vector parameterization is $r\left(t\right)=(3+2 t, 1+5 t)$

Therefore, the required equation is $r\left(t\right)=(3+2 t, 1+5 t)$

Part (b) Step 1: Calculation

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 parameterization is

$r\left(t\right)=(3+2 t, 1+5 t)$

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

Then, $r\left(t\right)=\left(x\right(t), y(t\left)\right)=(3+2 t, 1+5 t)$

Thus, the parametric equations are $x\left(t\right)=3+2 t, y\left(t\right)=1+5 t$

Therefore, the answer is $x\left(t\right)=3+2 t, y\left(t\right)=1+5 t$

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 $x\left(t\right)=3+2 t, y\left(t\right)=1+5 t$

Take $x\left(t\right)=3+2 t$

$x=3+2 t$

On both sides of the equation, add $-3$

$$\begin{gathered} x-3=3+2t-3 \\ x-3=342t-\backslash \mathrm{not}3 \\ x-3=2t \end{gathered}$$

Divide by two on both sides of the equation.

$\frac{x-3}{2}=\frac{2t}{2}$

$\frac{x-3}{2}=t\dots \dots \left(1\right)$

Take $y\left(t\right)=1+5 t$

$y=1+5 t$

Part (c) Step 2: Calculation

On both sides of the equation, add $-1$

$$\begin{gathered} y-1=1+5t-1 \\ y-1=\lambda +5t-\lambda \\ y-1=5t \end{gathered}$$

Divide by five on both sides of the equation.

$$\begin{gathered} \frac{y-1}{5}=\frac{5t}{5} \\ \frac{y-1}{5}=t\dots \dots \left(2\right) \end{gathered}$$

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

Thus,

$\frac{x-3}{2}=\frac{y-1}{5}=t$

Thus, the symmetric equations are $\frac{x-3}{2}=\frac{y-1}{5}$

Therefore, the required answer is $\frac{x-3}{2}=\frac{y-1}{5}$