/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Sketch the line segment represen... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch the line segment represented by each vector equation. $$ \begin{array}{l}{\text { (a) } \mathbf{r}=(1-t)(\mathbf{i}+\mathbf{j})+t \mathbf{k} ; 0 \leq t \leq 1} \\ {\text { (b) } \mathbf{r}=(1-t)(\mathbf{i}+\mathbf{j}+\mathbf{k})+t(\mathbf{i}+\mathbf{j}) ; 0 \leq t \leq 1}\end{array} $$

Short Answer

Expert verified
For (a), sketch from (1,1,0) to (0,0,1); for (b), sketch from (1,1,1) to (1,1,0).

Step by step solution

01

Identify the Start and End Points

For each vector equation, identify the vectors that correspond to \( t = 0 \) and \( t = 1 \), as these will represent the endpoints of the line segment in 3D space.
02

Compute the Endpoints for Equation (a)

For \( ext{(a)} \ \mathbf{r}=(1-t)(f{i}+f{j})+t f{k} \):- When \( t = 0 \), \( \mathbf{r} = 1 \cdot (f{i}+f{j}) + 0 \cdot f{k} = f{i}+f{j} \).- When \( t = 1 \), \( \mathbf{r} = 0 \cdot (f{i}+f{j}) + 1 \cdot f{k} = f{k} \). Thus, the line segment is between the points \( (1, 1, 0) \) and \( (0, 0, 1) \).
03

Compute the Endpoints for Equation (b)

For \( ext{(b)} \ \mathbf{r}=(1-t)(f{i}+f{j}+f{k})+t (f{i}+f{j}) \):- When \( t = 0 \), \( \mathbf{r} = 1 \cdot (f{i}+f{j}+f{k}) + 0 \cdot (f{i}+f{j}) = f{i}+f{j}+f{k} \).- When \( t = 1 \), \( \mathbf{r} = 0 \cdot (f{i}+f{j}+f{k}) + 1 \cdot (f{i}+f{j}) = f{i}+f{j} \).Thus, the line segment is between the points \( (1, 1, 1) \) and \( (1, 1, 0) \).
04

Sketch the Line Segments

Draw the Cartesian coordinate system and plot each endpoint found in the previous steps:- For equation (a), draw the line segment connecting \( (1, 1, 0) \) to \( (0, 0, 1) \).- For equation (b), draw the line segment connecting \( (1, 1, 1) \) to \( (1, 1, 0) \).Label each line segment clearly on the sketch.
05

Review and Verify

Double-check that the endpoints and vector equations match the segments sketched. Verify the linear nature by confirming the equations do not create curves, ensuring just straight line segments are plotted.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Line Segments
A line segment is a part of a line that is bounded by two distinct endpoints. Unlike a line, which extends infinitely in both directions, a line segment has a definite beginning and end. To represent a line segment in a mathematical form, especially in vector notation, we use parameters that determine its start and end points.

Understanding line segments is crucial in geometry because it helps you define exact portions of lines that you want to evaluate. When dealing with vector equations, finding the part of geometry that these vectors represent as line segments is essential. This is because vectors often describe movement or static position between two points in a defined interval, usually from point A to point B.

In practical terms, identifying and sketching line segments between endpoints derived from equations allows us to visualize mathematical concepts easily.
3D Coordinate Systems
3D coordinate systems allow us to represent objects in three-dimensional space by using three axes—typically labeled x, y, and z. Each point in this space can be defined with a set of three coordinates (x, y, z), which represent its position relative to these axes.

This system expands upon the two-dimensional coordinate system by introducing the z-axis, which allows for better representation of objects that have depth. Using 3D coordinate systems is common in physics, engineering, and computer graphics among other fields.

When plotting line segments that you've derived from vector equations, the 3D coordinate system helps translate these abstract concepts into visual representations. Proper understanding of this coordinate system allows for accurate drawing and interpretation of vectors and the geometric shapes they form, like line segments.
Endpoints Identification
Endpoints are the two points that define a line segment in space. For a given vector equation, determining the endpoints involves calculating the position of these points when the parameter equals the minimum and maximum values in its range, usually 0 and 1.

To find the endpoints of a vector equation, substitute these parameter values into the equation. For example, if the parameter is \( t \), calculate the endpoint positions at \( t = 0 \) and \( t = 1 \).

By finding these positions, you identify the precise start and finish of a line segment. Without these values, you wouldn't be able to accurately plot where these lines start and end in any dimension. Correct identification helps accurately plot line segments within a coordinate system, allowing viewers to understand the spatial relationships described by the vector equations.
Vector Notation
Vector notation is a mathematical representation used to describe a vector quantity. Vectors have both magnitude and direction, making them ideal for representing objects like line segments in space. In mathematics, vector notation helps to encapsulate movement or force in a direct way.

Typically, vectors are written in terms of unit vectors along the coordinate axes, such as \( \mathbf{i}, \mathbf{j}, \text{ and } \mathbf{k}, \) each representing a unit vector in the direction of the x, y, and z axes respectively. Vector equations can thus be expressed as combinations of these unit vectors along with scalar parameters.

Using vector notation allows for expressions that clearly convey direction and magnitudes in a compact form. It simplifies calculations and conveys enough information to fully describe physical systems in both theoretical problems and practical applications, like sketching line segments in physics and engineering contexts.

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Most popular questions from this chapter

Find the velocity, speed, and acceleration at the given time t of a particle moving along the given curve. $$ x=1+3 t, y=2-4 t, z=7+t ; t=2 $$

Suppose that a particle vibrates in such a way that its position function is \(\mathbf{r}(t)=16 \sin \pi t \mathbf{i}+4 \cos 2 \pi t \mathbf{j}\), where distance is in millimeters and \(t\) is in seconds. (a) Find the velocity and acceleration at time \(t=1\) s. (b) Show that the particle moves along a parabolic curve. (c) Show that the particle moves back and forth along the curve.

Find the displacement and the distance traveled over the indicated time interval. $$ \mathbf{r}=e^{t} \mathbf{i}+e^{-t} \mathbf{j}+\sqrt{2} t \mathbf{k} ; \quad 0 \leq t \leq \ln 3 $$

As illustrated in the accompanying figure, a train is traveling on a curved track. At a point where the train is traveling at a speed of \(132 \mathrm{ft} / \mathrm{s}\) and the radius of curvature of the track is \(3000 \mathrm{ft}\), the engineer hits the brakes to make the train slow down at a constant rate of \(7.5 \mathrm{ft} / \mathrm{s}^{2}\). (a) Find the magnitude of the acceleration vector at the instant the engineer hits the brakes. (b) Approximate the angle between the acceleration vector and the unit tangent vector \(\mathbf{T}\) at the instant the engineer hits the brakes.

Assume that \(s\) is an arc length parameter for a smooth vector-valued function \(\mathbf{r}(s)\) in 3 -space and that \(d \mathbf{T} / d s\) and \(d \mathbf{N} / d s\) exist at each point on the curve. (This implies that \(d \mathbf{B} / d s\) exists as well, since \(\mathbf{B}=\mathbf{T} \times \mathbf{N} .)\) (a) Show that \(d \mathbf{B} / d s\) is perpendicular to \(\mathbf{B}(s) .\) (b) Show that \(d \mathbf{B} / d s\) is perpendicular to \(\mathbf{T}(s)\). [Hint: Use the fact that \(\mathbf{B}(s)\) is perpendicular to both \(\mathbf{T}(s)\) and \(\mathbf{N}(s),\) and differentiate \(\mathbf{B} \cdot \mathbf{T} \text { with respect to } s .]\) (c) Use the results in parts (a) and (b) to show that \(d \mathbf{B} / d s\) is a scalar multiple of \(\mathbf{N}(s) .\) The negative of this scalar is called the torsion of \(\mathbf{r}(s)\) and is denoted by \(\tau(s) .\) Thus, $$\frac{d \mathbf{B}}{d s}=-\tau(s) \mathbf{N}(s)$$ (d) Show that \(\tau(s)=0\) for all \(s\) if the graph of \(\mathbf{r}(s)\) lies in a plane. [Note: For reasons that we cannot discuss here, the torsion is related to the "twisting" properties of the curve, and \(\tau(s)\) is regarded as a numerical measure of the tendency for the curve to twist out of the osculating plane. \(]\)

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