Problem 28 - Precalculus Mathematics for Calculus

Question

A description of a line is given. Find parametric equations for the line. The line crosses the \(x\) -axis where \(x=-2\) and crosses the \(z\) -axis where \(z=10\)

Step-by-Step Solution

Answer

Parametric equations: \(x = -2 + 2t\), \(y = 0\), \(z = 10t\).

Step 1: Identify the Points on the Line

The line crosses the \(x\)-axis at \((-2, 0, 0)\) and it crosses the \(z\)-axis at \((0, 0, 10)\). These are the two points through which the line passes.

Step 2: Find the Direction Vector

To find the direction vector \(\mathbf{d}\), subtract the coordinates of the first point from the coordinates of the second point. This results in \(\mathbf{d} = (0-(-2), 0-0, 10-0) = (2, 0, 10)\).

Step 3: Write the Parametric Equations

The parametric equations of the line are derived using the point-direction form. Using the point \((-2, 0, 0)\) and direction vector \((2, 0, 10)\), the parametric equations are: \[x = -2 + 2t\] \[y = 0\] \[z = 0 + 10t\], where \(t\) is the parameter.

Key Concepts

Direction Vector

The direction vector is a crucial concept when working with lines in three dimensions. In simple terms, a direction vector provides a sense of direction for a line. It's like an arrow that points along the line, showing you which way it goes. This vector is determined by taking two points on the line and subtracting the coordinates of one point from the coordinates of the other. This difference gives you the change in x, y, and z that the line undergoes from one point to another.

• The direction vector is represented as \((d_x, d_y, d_z)\), where each value is the change in the respective coordinate direction.
• The direction vector is not necessarily a unit vector; it only needs to indicate direction.
• In the provided exercise, we computed the direction vector as \((2, 0, 10)\), by calculating the differences between the points \((-2, 0, 0)\) and \((0, 0, 10)\).

Understanding how to find and use the direction vector forms the foundation for writing equations that describe the line.

Point-Direction Form

The point-direction form is a method to express the equation of a line using a known point on the line and a direction vector.
This form is especially useful in three-dimensional geometry as it simplifies the process of writing equations for lines. The formula for the point-direction form is as follows:

• Given a point \((x_0, y_0, z_0)\) on the line and a direction vector \((d_x, d_y, d_z)\), the equations are:
• \(x = x_0 + d_xt\)
• \(y = y_0 + d_yt\)
• \(z = z_0 + d_zt\)

Here, \(t\) is a parameter that varies over all real numbers, effectively tracing out every point on the line.
In our exercise, starting from point \((-2, 0, 0)\) and direction vector \((2, 0, 10)\), we formed the parametric equations:

• \(x = -2 + 2t\)
• \(y = 0\)
• \(z = 0 + 10t\)

This form gives a clear structure to describe any line in space.

Line in Three Dimensions

Describing a line in three dimensions can seem complicated at first, but understanding the building blocks like direction vectors and point-direction form simplifies the process. A line in space is defined by how it extends in three-dimensional coordinates: x, y, and z.
Unlike in two dimensions where a line can be expressed simply with a slope and intercept, in three dimensions, the line's orientation is given by the direction vector. The line is fully described by the set of parametric equations we devised from the point-direction form.

• The direction vector dictates the line's path.
• A specific point on the line helps anchor one end of the line.
• With these two elements, the line can be expressed as a set of equations that map out its path indefinitely.

For students, understanding a line in three dimensions involves visualizing how the line travels through space, passing through and extending beyond the given points. Each equation corresponds to movement along one of the axes, together describing the full trajectory of the line.