Question

Find the symmetric equations and the parametric equations of a line, and/or the equation of the plane satisfying the following given conditions. Line through (5,-4,2) and parallel to the line $\mathbf{r}=\mathbf{i}-\mathbf{j}+(5\mathbf{i}-2\mathbf{j}+\mathbf{k})t$.

Short answer

Parametric: $x=5+5t$, $y=-4-2t$, $z=2+t$. Symmetric: \ \frac{x - 5}{5} = \frac{y + 4}{-2} = z - 2 \.

Step-by-step solution

Understand the Direction Vector

The given line's equation is $\mathbf{r}=\mathbf{i}-\mathbf{j}+(5\mathbf{i}-2\mathbf{j}+\mathbf{k})t$. This can be rewritten as $\mathbf{r}=(1,-1,0)+t(5,-2,1)$. The direction vector of the line is $(5,-2,1)$.

Identify the Point and Direction of the New Line

The new line passes through the point $(5,-4,2)$ and is parallel to the direction vector $(5,-2,1)$.

Write the Parametric Equations

Using the point $(5,-4,2)$ and the direction vector $(5,-2,1)$, the parametric equations are: $$x=5+5t$$$$y=-4-2t$$$$z=2+t$$

Write the Symmetric Equations

To find the symmetric equations, solve the parametric equations for t:$$t=(x-5)/5$$$$t=(y+4)/-2$$$$t=(z-2)/1$$Setting these equal gives the symmetric equations:$$\frac{x-5}{5}=\frac{y+4}{-2}=z-2$$

Key concepts

Symmetric Equations

Symmetric equations are a way to represent a line in three-dimensional space. They are derived from parametric equations by eliminating the parameter, usually represented by $t$. For symmetric equations, we begin by solving each of the parametric equations for the parameter $t$ and then equating them to form a single expression.

For example, given the parametric equations:
$x=5+5t$
$y=-4-2t$
$z=2+t$
We solve each for $t$:
$t=\frac{x-5}{5}$
$t=\frac{y+4}{-2}$
$t=z-2$
Setting these equations equal to each other gives the symmetric equations:
$$\frac{x-5}{5}=\frac{y+4}{-2}=z-2$$
Simplifying these equations to find the parameter $t$ ensures that we accurately describe the line through a common equation.

Direction Vector

The direction vector of a line indicates the direction the line is pointing in three-dimensional space. It is a vector that, when added to a point on the line, gives another point on the line by scaling it with a parameter, typically $t$.

Given a line described parametrically as:
$\mathbf{\text{r}}=\mathbf{\text{a}}+\mathbf{\text{d}}t$
Where $\mathbf{\text{a}}$ is a point on the line and $\mathbf{\text{d}}$ is the direction vector. The direction vector $\mathbf{\text{d}}$ is essentially the vector formed by the coefficients of $t$ in the parametric equation.

In the problem, the given line is:
$\mathbf{\text{r}}=(1,-1,0)+t(5,-2,1)$
Hence, the direction vector is $(5,-2,1)$. For the new line through the point $(5,-4,2)$ and parallel to the given line, the direction vector remains $(5,-2,1)$.

Parametric Equations

Parametric equations of a line express the coordinates of points on the line as functions of a single parameter, usually $t$. They show how the coordinates change with respect to the parameter.

For a line passing through a point $(x_0,y_0,z_0)$ with a direction vector $(a,b,c)$, the parametric equations can be written as:
$x=x_0+at$
$y=y_0+bt$
$z=z_0+ct$
In our problem, the line passes through the point $(5,-4,2)$ with direction vector $(5,-2,1)$. Thus, the parametric equations are:
$x=5+5t$
$y=-4-2t$
$z=2+t$
These equations show the position of any point on the line as $t$ varies.