Problem 59
Question
In Problems \(59-62\), find the parametric equation of the line in \(x-\) \(y-z\) space that goes through the given points. \((5,4,-1)\) and \((2,0,3)\)
Step-by-Step Solution
Verified Answer
The parametric equations are: \(x = 5 - 3t\), \(y = 4 - 4t\), and \(z = -1 + 4t\).
1Step 1: Understand the Parametric Equation for a Line
In three-dimensional space, a line passing through a point \((x_0, y_0, z_0)\) with direction vector \((a, b, c)\) can be described by parametric equations:\[x = x_0 + at, \, y = y_0 + bt, \, z = z_0 + ct\]where \(t\) is a parameter.
2Step 2: Determine the Direction Vector
To find the direction vector \((a, b, c)\), subtract the coordinates of the first point from the second point. For points \((5, 4, -1)\) and \((2, 0, 3)\), the direction vector is:\[(2 - 5, \, 0 - 4, \, 3 - (-1)) = (-3, \, -4, \, 4)\]
3Step 3: Write the Parametric Equations
Using the point \((5, 4, -1)\) and the direction vector \((-3, -4, 4)\), the parametric equations are:- \(x = 5 - 3t\)- \(y = 4 - 4t\)- \(z = -1 + 4t\)
Key Concepts
Direction Vector3D SpaceLine Equations
Direction Vector
The direction vector is a fundamental concept when dealing with lines in geometry, especially in 3D space. It essentially tells us the direction in which a line is oriented. If you're tasked with finding a line that passes through two points in space, the direction vector is calculated by taking the difference between the coordinates of these two points.
For example, if we have two points \(A(5, 4, -1)\) and \(B(2, 0, 3)\), we would find the direction vector by subtracting the coordinates of \(A\) from those of \(B\).
For example, if we have two points \(A(5, 4, -1)\) and \(B(2, 0, 3)\), we would find the direction vector by subtracting the coordinates of \(A\) from those of \(B\).
- Direction along x: \(2 - 5 = -3\)
- Direction along y: \(0 - 4 = -4\)
- Direction along z: \(3 - (-1) = 4\)
3D Space
Understanding 3D space is vital for solving problems involving parametric equations of lines. In mathematical terms, 3D space is depicted with three axes, usually labeled x, y, and z. These axes are perpendicular to each other, helping to visualize the position and orientation of points and lines in this expansive environment.
The points in 3D space are given by ordered triples (x, y, z), representing positions along the x-axis, y-axis, and z-axis respectively. Unlike 2D, where a point only needs an x and a y coordinate, 3D incorporates depth with the z-coordinate. This allows us to model more realistic spaces and movements occurring in real life.
For students, getting familiar with drawing and imagining 3D space is often the first step in tackling parametric equations and understanding the vast and exciting world of 3D geometry.
The points in 3D space are given by ordered triples (x, y, z), representing positions along the x-axis, y-axis, and z-axis respectively. Unlike 2D, where a point only needs an x and a y coordinate, 3D incorporates depth with the z-coordinate. This allows us to model more realistic spaces and movements occurring in real life.
For students, getting familiar with drawing and imagining 3D space is often the first step in tackling parametric equations and understanding the vast and exciting world of 3D geometry.
Line Equations
Line equations in 3D space can appear intimidating, but breaking them down into parametric equations often simplifies the problem. In the context of 3D space, a line can be represented using a point it passes through and a direction vector that defines its orientation.
The parametric form of a line equation is expressed as:
The parametric form of a line equation is expressed as:
- \(x = x_0 + at\)
- \(y = y_0 + bt\)
- \(z = z_0 + ct\)
- \(x = 5 - 3t\)
- \(y = 4 - 4t\)
- \(z = -1 + 4t\)
Other exercises in this chapter
Problem 58
$$ \begin{array}{l} \text { In Problems , find the eigenvalues } \lambda_{1} \text { and } \lambda_{2} \text { for each matrix }\\\ A \end{array} $$ $$ A=\left[
View solution Problem 59
Use the determinant to find the inverse of \(A\) $$ A=\left[\begin{array}{rr} 2 & 1 \\ 3 & -1 \end{array}\right] $$
View solution 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