91Ó°ÊÓ

Parametric equations are a way to describe a line using a set of equations that capture each coordinate as a function of a parameter, typically denoted by \( t \). In our example, the line is defined by the equations \( x = 5 + 3t \), \( y = 5 + 4t \), and \( z = -3 - 5t \). Each coordinate (

• \( x \) coordinate increases by 3 for every unit increase in \( t \)
• \( y \) coordinate increases by 4 for every unit increase in \( t \)
• \( z \) coordinate decreases by 5 for every unit increase in \( t \).

Parametric equations are useful because they allow us to express a line in three dimensions, making it clear what the point's trajectory would look like as \( t \) changes. This method sets the foundation for analyzing and solving geometry problems involving lines.

Vector projection is an essential concept for finding the shortest distance between a point and a line. To understand it, consider two vectors: a vector \( \textbf{AP} \) from the point to a point on the line and the direction vector \( \textbf{d} \) of the line. The projection of \( \textbf{AP} \) on \( \textbf{d} \) determines how much of \( \textbf{AP} \) aligns with \( \textbf{d} \).
To calculate the projection, we use the formula:\[\text{proj}_d(\textbf{AP}) = \frac{\textbf{AP} \cdot \textbf{d}}{\textbf{d} \cdot \textbf{d}} \textbf{d}\]The dot product \( \textbf{AP} \cdot \textbf{d} \) finds the component of \( \textbf{AP} \) along \( \textbf{d} \), while \( \textbf{d} \cdot \textbf{d} \) normalizes this projection. The resulting vector tells us how far \( \textbf{AP} \) extends in the direction of \( \textbf{d} \), and subtracting this result from \( \textbf{AP} \) helps locate the orthogonal vector, which is crucial for calculating the distance from the point to the line.

Calculating the distance from a point to a line in three-dimensional space involves understanding both the vector geometry and projections. Once the orthogonal vector is found by subtracting the projection of \( \textbf{AP} \) from \( \textbf{AP} \) itself, the task is to find its magnitude:
For the vector \( \textbf{B} = \langle 2, 1, 2 \rangle \), the magnitude is calculated as:\[\| \textbf{B} \| = \sqrt{2^2 + 1^2 + 2^2}\]Breaking it down further:

• \(2^2 = 4\)
• \(1^2 = 1\)
• Another \(2^2 = 4\)

Adding these will give:\[\sqrt{4 + 1 + 4} = \sqrt{9} = 3\]This result, \(3\), represents the shortest distance from the point to the line, emphasizing the importance of orthogonal vectors in distance measurements in vector geometry.