Chapter 12: Problem 15
(a) Find the slope of the line in 2 -space that is represented by the vector equation \(\mathbf{r}=(1-2 t) \mathbf{i}-(2-3 t) \mathbf{j}\). (b) Find the coordinates of the point where the line $$ \mathbf{r}=(2+t) \mathbf{i}+(1-2 t) \mathbf{j}+3 t \mathbf{k} $$ intersects the \(x z\) -plane.
Short Answer
Expert verified
(a) Slope = \( \frac{3}{2} \). (b) Intersection at \( \left( \frac{5}{2}, 0, \frac{3}{2} \right) \).
Step by step solution
01
Determine the vector direction
The vector equation given is \( \mathbf{r} = (1-2t)\mathbf{i} - (2-3t)\mathbf{j} \). We identify the direction vector \( \mathbf{d} = \langle -2, -3 \rangle \) from the terms involving \( t \).
02
Calculate the slope from the vector direction
For a line in 2-space, the slope \( m \) can be calculated using the components of the direction vector \( \mathbf{d} = \langle -2, -3 \rangle \): \[ m = \frac{\text{change in } y}{\text{change in } x} = \frac{-3}{-2} = \frac{3}{2} \].
03
Set up the intersection condition for the xz-plane
The line's vector equation is \( \mathbf{r} = (2+t)\mathbf{i} + (1-2t)\mathbf{j} + 3t\mathbf{k} \). In the \( xz \)-plane, the \( y \)-coordinate must be zero. Therefore, set the \( y \)-component equal to zero, yielding the equation: \( 1 - 2t = 0 \).
04
Solve for the parameter t
Solve the equation \( 1 - 2t = 0 \) to find the value of \( t \): \[ 1 - 2t = 0 \] \[ 2t = 1 \] \[ t = \frac{1}{2} \].
05
Find the intersection coordinates
Substitute \( t = \frac{1}{2} \) into the vector equation to find the intersection coordinates: - \( x = 2 + \frac{1}{2} = \frac{5}{2} \) - \( y = 0 \) (as expected in the \( xz \)-plane) - \( z = 3 \times \frac{1}{2} = \frac{3}{2} \) Therefore, the intersection point is \( \left( \frac{5}{2}, 0, \frac{3}{2} \right) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Slope of a Line
The slope of a line is a measure of its steepness or incline. In 2D space, the slope is typically denoted by the symbol \( m \), and it describes how much the line rises or falls as it moves along the x-axis. In general terms, the slope is given by the change in the y-coordinate (vertical change) divided by the change in the x-coordinate (horizontal change). This can be expressed as:\[ m = \frac{\Delta y}{\Delta x} \]In the context of vector calculus, we often derive the slope from a vector equation describing the line. For example, given the vector equation \( \mathbf{r} = (1-2t)\mathbf{i} - (2-3t)\mathbf{j} \), the direction vector \( \mathbf{d} = \langle -2, -3 \rangle \) indicates the direction of the line.
- Change in y: The coefficient of \( \mathbf{j} \) is \(-3\).
- Change in x: The coefficient of \( \mathbf{i} \) is \(-2\).
- Slope \( m = \frac{-3}{-2} = \frac{3}{2} \).
Vector Equation
A vector equation is a mathematical way to describe a line or curve in space using vectors. It allows us to track the position of points along the line as a function of a parameter, typically denoted by \( t \). The general form of a vector equation for a line can be written as \( \mathbf{r} = \mathbf{a} + t \mathbf{d} \), where:
- \( \mathbf{a} \) is a position vector to a specific known point on the line.
- \( \mathbf{d} \) is the direction vector for the line, giving its orientation.
- \( t \) is a scalar parameter that signifies how far and in what direction to move along the line from \( \mathbf{a} \).
- \( \mathbf{a} = \langle 2, 1, 0 \rangle \), is an initial point on the line.
- \( \mathbf{d} = \langle 1, -2, 3 \rangle \) represents the direction in which the line extends.
Intersection of Planes
The intersection of planes refers to finding a point or line where two planes meet. In a 3D space context, determining where a line intersects another plane is key to solving many geometric and algebraic problems. Intersection problems typically involve finding the value of the parameter \( t \) that satisfies a certain condition.For instance, consider the line defined by the equation \( \mathbf{r} = (2+t) \mathbf{i} + (1-2t) \mathbf{j} + 3t \mathbf{k} \) intersecting the \( xz \)-plane.
- The \( xz \)-plane is characterized by its \( y \)-coordinate being zero (i.e., \( y = 0 \)).
- The condition for intersection is setting the \( y \)-component of the vector equation to zero: \( 1 - 2t = 0 \).
- \( x = 2 + \frac{1}{2} = \frac{5}{2} \)
- \( y = 0 \)
- \( z = 3 \cdot \frac{1}{2} = \frac{3}{2} \)