Problem 61
Question
Find the parametric equation of the line in \(x-y-z\) space that goes through the given points. \((2,-3,1)\) and \((-5,2,1)\)
Step-by-Step Solution
Verified Answer
The parametric equation is \(x = 2 - 7t\), \(y = -3 + 5t\), \(z = 1\).
1Step 1: Determine the Direction Vector
To find the parametric equation of the line, start by finding the direction vector. Subtract the coordinates of the first point from the second point:\[\vec{v} = (-5 - 2, 2 - (-3), 1 - 1) = (-7, 5, 0)\]
2Step 2: Write the Parametric Equation
A parametric equation of a line can be written as:\[(x, y, z) = (x_0, y_0, z_0) + t\cdot(a, b, c)\]where \((x_0, y_0, z_0)\) is a point on the line (let's use \((2, -3, 1)\)), and \((a, b, c)\) is the direction vector \((-7, 5, 0)\).
3Step 3: Substitute the Values
Substitute the known values into the parametric equation:\[x = 2 - 7t\]\[y = -3 + 5t\]\[z = 1\]where \(t\) is the parameter. This gives the parametric equations of the line through the points.
Key Concepts
Direction VectorLine in 3D SpacePoints in 3D Space
Direction Vector
In 3D space, a direction vector helps indicate the direction a line is heading. Imagine it like an arrow pointing towards where the line extends. To find this vector, we subtract the coordinates of two points on the line. In this specific example, the points given are \((2, -3, 1)\) and \((-5, 2, 1)\). The formula to calculate the direction vector is:
Plug in the values:\[\vec{v} = (-5 - 2, 2 - (-3), 1 - 1) = (-7, 5, 0)\]
This resulting vector, \((-7, 5, 0)\), tells us that the line is moving 7 units backward in the x-direction, moving 5 units forward in the y-direction, and stays constant in the z-direction.
- Subtract each coordinate of the first point from the corresponding coordinate of the second point.
Plug in the values:\[\vec{v} = (-5 - 2, 2 - (-3), 1 - 1) = (-7, 5, 0)\]
This resulting vector, \((-7, 5, 0)\), tells us that the line is moving 7 units backward in the x-direction, moving 5 units forward in the y-direction, and stays constant in the z-direction.
Line in 3D Space
Understanding a line in 3D space requires a slightly different approach compared to 2D space. A line is defined by a point it passes through and the direction vector.
The general form of a parametric equation for a line in 3D is: \[(x, y, z) = (x_0, y_0, z_0) + t\cdot(a, b, c)\]
Here, \((x_0, y_0, z_0)\) is a known point on the line, and \((a, b, c)\) is the direction vector calculated earlier. In our case, the sample point is \((2, -3, 1)\) and the direction vector is \((-7, 5, 0)\).
These equations mean that for each increase of \(t\), the point moves \(-7t\) in the x-direction, \(+5t\) in the y-direction, and remains constant in the z-direction.
- The parametric equation helps describe this line in terms of a parameter, usually denoted as \(t\).
The general form of a parametric equation for a line in 3D is: \[(x, y, z) = (x_0, y_0, z_0) + t\cdot(a, b, c)\]
Here, \((x_0, y_0, z_0)\) is a known point on the line, and \((a, b, c)\) is the direction vector calculated earlier. In our case, the sample point is \((2, -3, 1)\) and the direction vector is \((-7, 5, 0)\).
- Substitute these values into the parametric equation to get the specific parametric form of the line.
- \(x = 2 - 7t\)
- \(y = -3 + 5t\)
- \(z = 1\)
These equations mean that for each increase of \(t\), the point moves \(-7t\) in the x-direction, \(+5t\) in the y-direction, and remains constant in the z-direction.
Points in 3D Space
Points in 3D space are defined by three coordinates: x, y, and z. They signify a specific location in a 3-dimensional grid. For two points to lie on the same line in 3D, there must be a vector pointing from one to the other, identifying a straight path between them.
In this exercise, we start with two points: \((2, -3, 1)\) and \((-5, 2, 1)\). These points are used to:
Notice that the z-value for both points is the same. This implicates that the line is parallel to the xy-plane at z=1, adding simplicity to understanding the line's behavior in space. Using the known points helps to accurately plot and define lines within the 3D coordinate system, crucial for many practical applications, like engineering or physics modeling.
In this exercise, we start with two points: \((2, -3, 1)\) and \((-5, 2, 1)\). These points are used to:
- Calculate the direction vector, indicating the line's direction.
- Formulate the parametric equation based on this vector and one of the points.
Notice that the z-value for both points is the same. This implicates that the line is parallel to the xy-plane at z=1, adding simplicity to understanding the line's behavior in space. Using the known points helps to accurately plot and define lines within the 3D coordinate system, crucial for many practical applications, like engineering or physics modeling.
Other exercises in this chapter
Problem 60
Find the eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\) for each matrix \(A\). $$A=\left[\begin{array}{rr}-1 & 4 \\ 0 & -2\end{array}\right]$$
View solution Problem 60
Write down the inverse of \(A\). $$ A=\left[\begin{array}{ll} 1 & 2 \\ 1 & 3 \end{array}\right] $$
View solution Problem 61
Find the eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\) for $$ A=\left[\begin{array}{ll} a & 0 \\ c & b \end{array}\right] $$
View solution Problem 61
Write down the inverse of \(A\). $$ A=\left[\begin{array}{rr} -1 & 4 \\ 5 & 0 \end{array}\right] $$
View solution