91Ó°ÊÓ

Parametric equations are a valuable tool in vector calculus that allow us to express the coordinates of a line in terms of a single parameter, often denoted as \( t \). This approach is particularly useful in three-dimensional space where lines do not automatically align with coordinate axes. For a line running through a point \( P(x_0, y_0, z_0) \) and guided by a direction vector \( \mathbf{v} = (a, b, c) \), the parametric equations are expressed as:

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

These equations essentially describe how you move along the line as \( t \) changes. In the exercise, the point \( P \) is \((2, 3, 0)\) and the direction vector is \( (0, 0, 1) \). Plugging these into the parametric form gives us:

• \( x = 2 \)
• \( y = 3 \)
• \( z = t \)

It's evident here that as \( t \) changes, only the \( z \)-value varies, meaning the line is vertical, running parallel to the z-axis, but fixed on \( x = 2 \) and \( y = 3 \).

Symmetric equations offer another means to describe the trajectory of a line, yet they require some conditions; specifically, the direction vector's components in all three dimensions should be non-zero. The symmetric form is derived by taking each component of the parametric equations and solving for \( t \):

• \( \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \)

However, in our scenario, the direction vector \( \mathbf{v} = (0, 0, 1) \) means that both \( a \) and \( b \) are zero. This leads to both the x and y coordinates being constant and independent of \( t \), thus, symmetric equations do not precisely apply:

• \( x = 2 \), not allowing for variation
• \( y = 3 \), similarly static

The lack of a valid symmetric form in this case reflects that the line is vertical, stretched across the z-axis. It's a limitation when a line is parallel to any axis, but still essential to understand the nature of the line being described.

In vector calculus, finding where a line intersects a plane is a common task, and for the xy-plane specifically, the intersection occurs where the \( z \)-coordinate of the line is zero. For parametric equations of any line, you solve for \( t \) in the equation \( z = t \), setting \( z \) to zero:

• For the given parametric form \( z = t \), the intersection occurs when \( t = 0 \).

Plugging \( t = 0 \) back into the equations for \( x \), \( y \), and \( z \) provides the coordinates of the intersection point:

• \( (x, y, z) = (2, 3, 0) \)

This tells us that the line crosses the xy-plane precisely at the initial point \( P \), which is indicative of the line having originated on that plane. Understanding these intersections helps visualize where and how geometric objects relate in three-dimensional space, an essential skill in vector calculus.