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Chapter 3: Problem 8

Use the given equation of a line to find a point on the line and a vector parallel to the line. \(\mathbf{x}=(1-t)(0,-5,1)\)

Short Answer

Expert verified
A point on the line is (0, -5, 1), and a vector parallel to the line is (0, -5, 1).

Step by step solution

01

Identify Parametric Equation Components

The given equation of the line is in the form \( \mathbf{x} = (1-t)(0,-5,1) \). This is a parametric equation where \( t \) is the parameter. Here, the components \((0, -5, 1)\) represent the direction vector of the line.
02

Determine a Vector Parallel to the Line

The vector parallel to the line can be directly taken from the equation. The direction vector given by \((0, -5, 1)\) is a vector that is parallel to the line. This vector provides the direction in which the line extends.
03

Calculate a Specific Point on the Line

To find a specific point on the line, we can choose a parameter value. Let's set \( t = 0 \). Substituting \( t = 0 \) into the equation, we get: \( \mathbf{x} = (1-0)(0,-5,1) = (0, -5, 1) \). Hence, the point \((0, -5, 1)\) lies on the line.

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Key Concepts

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

Vector Mathematics
Vector mathematics is a branch of mathematics that deals primarily with vectors, which are geometric entities characterized by a magnitude and a direction. In the context of three-dimensional space, vectors are often represented in component form, such as \(\mathbf{v} = (a,b,c)\). This representation helps to highlight each directional influence along the x, y, and z axes.

  • Magnitude of a Vector: The magnitude or length of a vector \(\mathbf{v}\) is denoted as \(|\mathbf{v}|\) and can be calculated using the formula \(|\mathbf{v}| = \sqrt{a^2 + b^2 + c^2}\).
  • Addition and Subtraction: Vectors can be added or subtracted component-wise. For example, given two vectors \(\mathbf{u} = (a_1, b_1, c_1)\) and \(\mathbf{v} = (a_2, b_2, c_2)\), their sum is \(\mathbf{u} + \mathbf{v} = (a_1 + a_2, b_1 + b_2, c_1 + c_2)\).
  • Scalar Multiplication: Multiplying a vector by a scalar changes its magnitude but not its direction, provided the scalar is positive. If \( ext{scalar} = k\), then \(k \mathbf{v} = (ka, kb, kc)\).
Understanding vectors is crucial in mathematics because they provide a way to model physical quantities, such as force and displacement, in physics. They also feature heavily in calculus and linear algebra, and are essential for describing lines, planes, and motions in space.
Direction Vector
A direction vector is a crucial concept in vector mathematics, serving as a pointer indicating the direction in which a line extends. The direction vector is key to identifying or specifying the orientation of a line in space.

  • Representation in Parametric Equations: In parametric equations of the form \(\mathbf{x} = \mathbf{p} + t\mathbf{d}\), \(\mathbf{d}\) is often the direction vector. This vector does not change with the parameter \(t\), and its role is to establish the line鈥檚 orientation.
  • Determination of Direction: The direction vector is derived from the coefficients of the line in its vector equation. In our exercise, \(\mathbf{x}=(1-t)(0,-5,1)\), the direction vector is clearly \((0, -5, 1)\), showing no movement along the x-axis, downward movement in the y-axis, and upwards movement in the z-axis.
The significance of the direction vector cannot be understated as it sets the trajectory of the line. This, in turn, can affect further geometrical constructions or analyses, such as finding points on the line, or computing angles between lines.
Point on a Line
A line in three-dimensional space can be represented using parametric equations, and it can pass through numerous points. To find a specific point on a line, we use the parametric equation of the line, which includes a parameter that can be varied to yield different points along the line.

  • Using the Parameter: In this context, setting a specific value for the parameter \(t\) allows us to identify a distinct point on the line. For example, the equation \(\mathbf{x}=(1-t)(0,-5,1)\) will let us find specific points for different \(t\) values.
  • Example: Setting \(t = 0\) in our exercise yields the point \((0, -5, 1)\), showing us a tangible position on the line. As \(t\) varies, so does the location of the calculated point, illustrating the infinite number of possible positions a line can pass through.
Grasping how to determine points on a line using parametric equations is a fundamental skill in geometry and vector calculus, allowing for diverse applications, from simple plotting to complex computational modeling.

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Most popular questions from this chapter

Find the area of the triangle in 3 -space that has the given vertices. $$P(1,-1,2), Q(0,3,4), R(6,1,8)$$

Suppose that \(\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})=3 .\) Find (a) \(\mathbf{v} \cdot(\mathbf{u} \times \mathbf{w})\) (b) \((\mathbf{u} \times \mathbf{w}) \cdot \mathbf{v}\) (c) \(\mathbf{v} \cdot(\mathbf{w} \times \mathbf{w})\)

Find a general solution of the system, and use that solution to find a general solution of the associated homogeneous system and a particular solution of the given system. \(\left[\begin{array}{cccc}9 & -3 & 5 & 6 \\ 6 & -2 & 3 & 1 \\ 3 & -1 & 3 & 14\end{array}\right]\left[\begin{array}{c}x_{1} \\ x_{2} \\ x_{3} \\\ x_{4}\end{array}\right]=\left[\begin{array}{r}4 \\ 5 \\\ -8\end{array}\right]\)

Show that there do not exist scalars \(c_{1}, c_{2},\) and \(c_{3}\) such that $$ c_{1}(-2,9,6)+c_{2}(-3,2,1)+c_{3}(1,7,5)=(0,5,4) $$

Find the distance between the given parallel planes. $$2 x-y-z=5 \text { and }-4 x+2 y+2 z=12$$
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