91Ó°ÊÓ

Symmetric equations provide a way to describe a line without using a parameter. Compared to parametric equations which use a variable, often denoted as \(t\), symmetric equations elegantly represent the same line by equating expressions derived from parametric forms. Suppose you have parametric equations for a line like \(x = 3 + t\), \(y = 1 + t\), and \(z = 5 + t\). Here, each component contains a \(t\) value that adjusts from the given point \((3,1,5)\) using the direction vector \(\langle 1,1,1 \rangle \). To derive the symmetric equation, we solve each parametric equation for \(t\). These are \(t = x - 3\), \(t = y - 1\), and \(t = z - 5\). The essence of symmetric equations is to link these expressions:

• \(x - 3 = y - 1\)
• \(y - 1 = z - 5\)
• Consequently, \(x - 3 = y - 1 = z - 5\)

This linked equation summarizes the line’s path through space, showing that the difference from the point \((3,1,5)\) follows the same pattern across all axes.

A direction vector is fundamental in defining the path of a line in three-dimensional space. If you imagine a line as an arrow, the direction vector is like the arrow’s shaft pointing in the line’s travel direction. When we refer to a vector such as \(\langle 1,1,1 \rangle\), we are talking about the steps needed in each axis direction to move along that line. The components of this vector \((1,1,1)\) indicate moving one unit along the \(x\), \(y\), and \(z\) axes simultaneously. A direction vector is crucial for multiple mathematical interpretations:

• It assists in forming parametric equations by scaling them via a parameter \(t\).
• Derives symmetric equations when used effectively.
• Acts as a tool in understanding the line’s derivation from a point \((3,1,5)\) into infinite space.

This vector, therefore, serves a dual function—it provides both direction and helps in transforming one-dimensional parametric expressions into robust symmetric equations.

Finding the intersection of a line with a plane involves identifying the point where the line crosses the plane’s surface. In our example, we find where the line intersects the \(xy\)-plane. The crucial idea here is that for the \(xy\)-plane, the \(z\)-coordinate is zero. Consequently, to find the intersection, you substitute this fact into the parametric equation of \(z\).Given \(z = 5 + t\), set this equation to zero:

• \(z = 0\) means \(5 + t = 0\).
• Solving this gives \(t = -5\).

Plugging \(t = -5\) back into \(x = 3 + t\) and \(y = 1 + t\) results in:

• \(x = 3 - 5 = -2\)
• \(y = 1 - 5 = -4\)

Thus, the intersection point is \((-2, -4, 0)\). This process shows how parametric equations can be translated into a real-world position where a line meets a two-dimensional plane.