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Chapter 11: Problem 15

Find parametric equations for the line through (-2,1,5) and (6,2,-3)

Short Answer

Expert verified
The parametric equations are \(x = -2 + 8t\), \(y = 1 + t\), \(z = 5 - 8t\).

Step by step solution

01

Identify Two Points on the Line

We have two points, \((-2, 1, 5)\) and \(6, 2, -3)\), through which the line passes. Let鈥檚 denote them as \((x_1, y_1, z_1) = (-2, 1, 5)\) and \((x_2, y_2, z_2) = (6, 2, -3)\).
02

Find the Direction Vector

The direction vector \( extbf{d} \) of the line can be found by subtracting the coordinates of the two points: \[ extbf{d} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) = (6 - (-2), 2 - 1, -3 - 5) = (8, 1, -8). \]
03

Write the Parametric Equations

Using the point \(-2, 1, 5\) and the direction vector \(8, 1, -8)\), we can write the parametric equations of the line as follows:\[ x = -2 + 8t \ y = 1 + t \ z = 5 - 8t, \] where \(t\) is a parameter.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direction Vector
A direction vector is crucial for defining a line in three-dimensional space. It describes the line's direction or the tangent along which it extends. Think of it as an arrow pointing through the line's path.
To find this vector, we subtract the components of one point from the other. In our example, we used the points
  • Point 1: \((-2, 1, 5)\)
  • Point 2: \((6, 2, -3)\)
The direction vector \(\textbf{d}\) is calculated as: \[\textbf{d} = (6 - (-2), 2 - 1, -3 - 5) = (8, 1, -8) \]
  • Each component tells how much the line extends in the respective axis direction: 8 units in \(x\), 1 unit in \(y\), and -8 units in \(z\).
  • This vector is essential in forming the parametric equation of a line.
Points on a Line
Understanding how to use points on a line is essential in forming parametric equations. You need at least two points to define a line. Each point represents a position in 3D space using coordinates \((x, y, z)\).
Let's take a look at how they were used in the exercise:
  • First Point: \((-2, 1, 5)\)
  • Second Point: \((6, 2, -3)\)
  • The coordinates \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) helped in computing the direction vector, which is instrumental in defining the line's path.
  • Even though the direction may differ, any point on the line can be expressed using parametric equations that incorporate these coordinates.
Knowing how to efficiently select and utilize these points simplifies your work when dealing with vector algebra or geometry problems.
3D Line Equations
Parametric equations provide a powerful way to describe lines in three dimensions. Unlike 2D lines, where an equation might be in the form \(y = mx + b\), 3D lines require a vector approach.
The parametric form is particularly useful as it succinctly captures both a point through which the line passes and its direction. Let's express it using our example:
  • Starting Point: \((-2, 1, 5)\)
  • Direction Vector: \((8, 1, -8)\)
Using these, the parametric equations for the line become:
  • \(x = -2 + 8t\)
  • \(y = 1 + t\)
  • \(z = 5 - 8t\)
Each of these equations expresses the line's path in the \(x\), \(y\), and \(z\) directions as \(t\) changes.
This format efficiently conveys how the line extends through space, giving a complete picture of its orientation and placement.

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Most popular questions from this chapter

Find the curvature \(\kappa,\) the unit tangent vector \(\mathbf{T},\) the unit normal vector \(\mathbf{N},\) and the binormal vector \(\mathbf{B}\) at \(t=t_{1}\). $$ x=7 \sin 3 t, y=7 \cos 3 t, z=14 t, t_{1}=\pi / 3 $$

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