Problem 60
Question
Find the parametric equation of the line in \(x-\) \(y-z\) space that goes through the given points. \((2,0,-3)\) and \((4,1,0)\)
Step-by-Step Solution
Verified Answer
The parametric equations are:
\[x = 2 + 2t, \quad y = t, \quad z = -3 + 3t.\]
1Step 1: Identify Direction Vector
To find the parametric equation of a line in space, we first must determine the direction vector. This vector can be found by subtracting the coordinates of the first point from the second point:\[\vec{d} = (4 - 2, 1 - 0, 0 + 3) = (2, 1, 3)\]
2Step 2: Define Parametric Equations
With the direction vector and one point, we can write the parametric equations. Generally, the parametric equations of a line that passes through a point \(x_0, y_0, z_0\) with direction vector \(a, b, c\) are \[x = x_0 + at\y = y_0 + bt\z = z_0 + ct\]
3Step 3: Write Parametric Equations for This Line
Using the point \(2, 0, -3\) and direction vector \(2, 1, 3\), we substitute to find the equations:\[x = 2 + 2t\y = 0 + 1t\z = -3 + 3t\]
4Step 4: Complete Parametric Equations
Thus, the parametric equations describing the line that goes through the given points are:\[x = 2 + 2t\y = t\z = -3 + 3t\]
Key Concepts
Direction Vector3D Coordinate GeometryLine in Space
Direction Vector
The concept of a direction vector is crucial when working with parametric equations in three-dimensional geometry. A direction vector essentially describes the direction and the "slope" of a line. When given two points, like \( (2, 0, -3) \) and \( (4, 1, 0) \), we determine the direction by subtracting corresponding coordinates of these points.
- Start with the x-coordinates: \( 4 - 2 = 2 \)
- Next, the y-coordinates: \( 1 - 0 = 1 \)
- Finally, the z-coordinates: \( 0 - (-3) = 3 \)
3D Coordinate Geometry
Three-dimensional coordinate geometry involves analyzing shapes, lines, and points in a space defined by three axes: x, y, and z. This makes it possible to describe and visualize objects and their relations in a space with depth, unlike the traditional 2D plane. In 3D geometry, understanding vectors and how they function can greatly simplify problems. Instead of viewing a line as a series of points, we think of it as a direction governed by its direction vector.
- The point coordinates, such as \( (2, 0, -3) \), give us the starting point in the space.
- The direction vector \( (2, 1, 3) \) then propels us in a particular direction within this space.
Line in Space
Describing a line in the three-dimensional space involves parametric equations, which rely on both a point on the line and a direction vector. These aspects ensure that every point on the line can be represented. The parametric equations for a line involve assigning each coordinate – x, y, and z – as a function of a parameter, often denoted as \( t \). For example, given \( (2, 0, -3) \) as a point and \( (2, 1, 3) \) as a direction vector, we form equations:\[ x = 2 + 2t \]\[ y = 0 + 1t \ (or \ y = t) \]\[ z = -3 + 3t \]This means that, by altering \( t \), you move along the line uniquely mapped out in the 3D space. Parametric equations have a distinctive advantage in that they offer a straightforward way to traverse the entirety of any line in space, allowing for flexible and practical use in various applications, from graphics to engineering design.
Other exercises in this chapter
Problem 59
$$ \begin{array}{l} \text { In Problems , find the eigenvalues } \lambda_{1} \text { and } \lambda_{2} \text { for each matrix }\\\ A \end{array} $$ $$ A=\left[
View solution Problem 60
Use the determinant to find the inverse of \(A\) $$ A=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right] $$
View solution Problem 60
$$ \begin{array}{l} \text { In Problems , find the eigenvalues } \lambda_{1} \text { and } \lambda_{2} \text { for each matrix }\\\ A \end{array} $$ $$ \text {
View solution Problem 61
Use the determinant to find the inverse of \(A\) $$ A=\left[\begin{array}{rr} -1 & 4 \\ 5 & 1 \end{array}\right] $$
View solution