Chapter 8: Problem 1
A vector equation is given as \(\vec{r}=\left(\frac{1}{2},-\frac{3}{4}\right)+s\left(\frac{1}{3}, \frac{1}{6}\right), s \in \mathbf{R} .\) Explain why \(\vec{m}=(-2,-1), \vec{m}=(2,1),\) and \(\vec{m}=\left(\frac{2}{7}, \frac{1}{7}\right)\) are acceptable direction vectors for this line.
Short Answer
Expert verified
All three vectors are scalar multiples of \( \left( \frac{1}{3}, \frac{1}{6} \right) \), hence they are valid direction vectors for the line.
Step by step solution
01
Understand Line Equation
The given vector equation describes a line as \( \vec{r} = \left( \frac{1}{2}, -\frac{3}{4} \right) + s\left( \frac{1}{3}, \frac{1}{6} \right), s \in \mathbf{R} \). Here, \( \left( \frac{1}{3}, \frac{1}{6} \right) \) is the direction vector of the line.
02
Understand Acceptable Direction Vectors
An acceptable direction vector for a line must be a scalar multiple of the given direction vector. This means any vector \( \vec{m} \) is acceptable if it can be expressed as \( k \left( \frac{1}{3}, \frac{1}{6} \right) \) for some scalar \( k \).
03
Verify \(-2, -1\) as Direction Vector
To check \( \vec{m} = (-2, -1) \), we verify if there exists a scalar \( k \) such that \( k \left( \frac{1}{3}, \frac{1}{6} \right) = (-2, -1) \). Solving \( k \cdot \frac{1}{3} = -2 \) gives \( k = -6 \) and \( k \cdot \frac{1}{6} = -1 \) also gives \( k = -6 \). Thus, \( \vec{m} = (-2, -1) \) is a valid direction vector.
04
Verify \(2, 1\) as Direction Vector
Check \( \vec{m} = (2, 1) \) using a scalar \( k \) such that \( k \left( \frac{1}{3}, \frac{1}{6} \right) = (2, 1) \). \( k \cdot \frac{1}{3} = 2 \) gives \( k = 6 \) and \( k \cdot \frac{1}{6} = 1 \) also gives \( k = 6 \). Thus, \( \vec{m} = (2, 1) \) is a valid direction vector.
05
Verify \(\frac{2}{7}, \frac{1}{7}\) as Direction Vector
For \( \vec{m} = \left( \frac{2}{7}, \frac{1}{7} \right) \), check \( k \left( \frac{1}{3}, \frac{1}{6} \right) = \left( \frac{2}{7}, \frac{1}{7} \right) \). \( k \cdot \frac{1}{3} = \frac{2}{7} \) gives \( k = \frac{6}{7} \) and \( k \cdot \frac{1}{6} = \frac{1}{7} \) also gives \( k = \frac{6}{7} \). Thus, \( \vec{m} = \left( \frac{2}{7}, \frac{1}{7} \right) \) is a valid direction vector.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
direction vector
In vector mathematics, understanding the concept of a direction vector is crucial for defining lines. A direction vector essentially guides the path on which the line extends. In a vector equation representing a line, like \( \vec{r} = \left( \frac{1}{2}, -\frac{3}{4} \right) + s\left( \frac{1}{3}, \frac{1}{6} \right), s \in \mathbf{R} \), the vector \( \left( \frac{1}{3}, \frac{1}{6} \right) \) is the direction vector. This means it controls the direction along which the line extends indefinitely in both positive and negative directions.
If you imagine placing this vector on a plane, it would point along the same path or line, regardless of its magnitude.
If you imagine placing this vector on a plane, it would point along the same path or line, regardless of its magnitude.
- Direction vectors can be scaled by any non-zero real number (scalar) without changing the direction they point towards.
- They are essentially the "backbone" of a line, dictating its slant and orientation on the graph.
scalar multiple
A scalar multiple in vector equations is a key concept that deals with scaling vectors. It decides the size or length of the vector without changing its direction. When working with a line equation, the direction vector can be multiplied by any scalar to produce another vector that is parallel to the original. This is because changing the magnitude of a vector (by scaling it) does not affect the path it follows—it only alters how far along that path it extends.
Let's break it down:
Let's break it down:
- If \( \vec{d} = \left( \frac{1}{3}, \frac{1}{6} \right) \) is the original direction vector, then any vector \( \vec{m} = k \cdot \vec{d} \) is a scalar multiple, where \( k \) is a real number (the scalar).
- For a vector \( \vec{m} \) to be an acceptable direction vector for a line, there must be a scalar \( k \) such that \( k \vec{d} = \vec{m} \). This means \( \vec{m} \) is parallel to \( \vec{d} \).
line equation
A line equation in vector form is a powerful tool in mathematics used to describe a straight line in a plane or space. Unlike other mathematical representations like slope-intercept or point-slope form, the vector form primarily focuses on the combination of position vectors and direction vectors.
In our example, the line equation \( \vec{r} = \left( \frac{1}{2}, -\frac{3}{4} \right) + s\left( \frac{1}{3}, \frac{1}{6} \right), s \in \mathbf{R} \) expresses how any point \( \vec{r} \) on the line can be found by summing the initial point vector \( \left( \frac{1}{2}, -\frac{3}{4} \right) \) and a scaled direction vector \( s\left( \frac{1}{3}, \frac{1}{6} \right) \).
This format showcases the line as a continuous stretch of points, emphasizing its limitless nature and helping visualize how various vectors can dictate its course and extension.
In our example, the line equation \( \vec{r} = \left( \frac{1}{2}, -\frac{3}{4} \right) + s\left( \frac{1}{3}, \frac{1}{6} \right), s \in \mathbf{R} \) expresses how any point \( \vec{r} \) on the line can be found by summing the initial point vector \( \left( \frac{1}{2}, -\frac{3}{4} \right) \) and a scaled direction vector \( s\left( \frac{1}{3}, \frac{1}{6} \right) \).
- The initial point vector sets a fixed starting position on the line.
- The term with the scalar \( s \) allows for moving along the line infinitely.
This format showcases the line as a continuous stretch of points, emphasizing its limitless nature and helping visualize how various vectors can dictate its course and extension.