Problem 1
Question
Find the parametric equations of the line through the given pair of points. \((1,-2,3),(4,5,6)\)
Step-by-Step Solution
Verified Answer
Parametric equations: \(x = 1 + 3t\), \(y = -2 + 7t\), \(z = 3 + 3t\).
1Step 1: Identify Direction Vector
To find the parametric equations of a line, we first need the direction vector. Given points are \((1, -2, 3)\) and \((4, 5, 6)\). The direction vector \(\mathbf{v}\) is calculated by subtracting the coordinates of the first point from the second point: \[ \mathbf{v} = (4 - 1, 5 - (-2), 6 - 3) = (3, 7, 3) \]
2Step 2: Write Parametric Equations
Next, use the point \((1, -2, 3)\) as a point on the line, and the vector \((3, 7, 3)\) as the direction vector. The parametric equations are expressed as: \[ x = 1 + 3t \] \[ y = -2 + 7t \] \[ z = 3 + 3t \] Here, \(t\) is the parameter and can be any real number.
Key Concepts
Direction VectorPoints in 3D SpaceLine through PointsVector Subtraction
Direction Vector
A direction vector is essential when defining a line in three-dimensional space using parametric equations. It tells us the direction in which the line extends. To find the direction vector, we subtract the coordinates of one given point from the other. For example, given the points
This vector signifies that for every movement from one point to another along the line, the x-coordinate increases by 3, the y-coordinate by 7, and the z-coordinate by 3.
Such calculations are crucial for understanding the geometry of the line in 3D coordinates.
- \((1, -2, 3)\)
- \((4, 5, 6)\)\
This vector signifies that for every movement from one point to another along the line, the x-coordinate increases by 3, the y-coordinate by 7, and the z-coordinate by 3.
Such calculations are crucial for understanding the geometry of the line in 3D coordinates.
Points in 3D Space
Points in 3D space are identified by coordinates, represented as \((x, y, z)\), where \(x\), \(y\), and \(z\) denote positions along the respective axis.
- Coordinate \(x\) determines the horizontal position.
- Coordinate \(y\) indicates the vertical position.
- Coordinate \(z\) dictates the depth at that position.
- \((1, -2, 3)\)
- \((4, 5, 6)\)\
Line through Points
A line in 3D space can be described in parametric form through given points. Such a description requires an initial point known as the position point and a direction vector.
A parametric equation represents each component (x, y, z) in terms of a parameter, usually denoted by \(t\). This equation is helpful when you need to express a line passing through a specific set of points. Using the points
A parametric equation represents each component (x, y, z) in terms of a parameter, usually denoted by \(t\). This equation is helpful when you need to express a line passing through a specific set of points. Using the points
- \((1, -2, 3)\)
- \((4, 5, 6)\)\
- \((3, 7, 3)\)
- x-coordinate: \(x = 1 + 3t\)
- y-coordinate: \(y = -2 + 7t\)
- z-coordinate: \(z = 3 + 3t\)
Vector Subtraction
Vector subtraction is a mathematical operation used to determine the vector between two points in space.
This operation is integral when finding the direction vector of a line. When subtracting two vectors, each of their corresponding components are subtracted.
For the points
This operation is integral when finding the direction vector of a line. When subtracting two vectors, each of their corresponding components are subtracted.
For the points
- \((1, -2, 3)\)
- \((4, 5, 6)\)\
- x-component: \(4 - 1 = 3\)
- y-component: \(5 - (-2) = 7\)
- z-component: \(6 - 3 = 3\)
- \((3, 7, 3)\)
Other exercises in this chapter
Problem 1
Name and sketch the graph of each of the following equations in three-space. $$ 4 x^{2}+36 y^{2}=144 $$
View solution Problem 1
In Problems 1-6, sketch the curve over the indicated domain for \(t\). Find \(\mathbf{v}, \mathbf{a}, \mathbf{T}\), and \(\kappa\) at the point where \(t=t_{1}
View solution Problem 1
In Problems \(1-8\), find the required limit or indicate that it does not exist. $$ \lim _{t \rightarrow 1}\left[2 t \mathbf{i}-t^{2} \mathbf{j}\right] $$
View solution Problem 1
Let \(\quad \mathbf{a}=-3 \mathbf{i}+2 \mathbf{j}-2 \mathbf{k}, \quad \mathbf{b}=-\mathbf{i}+2 \mathbf{j}-4 \mathbf{k}, \quad\) and \(\mathbf{c}=7 \mathbf{i}+3
View solution