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Parametric equations are a set of equations that express the coordinates of the points making up a geometric object as functions of a variable, often called a parameter. In the context of lines, these equations are used to describe the position of points along the line as the parameter changes.

For a line in three dimensions, the parametric equations take the form:

• \( x = x_0 + at \)
• \( y = y_0 + bt \)
• \( z = z_0 + ct \)

Here, \( (x_0, y_0, z_0) \) is a specific point on the line (often called a base point) and \( (a, b, c) \) are the direction ratios determining the line's orientation. The variable \( t \) is the parameter that "moves" the point along the line. Each value of \( t \) generates a new point on the line.

In the given exercise, our line's parametric equations were translated into \( x = 0 \), \( y = 2 \), \( z = 3 + 2t \). So, no matter how \( t \) changes, \( x \) and \( y \) remain constant, while \( z \) varies linearly.

The direction vector is a foundational element in understanding the vector equation of a line. It specifies the orientation of the line in three-dimensional space. For a line, the direction vector is parallel to all points on the line, meaning it guides the path that the line follows.

In the vector equation \[ \mathbf{r} = \mathbf{r}_0 + t \mathbf{d} \]\( \mathbf{d} \) represents the direction vector of the line. This vector remains constant because it only describes the path direction, not the position or length of the line. It plays a critical role in finding parametric equations because the coefficients of the parameter \( t \) in those equations are derived directly from this vector.

In our given problem, the original line had the equation \[ \mathbf{r} = \mathbf{i} - 2\mathbf{j} + 2t\mathbf{k}, \]which gave the direction vector \( \mathbf{d} = 0\mathbf{i} + 0\mathbf{j} + 2\mathbf{k} \). Thus, the new line, being parallel, shares the same direction vector \( \mathbf{d} \). This tells us the line will move "upwards" along the \( z \)-axis by "2 units per unit of \( t \)."

A line that passes through a specific point in three-dimensional space can be described using the point-direction form of the line equation. This method combines a starting point on the line with a direction vector, providing a complete description of the line's path.

The vector equation for a line through a given point \( (x_0, y_0, z_0) \) is written as: \[ \mathbf{r} = \mathbf{r}_0 + t \mathbf{d}, \]where \( \mathbf{r}_0 = x_0\mathbf{i} + y_0\mathbf{j} + z_0\mathbf{k} \) is the position vector marking the starting point, and \( \mathbf{d} \) is the direction vector. The parameter \( t \) is used to "travel" along the line.

Using the example from the exercise, the line intersects the point (0, 2, 3), so the position vector \( \mathbf{r}_0 \) becomes \( 0\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \). Controlling \( t \), we shift along the direction vector, replicating the orientation of the referred original line, which was parallel. This gives the new line's vector equation \[ \mathbf{r} = (0\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) + t(0\mathbf{i} + 0\mathbf{j} + 2\mathbf{k}). \]