91Ó°ÊÓ

To express the line in vector form, we use the equation of a line in three-dimensional space. This equation is written as:
\( \mathbf{r}(t) = \mathbf{r}_0 + t \mathbf{v} \)
Here, \( \mathbf{r}(t) \) represents the position vector of any point on the line, \( \mathbf{r}_0 \) is the position vector of a known point on the line, and \( \mathbf{v} \) is the direction vector. The parameter \( t \) can be any real number.
In our case, the point \( P = (2, 3, -2) \) gives us \( \mathbf{r}_0 \) as \( \begin{pmatrix} 2 \ 3 \ -2 \end{pmatrix} \). The direction vector \( \mathbf{v} = (5, 4, -3) \) is given by \( \begin{pmatrix} 5 \ 4 \ -3 \end{pmatrix} \). Therefore, we can write the vector form of the line as:
\(\mathbf{r}(t) = \begin{pmatrix} 2 \ 3 \ -2 \end{pmatrix} + t \begin{pmatrix} 5 \ 4 \ -3 \end{pmatrix}\).
This form is particularly useful because it clearly shows the direction in which the line extends and requires only a point and a direction vector.

The parametric form of a line offers another way to represent a line in space. In this form, each coordinate is written as a function of a single parameter, typically denoted by \( t \).
For a line passing through \( P = (2, 3, -2) \) and parallel to the vector \( \mathbf{v} = (5, 4, -3) \), the parametric equations are found by writing:

• \( x = 2 + 5t \)
• \( y = 3 + 4t \)
• \( z = -2 - 3t \)

Here, \( t \) represents any real number. These equations decompose the line into its x, y, and z components.
The main advantage of parametric equations is that they are straightforward and allow us to compute the coordinates of any point on the line for a given value of \( t \). This form is especially useful in applications where the line's position at specific intervals is needed, such as in computer graphics and physics simulations.

The symmetric form of a line involves expressing the parametric equations in a way that eliminates the parameter \( t \). This results in a single equation which implicitly contains the parameter. For our line through \( P = (2, 3, -2) \) and parallel to the vector \( \mathbf{v} = (5, 4, -3) \), we start from the parametric forms:

• \( x = 2 + 5t \)
• \( y = 3 + 4t \)
• \( z = -2 - 3t \)

We then isolate \( t \) in each equation:

• \( t = \frac{x - 2}{5} \)
• \( t = \frac{y - 3}{4} \)
• \( t = \frac{z + 2}{-3} \)

By setting these equal to each other, we get the symmetric form of the line:
\(\frac{x - 2}{5} = \frac{y - 3}{4} = \frac{z + 2}{-3}\).
The symmetric form is convenient when trying to establish if a point lies on the line, as it uses a single relationship to link x, y, and z coordinates. Its simplicity is beneficial in geometric interpretations and some analytical methods.