Question

Find the symmetric equations and the parametric equations of a line, and/or the equation of the plane satisfying the following given conditions. Line through (1,-1,-5) and (2,-3,-3)

Short answer

Parametric: x = 1 + t, y = -1 - 2t, z = -5 + 2t.Symmetric: \[ \frac{x - 1}{1} = \frac{y + 1}{-2} = \frac{z + 5}{2} \].

Step-by-step solution

- Find the direction vector

To determine the direction vector \vec{d}\ of the line, subtract the coordinates of the two given points. Specifically, if the line passes through points A(1, -1, -5) and B(2, -3, -3), then the direction vector \vec{d}\ is \[ \vec{d} = B - A = (2-1, -3 -(-1), -3 -(-5)) = (1, -2, 2) \]

- Write the parametric equations

Using point A(1, -1, -5) and the direction vector \vec{d} = (1, -2, 2), write the parametric equations of the line. The general forms are: \[(x, y, z) = (x_0, y_0, z_0) + t(a, b, c)\]where (x_0, y_0, z_0) is a point on the line and (a, b, c) is the direction vector: \[x = 1 + t\]\[y = -1 - 2t\]\[z = -5 + 2t\]

- Write the symmetric equations

To convert the parametric equations to symmetric form, solve each parametric equation for t and set them equal to each other: \[t = x - 1\]\[t = \frac{y + 1}{-2}\]\[t = \frac{z + 5}{2}\]Therefore, the symmetric equations are: \[\frac{x - 1}{1} = \frac{y + 1}{-2} = \frac{z + 5}{2}\]

Key concepts

Direction Vector

To find the equation of a line in 3D, one of the first steps is to determine the direction vector. The direction vector shows the direction in which the line is pointing. For a line passing through points A(1, -1, -5) and B(2, -3, -3), the direction vector \(\text{\vec{d}}\) is found by subtracting the coordinates of point A from point B. Thus, \(\text{\vec{d}} = B - A = (2 - 1, -3 -(-1), -3 -(-5)) = (1, -2, 2)\). This vector, (1, -2, 2), will be used in further calculations when deriving parametric and symmetric equations.

Parametric Equations

Once the direction vector is known, parametric equations can describe the line in 3D. A parametric equation uses a parameter \(t\) to express the coordinates of any point on the line. Using a point on the line, say A(1, -1, -5), and the direction vector (1, -2, 2), the parametric equations are:

• \(x = 1 + t\)
• \(y = -1 - 2t\)
• \(z = -5 + 2t\)

Here \(t\) is any real number. As \(t\) changes, (x, y, z) traces out the line in three-dimensional space. These equations offer a straightforward way to generate the coordinates of points on the line.

Symmetric Equations

Symmetric equations eliminate the parameter \(t\) by setting each parametric equation equal to \(t\) and then equating them. From the parametric equations:

• \(x = 1 + t\)
• \(y = -1 - 2t\)
• \(z = -5 + 2t\)

We solve each for \(t\):

• \(t = x - 1\)
• \(t = \frac{y + 1}{-2}\)
• \(t = \frac{z + 5}{2}\)

By setting these expressions for \(t\) equal to each other, we get the symmetric equations: \(\frac{x - 1}{1} = \frac{y + 1}{-2} = \frac{z + 5}{2}\). These equations describe the same line as the parametric equations, but without using the parameter \((t)\).

3D Geometry

Understanding lines in 3D involves familiarity with 3D geometry, which deals with points, lines, planes, and their interactions in three-dimensional space. When you're working with lines in 3D, it's essential to visualize how they behave compared to 2D lines. For example, lines in 3D can intersect, be parallel, or skew (i.e., they do not intersect and are not parallel). These concepts extend our 2D intuitions to a space where three coordinates (x, y, z) define each position. Every new dimension adds complexity but also provides new opportunities for describing shapes and forms in a space more similar to our real-life experiences.

Line Through Two Points

To describe a line that passes through two points in 3D, we need both the direction vector and a point on the line. Given two points, A and B:

• A(1, -1, -5)
• B(2, -3, -3)

The direction vector \(\text{\vec{d}}\) obtained by subtracting these coordinates is \((1, -2, 2)\). With either A or B as the reference point, we can write the parametric and symmetric equations as explained earlier. Finding the line through two points is fundamental in geometry and linear algebra. It often serves as the basis for further analysis, such as calculating distances between points or angles between lines and planes.