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\(2-5\) Find a vector equation and parametric equations for the line. The line through the point \((6,-5,2)\) and parallel to the vector \(\left\langle 1,3,-\frac{2}{3}\right\rangle\)

Short Answer

Expert verified
The vector equation is \( \mathbf{r}(t) = \langle 6 + t, -5 + 3t, 2 - \frac{2}{3}t \rangle \) and parametric equations are \( x = 6 + t \), \( y = -5 + 3t \), \( z = 2 - \frac{2}{3}t \).

Step by step solution

01

Understand Vector Equation

A vector equation for a line in three-dimensional space through a point \( (x_0, y_0, z_0) \) with a direction vector \( \langle a, b, c \rangle \) is written as \( \mathbf{r}(t) = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle \). The parameter \( t \) is a real number.
02

Identify Given Information

The given point is \( (6, -5, 2) \) and the direction vector is \( \langle 1, 3, -\frac{2}{3} \rangle \). These will be used in the equation for the line.
03

Write the Vector Equation

Using the formula \( \mathbf{r}(t) = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle \), substitute \( (6, -5, 2) \) as the point and \( \langle 1, 3, -\frac{2}{3} \rangle \) as the direction vector: \[ \mathbf{r}(t) = \langle 6, -5, 2 \rangle + t \langle 1, 3, -\frac{2}{3} \rangle. \] Thus, the vector equation becomes \[ \mathbf{r}(t) = \langle 6 + t, -5 + 3t, 2 - \frac{2}{3}t \rangle. \]
04

Derive Parametric Equations

From the vector equation \( \mathbf{r}(t) = \langle 6 + t, -5 + 3t, 2 - \frac{2}{3}t \rangle \), we can write the parametric equations as follows: \( x(t) = 6 + t \), \( y(t) = -5 + 3t \), \( z(t) = 2 - \frac{2}{3}t \).

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Key Concepts

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

Parametric Equations
Parametric equations are a way to express a set of related quantities as explicit functions of an independent variable, often called a parameter. In the context of a line in three-dimensional space, these equations describe the coordinates of any point on the line in terms of a single parameter, typically denoted as \( t \).

For a line with a vector equation \( \mathbf{r}(t) = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle \), the parametric equations take the form:
  • \( x(t) = x_0 + at \)
  • \( y(t) = y_0 + bt \)
  • \( z(t) = z_0 + ct \)
These equations allow us to find specific points on the line by substituting different values for \( t \). Each value of \( t \) corresponds to a unique point on the line. In the exercise, substituting \( t \) in the parametric equations \( x(t) = 6 + t \), \( y(t) = -5 + 3t \), and \( z(t) = 2 - \frac{2}{3}t \) allows us to locate any point on the line passing through the specified point and direction. This representation is particularly useful in physics and engineering, where motion along a path is analyzed.
Line in Three-Dimensional Space
Lines in three-dimensional space can be more complex to visualize and compute than in two-dimensional scenarios. However, using vector notation simplifies this greatly. A line in three-dimensional space can be thought of as an infinite set of points that extends in two directions without ever bending or curving.

In mathematical terms, to define such a line, you need:
  • A point through which the line passes. This gives a starting reference point in space.
  • A direction vector that indicates the path the line takes from that point.
With these, you can use a vector equation to precisely describe the line. The line is then described by the vector \( \mathbf{r}(t) = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle \), where \( t \) varies over all real numbers.

In the example provided, the point \( (6, -5, 2) \) serves as the reference point, and the vector \( \langle 1, 3, -\frac{2}{3} \rangle \) describes the direction in which the line extends. With these components, you can easily find any point on the line by simply varying \( t \).
Direction Vector
The direction vector of a line in three-dimensional space plays a crucial role in determining the orientation of the line. It is essentially a vector that points in the direction the line travels.

In vector equations, this direction vector is denoted as \( \langle a, b, c \rangle \). It tells us how far the line will move along the \( x \), \( y \), and \( z \) axes when the parameter \( t \) changes by one unit. Specifically, for the exercise you are working on, the direction vector is \( \langle 1, 3, -\frac{2}{3} \rangle \).

  • The first component, \( 1 \), indicates a movement of 1 unit in the x-direction as \( t \) increases by 1.
  • The second component, \( 3 \), indicates a movement of 3 units in the y-direction.
  • The third component, \(-\frac{2}{3}\), indicates a movement in the z-direction, specifically descending \( \frac{2}{3} \) of a unit.
These components define a unique path through space, and by scaling the direction vector using the parameter \( t \), you effectively trace a line extending indefinitely in both directions.
Point-Direction Form
The point-direction form is a way to write the equation of a line using a specific point and a direction vector. This form is very useful as it directly shows where the line starts and how it moves.

A line in point-direction form is represented using a vector equation \( \mathbf{r}(t) = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle \). In this format:
  • \( \langle x_0, y_0, z_0 \rangle \) is a position vector pointing to a known point on the line, acting like an anchor.
  • \( \langle a, b, c \rangle \) is the direction vector that determines the line's direction.
By combining these two components, the point-direction form provides a simple yet powerful method to describe a line in space. It allows you to determine any point on the line by substituting various values for the parameter \( t \).

In the specific exercise example, the line is anchored at the point \( (6, -5, 2) \), and as \( t \) varies, the direction vector \( \langle 1, 3, -\frac{2}{3} \rangle \) guides the line's trajectory through three-dimensional space.

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