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Find parametric equations for the line that passes through the point \(P\) and is parallel to the vector \(\mathbf{v}\). $$P(1,1,1), \quad \mathbf{v}=\mathbf{i}-\mathbf{j}+\mathbf{k}$$

Short Answer

Expert verified
The parametric equations are \( x = 1 + t, y = 1 - t, z = 1 + t \).

Step by step solution

01

Understanding the line equation

The line has a point it passes through, namely point \(P(1, 1, 1)\). The line is also parallel to the vector \( \mathbf{v} = \mathbf{i} - \mathbf{j} + \mathbf{k} \). This vector can be rewritten as \( (1, -1, 1) \). We will use this information to derive the parametric equations for the line.
02

Write Parametric Equations

A parametric equation for a line can be written in the form \( (x, y, z) = (x_0, y_0, z_0) + t(a, b, c) \), where \( t \) is a parameter, \( (x_0, y_0, z_0) \) is a point on the line, and \( (a, b, c) \) is a direction vector. For our problem, \( (x_0, y_0, z_0) = (1, 1, 1) \) and \( (a, b, c) = (1, -1, 1) \).
03

Formulate the Parametric Equations

Substitute the point and the direction vector into the parametric equation: \( (x, y, z) = (1, 1, 1) + t(1, -1, 1) \). This gives the equations:1. \( x = 1 + t \)2. \( y = 1 - t \)3. \( z = 1 + t \).These equations describe the line in parametric form.

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Key Concepts

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

Vector Equation of a Line
Vectors are powerful tools in geometry that help us describe lines efficiently. When we talk about the vector equation of a line, we refer to an expression that uses vectors to define every point on a line. The general form is:
  • \( \mathbf{r} = \mathbf{r_0} + t \mathbf{v} \)
Here, \( \mathbf{r} \) represents the position vector of a point on the line, \( \mathbf{r_0} \) is the position vector of a given point through which the line passes, and \( \mathbf{v} \) is the direction vector of the line. The parameter \( t \) can take any real number value, allowing us to trace the entire line as we change \( t \).

In the case of the original exercise, the line passes through point \( P(1, 1, 1) \). Therefore, the vector equation is derived as:
  • \( (x, y, z) = (1, 1, 1) + t(1, -1, 1) \)
This equation states that for every value of \( t \), you can find a corresponding point on the line by substituting \( t \) to get the coordinates \( (x, y, z) \).
Direction Vector
A direction vector is key to understanding how a line extends through space. It tells us the orientation of the line, enabling it to be identified uniquely. For a line to be defined in three-dimensional space, a direction vector provides a direction from a given point and determines how we "move" along the line.
The direction vector is denoted as \( \mathbf{v} = (a, b, c) \) in vector equations. It defines the slope or direction in which the line stretches infinitely. In our exercise, \( \mathbf{v} = (1, -1, 1) \), showing that
  • the line increases by 1 unit in the x-direction,
  • decreases by 1 unit in the y-direction, and
  • increases by 1 unit in the z-direction per unit of \( t \).
Understanding and correctly interpreting this vector is crucial for plotting or visualizing the line accurately in space.
Coordinate Geometry
Coordinate geometry, also known as analytical geometry, provides a link between algebra and geometry through graphs, coordinate points, and equations. This method involves using coordinates to precisely describe geometric figures and their properties.
In three-dimensional coordinate geometry, a point in space is represented as \( (x, y, z) \), and a line can be described using parametric equations derived from a vector equation. For example:
  • The parametric equations derived in the exercise are \( x = 1 + t \), \( y = 1 - t \), and \( z = 1 + t \).
These equations allow us to compute the position of points along the line for any value of \( t \), thus providing a dynamic tool to describe lines.
The foundational concept of coordinate geometry assists in solving various geometric problems, analyzing shapes, and understanding spatial relationships efficiently.

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Most popular questions from this chapter

Find the magnitude and direction (in degrees) of the vector. $$\mathbf{v}=\left\langle-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right\rangle$$

The lengths of two vectors a and \(b\) and the angle \(\theta\) between them are given. Find the length of their cross product, \(|\mathbf{a} \times \mathbf{b}|\). $$|\mathbf{a}|=6, \quad|\mathbf{b}|=\frac{1}{2}, \quad \theta=60^{\circ}$$

Two vectors a and b are given. (a) Find a vector perpendicular to both a and b. (b) Find a unit vector perpendicular to both a and b. $$\mathbf{a}=3 \mathbf{j}+5 \mathbf{k}, \quad \mathbf{b}=-\mathbf{i}+2 \mathbf{k}$$

Find \(|\mathbf{u}|,|\mathbf{v}|,|2 \mathbf{u}|,\left|\frac{1}{2} \mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|,\) and \(|\mathbf{u}|-|\mathbf{v}|\) $$\mathbf{u}=\langle- 6,6\rangle, \quad \mathbf{v}=\langle- 2,-1\rangle$$

A tetrahedron is a solid with four triangular faces, four vertices, and six edges, as shown in the figure. In a regular tetrahedron, the edges are all of the same length. Consider the tetrahedron with vertices \(A(1,0,0), B(0,1,0), C(0,0,1),\) and \(D(1,1,1)\) (a) Show that the tetrahedron is regular. (b) The center of the tetrahedron is the point \(E\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\) (the "average" of the vertices). Find the angle between the vectors that join the center to any two of the vertices (for instance, \(\langle A E B\) ). This angle is called the central angle of the tetrahedron. NOTE: In a molecule of methane \(\left(\mathrm{CH}_{4}\right)\) the four hydrogen atoms form the vertices of a regular tetrahedron with the carbon atom at the center. In this case chemists refer to the central angle as the bond angle. In the figure, the tetrahedron in the exercise is shown, with the vertices labeled \(H\) for hydrogen, and the center labeled \(C\) for carbon. (figure cannot copy)

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