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For the following exercises, find the vector and parametric equations of the line with the given properties.The line that passes through point \((2,-3,7)\) that is parallel to vector \(\langle 1,3,-2\rangle\)

Short Answer

Expert verified
Vector equation: \( \mathbf{r}(t) = \langle 2, -3, 7 \rangle + t\langle 1, 3, -2 \rangle \); Parametric equations: \( x = 2 + t \), \( y = -3 + 3t \), \( z = 7 - 2t \).

Step by step solution

01

Understand the Vector Equation of a Line

A vector equation of a line in space can be represented as \( \mathbf{r}(t) = \mathbf{r_0} + t\mathbf{v} \), where \( \mathbf{r_0} \) is a point on the line, \( \mathbf{v} \) is a direction vector, and \( t \) is a parameter.
02

Identify Given Values

From the problem, the point through which the line passes is \((2, -3, 7)\), which is \( \mathbf{r_0} \). The direction vector parallel to the line is \( \langle 1, 3, -2 \rangle \), which is \( \mathbf{v} \).
03

Write the Vector Equation

Using the values from Step 2, substitute into the general form of the vector equation: \( \mathbf{r}(t) = \langle 2, -3, 7 \rangle + t\langle 1, 3, -2 \rangle \).
04

Write the Parametric Equations

The vector equation \( \mathbf{r}(t) = \langle 2, -3, 7 \rangle + t\langle 1, 3, -2 \rangle \) can be split into parametric equations for each component: \( x = 2 + t \), \( y = -3 + 3t \), \( z = 7 - 2t \).

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Key Concepts

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

Understanding Parametric Equations
Parametric equations are a powerful tool for representing geometric objects like lines, curves, and surfaces in space. Instead of describing a line explicitly with a single equation, parametric equations express the coordinates of points on the line as functions of a single parameter, usually denoted as \( t \).

For a line in three-dimensional space, we use three separate equations to describe the \( x \), \( y \), and \( z \) coordinates in terms of \( t \).
For instance, given a vector equation of the form \( \mathbf{r}(t) = \langle 2, -3, 7 \rangle + t \langle 1, 3, -2 \rangle \), we can derive the parametric equations:
  • \( x = 2 + t \)
  • \( y = -3 + 3t \)
  • \( z = 7 - 2t \)

These parametric equations allow us to plug in different values of \( t \) to find specific points on the line. By altering \( t \), we can move to different spots along the direction of the line, showcasing the line's trajectory through space.
Exploring the Direction Vector
The direction vector plays a crucial role in defining the line's orientation in space. Imagine it like a guiding arrow that points the way the line extends. In the context of a vector equation \( \mathbf{r}(t) = \mathbf{r_0} + t \mathbf{v} \), this vector \( \mathbf{v} \) tells us the direction and the sense (i.e., forward or backward) of the line.

For the exercise example, the direction vector is \( \langle 1, 3, -2 \rangle \). This indicates that:
  • The line moves one unit in the positive \( x \)-direction for every unit of \( t \).
  • It advances three units in the positive \( y \)-direction.
  • Concurrently, it shifts two units backward in the \( z \)-direction.

This vector is fundamental because it ensures that the line can be accurately positioned and drawn in space, stretching infinitely in the direction of \( \mathbf{v} \) through the point where it begins.
Line Through a Point
When constructing a line in vector form, one of the essential aspects is having a starting point. This is often referred to as the position vector or a point on the line. In our vector equation \( \mathbf{r}(t) = \mathbf{r_0} + t \mathbf{v} \), the point \( \mathbf{r_0} \) is the fixed point through which the line passes.

In our example, the line runs through the point \( (2, -3, 7) \). This means if \( t = 0 \), then \( \mathbf{r}(0) = \langle 2, -3, 7 \rangle \). This is our starting location on the line.

Starting from this fixed point, changing \( t \) means we are exploring different locations on the line, all dictated by the direction vector \( \mathbf{v} \). This point is critical because it anchors the line, giving it a defined position in the three-dimensional space before extending infinitely in both directions. Without it, the direction vector alone cannot place the line uniquely on the coordinate grid.

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