Problem 1
Question
In Problems, find a vector equation for the line through the given points. $$ (1,2,1),(3,5,-2) $$
Step-by-Step Solution
Verified Answer
The vector equation is \( \mathbf{r} = (1, 2, 1) + t(2, 3, -3) \).
1Step 1: Identify Points
The problem provides two points through which the line passes: \( P_1 = (1, 2, 1) \) and \( P_2 = (3, 5, -2) \).
2Step 2: Find the Direction Vector
Calculate the direction vector \( \mathbf{d} \) by subtracting the coordinates of \( P_1 \) from \( P_2 \): \( \mathbf{d} = (3-1, 5-2, -2-1) = (2, 3, -3) \).
3Step 3: Create the Vector Equation
The vector equation of a line can be written as \( \mathbf{r} = \mathbf{r_0} + t\mathbf{d} \), where \( \mathbf{r_0} \) is a position vector to a point on the line (we can use \( P_1 = (1, 2, 1) \)) and \( \mathbf{d} \) is the direction vector. Thus, the vector equation becomes \( \mathbf{r} = (1, 2, 1) + t(2, 3, -3) \).
Key Concepts
Direction VectorPosition VectorThree-Dimensional Coordinate Geometry
Direction Vector
When dealing with lines in three-dimensional space, the direction vector is a central concept. It essentially tells us the direction in which the line extends. To find this vector, we use two points on the line. Consider the points given in the exercise: \( P_1 = (1, 2, 1) \) and \( P_2 = (3, 5, -2) \). The direction vector \( \mathbf{d} \) is obtained by subtracting the coordinates of \( P_1 \) from \( P_2 \):
The beauty of the direction vector is its simplicity. Regardless of where you start on the line, multiplying this vector by any scalar \( t \) will point you somewhere else along the line. The line literally stretches in the direction dictated by \( \mathbf{d} \). It helps create the vector equation of the line in three-dimensional geometry.
- This gives us \( \mathbf{d} = (3-1, 5-2, -2-1) = (2, 3, -3) \).
The beauty of the direction vector is its simplicity. Regardless of where you start on the line, multiplying this vector by any scalar \( t \) will point you somewhere else along the line. The line literally stretches in the direction dictated by \( \mathbf{d} \). It helps create the vector equation of the line in three-dimensional geometry.
Position Vector
The position vector is our anchor point in describing a line in space. It represents a specific point on the line, usually one of the provided points. For making things simple and straight, we often choose the first point given in an exercise.
In the case of the given exercise, we have: \( P_1 = (1, 2, 1) \). The resulting position vector, \( \mathbf{r_0} = (1, 2, 1) \), gives us a starting point or a fixed location on the line.
In the case of the given exercise, we have: \( P_1 = (1, 2, 1) \). The resulting position vector, \( \mathbf{r_0} = (1, 2, 1) \), gives us a starting point or a fixed location on the line.
- It ensures that our line equation is grounded on an actual point in space rather than floating unpredictably.
- Crucially, the position vector defines a direction from the origin \((0, 0, 0)\) to this point \((1, 2, 1)\).
Three-Dimensional Coordinate Geometry
Three-dimensional coordinate geometry allows us to explore spaces beyond the flat confines of two-dimensional planes. When plotting lines in this space, we describe them using vector equations. This is a standard approach, as vectors easily translate movements and directions across three axes.
In three-dimensional geometry, each point is specified using three coordinates \((x, y, z)\). For example, the points \((1, 2, 1)\) and \((3, 5, -2)\) from the exercise represent specific locations in space.
In three-dimensional geometry, each point is specified using three coordinates \((x, y, z)\). For example, the points \((1, 2, 1)\) and \((3, 5, -2)\) from the exercise represent specific locations in space.
- The vector equation \( \mathbf{r} = \mathbf{r_0} + t\mathbf{d} \) conveniently represents a line.
- Here, \( \mathbf{r_0} \) is our position vector, and \( \mathbf{d} \) is our direction vector.
- By using a parameter \( t \), the line can extend infinitely in both directions.
Other exercises in this chapter
Problem 1
In Problems 1-10, determine whether the given set is a vector space. If not, give at least one axiom that is not satisfied. Unless stated to the contrary, assum
View solution Problem 1
Verify that the basis \(B\) for the given vector space is orthonormal. Use Theorem \(7.7 .1\) to find the coordinates of the vector \(\mathbf{u}\) relative to t
View solution Problem 1
In Problems \(1-12, \mathbf{a}=\langle 2,-3,4\rangle, \mathbf{b}=\langle-1,2,5\rangle\), and \(\mathbf{c}=\langle 3,6,-1\rangle .\) Find the indicated scalar or
View solution