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Chapter 2: Problem 13

Find a parametrization of the line in \(\mathbb{R}^{3}\) that passes through the point \(\mathbf{a}=(1,2,3)\) and is parallel to \(\mathbf{v}=(4,5,6)\)

Short Answer

Expert verified
\( \mathbf{r}(t) = (1 + 4t, 2 + 5t, 3 + 6t) \)

Step by step solution

01

Understand the line parametrization in 3D

A line in three-dimensional space can be parametrized as \( \mathbf{r}(t) = \mathbf{a} + t\mathbf{v} \), where \( \mathbf{a} \) is a point on the line and \( \mathbf{v} \) is the direction vector. The parameter \( t \) is a scalar that varies over all real numbers.
02

Identify the given elements

We are given the point \( \mathbf{a} = (1, 2, 3) \) and the direction vector \( \mathbf{v} = (4, 5, 6) \). These will be used to construct the parametrization expression.
03

Construct the parametrization

Substitute the given point and direction vector into the parametrization formula. The resulting expression is \( \mathbf{r}(t) = (1, 2, 3) + t(4, 5, 6) \).
04

Simplify the expression

Expand the expression to write the parametrization in component form: \( \mathbf{r}(t) = (1 + 4t, 2 + 5t, 3 + 6t) \). This is the parametrization of the line in \( \mathbb{R}^3 \).

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Key Concepts

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

Direction Vector
In the context of three-dimensional geometry, a direction vector is essential in defining a line. Think of it as the compass needle that guides the path along which the line extends.
This vector, typically denoted as \( \mathbf{v} \), indicates the direction of the line, and it plays a crucial role when parametrizing a line in space.
  • It tells you how the line stretches in space along each axis.
  • The vector components show how much the line moves in the x, y, and z directions with each step of change in the parameter \( t \).
For example, in our exercise, the direction vector \( \mathbf{v} = (4, 5, 6) \) suggests that as \( t \) changes, the line moves 4 units in the x direction, 5 units in the y direction, and 6 units in the z direction.
Without the direction vector, you wouldn't know which direction your line is heading!
Point in Space
A point in space, represented as \( \mathbf{a} = (x_1, y_1, z_1) \), serves a foundational role when defining a line in three-dimensional space.
This point is a specific location that your line will pass through. It works in tandem with the direction vector to anchor the line to a specific starting point in space.
  • You can think of the point \( \mathbf{a} \) as the fixed point where the line touches "reality," giving it a concrete position.
  • In the equation for line parametrization, this point gives the initial position of the line before any scaling or direction is considered via the parameter \( t \).
In the discussed example, the point \( \mathbf{a} = (1, 2, 3) \) is where the line definitively passes through as you apply the line's parameterization formula.
This point ensures that every resulting coordinate on the line is traced back to starting here.
Line in Three-Dimensional Space
Defining a line in three-dimensional space involves bringing together the concepts of a point and a direction vector through the use of parametrization.
This creates a mathematical representation that describes an infinite set of points lying on the line as determined by changing a single variable, usually \( t \).
  • The standard formula to define such a line is \( \mathbf{r}(t) = \mathbf{a} + t\mathbf{v} \).
  • Here, \( \mathbf{a} \) represents the starting point, while \( \mathbf{v} \) indicates the direction.
  • By varying \( t \), you can obtain every possible point on the line in space.
For our given problem, the parametrization \( \mathbf{r}(t) = (1, 2, 3) + t(4, 5, 6) \) means that you can substitute any real number for \( t \) to get a point on the line.
As \( t \) shifts, the position changes along the vector's direction, effectively plotting an infinite number of points on the line.

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Most popular questions from this chapter

\(\alpha\) is a path in \(\mathbb{R}^{3}\) with velocity \(\mathbf{v}(t),\) speed \(v(t),\) acceleration \(\mathbf{a}(t),\) and Frenet vectors \(\mathbf{T}(t), \mathbf{N}(t),\) and \(\mathbf{B}(t)\). You may assume that \(v(t) \neq 0\) and \(\mathbf{T}^{\prime}(t) \neq 0\) for all \(t\) so that the Frenet vectors are defined. Prove the following statements. (a) \(\mathbf{v}=v \mathbf{T}\) (b) \(\mathrm{a}=v^{\prime} \mathbf{T}+\kappa v^{2} \mathbf{N}\) (Hence \(v^{\prime}\) is the tangential component of acceleration and \(\kappa v^{2}\) the normal component.)

Find a parametrization of the line in \(\mathbb{R}^{3}\) that passes through the points \(\mathbf{a}=(1,0,0)\) and \(\mathbf{b}=(0,0,1)\).

Sketch the curve parametrized by the given path. Indicate on the curve the direction in which it is being traversed. $$ \alpha: \mathbb{R} \rightarrow \mathbb{R}^{3}, \alpha(t)=(\cos t, \sin t,-t) $$

Sketch the curve parametrized by the given path. Indicate on the curve the direction in which it is being traversed. $$ \alpha: \mathbb{R} \rightarrow \mathbb{R}^{2}, \alpha(t)=\left(e^{t}, e^{-t}\right) $$

Prove the product rule for cross products: If \(\alpha\) and \(\beta\) are differentiable paths in \(\mathbb{R}^{3},\) then: $$ (\alpha \times \beta)^{\prime}=\alpha^{\prime} \times \beta+\alpha \times \beta^{\prime} $$
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