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Chapter 4: Problem 7

Find the vector equation and parametric equations for the line through the point (-7,10,-6) with a direction vector \(\vec{d}=\left[\begin{array}{l}1 \\\ 1 \\ 3\end{array}\right]\).

Short Answer

Expert verified
Vector equation: \(\textbf{r} = [-7, 10, -6] + t[1, 1, 3]\) Parametric equations: \[ x = -7 + t \, y = 10 + t \, z = -6 + 3t \]

Step by step solution

01

- Determine the vector equation

To find the vector equation of a line, use the formula \(\textbf{r} = \textbf{r}_0 + t\textbf{d} \) where \(\textbf{r}_0 \) is a position vector to a point on the line and \(\textbf{d} \) is the direction vector. Here, \(\textbf{r}_0 = [-7, 10, -6] \) and \(\textbf{d} = [1, 1, 3] \). Thus, the vector equation is \(\textbf{r} = [-7, 10, -6] + t[1, 1, 3] \).
02

- Separate into components

The vector equation \(\textbf{r} = [-7, 10, -6] + t[1, 1, 3] \) can be separated into its component form as: \[ x = -7 + t \] \[ y = 10 + t \] \[ z = -6 + 3t \]
03

- Write the parametric equations

Based on the separated components, the parametric equations for the line are: \[ x = -7 + t \] \[ y = 10 + t \] \[ z = -6 + 3t \]

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Key Concepts

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

Direction Vector
Understanding the direction vector is crucial when working with vector equations of a line. In vector notation, a direction vector gives the line its orientation in space. It tells us where the line is pointing.
For the given problem, the direction vector \( \vec{d} = \left[1, 1, 3\right] \) indicates that:
  • The x-coordinate increases by 1 unit for every step along the line.
  • The y-coordinate also increases by 1 unit simultaneously.
  • The z-coordinate increases by 3 units, making the line rise more steeply in the z-direction.
To see this in equation form, we use the direction vector in combination with a point through which the line passes, which leads us to the vector and parametric equations.
Parametric Equations
The parametric equations are derived from separating the components of the vector equation. These equations express the coordinates of any point on the line as functions of a parameter, often denoted as \( t \).
For the given problem:
  • The x-coordinate is given by \( x = -7 + t \).
  • The y-coordinate is given by \( y = 10 + t \).
  • The z-coordinate is given by \( z = -6 + 3t \).
The parameter \( t \) ranges over all real numbers \( t \in \mathbb{R} \), representing each and every point on the line as \( t \) changes. This way, the parametric equations clearly show how each coordinate depends on the same parameter.
Linear Algebra
Linear algebra plays a key role in understanding vector equations and parametric equations. It provides the framework for solving problems related to lines, planes, and other geometric objects in multidimensional space.
By transforming points and vectors in forms that are easy to manipulate, linear algebra helps in:
  • Determining the position and direction of lines.
  • Solving systems of linear equations.
  • Working with coordinate transformations.
In this exercise, using linear algebra principles, we combined a point \( -7, 10, -6 \) and a direction vector \( \left[1, 1, 3\right] \) to derive the equations of the line. These equations make understanding the spatial relationships straightforward and accessible.

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Most popular questions from this chapter

Suppose \(V\) and \(W\) both have dimension equal to 7 and they are subspaces of \(\mathbb{R}^{10} .\) What are the possibilities for the dimension of \(V \cap W ?\) Hint: Remember that a linear independent set can be extended to form a basis.

If you have 5 vectors in \(\mathbb{R}^{5}\) and the vectors are linearly independent, can it always be concluded they span \(\mathbb{R}^{5} ?\) Explain.

Let \(H\) denote span \(\left\\{\left[\begin{array}{l}5 \\ 1 \\ 1 \\\ 4\end{array}\right],\left[\begin{array}{r}14 \\ 3 \\ 2 \\\ 8\end{array}\right],\left[\begin{array}{r}38 \\ 8 \\ 6 \\\ 24\end{array}\right],\left[\begin{array}{r}47 \\ 10 \\ 7 \\\ 28\end{array}\right],\left[\begin{array}{r}10 \\ 2 \\ 3 \\\ 12\end{array}\right]\right\\} .\) Find the dimension of \(H\) and determine a basis.

Let \(\vec{v}\) be a vector and let \(\vec{n}\) be a normal vector for a plane through the origin. Find the equation of the line through the point determined by \(\vec{v}\) which has direction vector \(\vec{n}\). Show that it intersects the plane at the point determined by \(\vec{v}-\operatorname{proj}_{\vec{n}} \vec{v}\). Hint: The line: \(\vec{v}+t \vec{n}\). It is in the plane if \(\vec{n} \bullet(\vec{v}+t \vec{n})=0\). Determine \(t .\) Then substitute in to the equation of the line.

An object moves 10 meters in the direction of \(\vec{j}+\vec{i}\). There are two forces acting on this object, \(\vec{F}_{1}=\vec{i}+2 \vec{j}+2 \vec{k},\) and \(\vec{F}_{2}=5 \vec{i}+2 \vec{j}-6 \vec{k} .\) Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?
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