Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

a. Show that \(\mathbf{n}=\left[\begin{array}{l}a \\ b\end{array}\right]\) is orthogonal to every vector along the line \(a x+b y+c=0\). b. Show that the shortest distance from \(P_{0}\left(x_{0}, y_{0}\right)\) to the line is \(\frac{\left|a x_{0}+b y_{0}+c\right|}{\sqrt{a^{2}+b^{2}}}\).

Short Answer

Expert verified
a. The vector \( \mathbf{n} = [a, b] \) is orthogonal to line vectors. b. The distance is \( \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \).

Step by step solution

01

Understanding Orthogonality Condition

A vector \( \mathbf{n} = [a, b] \) is orthogonal to any vector along the line \( ax + by + c = 0 \) if the dot product of \( \mathbf{n} \) with the direction vector of the line is zero.
02

Direction of Line Vector Calculation

The direction vector of the line \( ax + by + c = 0 \) can be represented as \( \mathbf{d} = [-b, a] \).
03

Proving Orthogonality

Compute the dot product of \( \mathbf{n} = [a, b] \) and \( \mathbf{d} = [-b, a] \): \\[\mathbf{n} \cdot \mathbf{d} = a(-b) + b(a) = -ab + ab = 0\]. Since the dot product is zero, \( \mathbf{n} \) is orthogonal to every vector along the line.
04

Formula for Distance from a Point to a Line

The shortest distance \( d \) from a point \( P_0(x_0, y_0) \) to a line \( ax + by + c = 0 \) can be calculated using the formula: \( d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}\).
05

Derivation of Distance Formula

The perpendicular distance from point \( P_0 \) to the line is calculated using the absolute value and square root to take into account both direction and normalization of the normal vector \( \mathbf{n} \), ensuring that the distance is always positive.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Dot Product
The dot product, also known as the scalar product, is a fundamental operation in vector algebra, providing a measure of how much one vector extends in the direction of another. To compute the dot product of two vectors \(\mathbf{a} = [a_1, a_2]\) and \(\mathbf{b} = [b_1, b_2]\), we use the formula: \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2\). This results in a scalar (a single number) rather than another vector.
When the dot product is zero, it means the vectors are orthogonal (perpendicular) to each other. This property is useful in determining orthogonality in geometry, particularly when analyzing lines and planes. Understanding this concept is crucial for solving problems where you need to determine whether two vectors are perpendicular.
Distance from a Point to a Line
Calculating the distance from a point to a line is a common task in geometry, especially in linear algebra and vector geometry. The formula for the shortest distance \(d\) from a point \(P_0(x_0, y_0)\) to a line \(ax + by + c = 0\) is given by: \(d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}\).
This formula encompasses several important geometric concepts:
  • **Absolute Value**: Ensures the distance is always non-negative, as distance cannot be negative.
  • **Normalization**: The denominator, \(\sqrt{a^2 + b^2}\), normalizes the vector \([a, b]\), converting it into a unit vector, which scales the distance appropriately.
Therefore, this formula efficiently calculates the minimum or perpendicular distance from a point to a line, which can be incredibly useful in many fields, including computer graphics and engineering.
Direction Vector
The direction vector is an essential element in understanding lines in vector geometry. For any line represented by the equation \(ax + by + c = 0\), its direction vector can be defined as \(\mathbf{d} = [-b, a]\).
The direction vector provides several benefits:
  • **Parallelism**: It gives us the direction along which the line extends.
  • **Orientation**: Helps pinpoint the orientation of the line in 2-dimensional space.
By finding a direction vector, you can better understand how a line positions itself in the coordinate plane. This vector is instrumental when assessing perpendicular relationships between lines, particularly through the dot product method.
Perpendicular Distance Formula
The perpendicular distance formula calculates the shortest distance from a point to a line, which often coincides with the concept of orthogonality. In the vector expression, this entails using a perpendicular or normal vector to derive the distance.
The formula \( \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \) ensures we're computing the perpendicular distance, taking careful account of:
  • **Positive Distance**: By incorporating an absolute value, we guarantee the result is a realistic positive measure.
  • **Efficient Computation**: It directly uses coefficients from the line's equation and the coordinates of the point, providing a straightforward computation.
Understanding the perpendicular distance formula is essential for geometry students. It simplifies the process of measuring the shortest path from a point to a line, which is a crucial skill in various mathematical and applied geometric problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that two lines in the plane with slopes \(m_{1}\) and \(m_{2}\) are perpendicular if and only if \(m_{1} m_{2}=-1 .[\) Hint : Example \(4.1 .11 .]\)

Show that if the diagonals of a parallelogram are perpendicular, it is necessarily a rhombus.

Assume \(\mathbf{u}\) and \(\mathbf{v}\) are nonzero vectors that are not parallel. Show that \(\mathbf{w}=\|\mathbf{u}\| \mathbf{v}+\|\mathbf{v}\| \mathbf{u}\) is a nonzero vector that bisects the angle between \(\mathbf{u}\) and \(\mathbf{v}\).

In each case, find a vector equation of the line. a. Passing through \(P(3,-1,4)\) and perpendicular to the plane \(3 x-2 y-z=0\) b. Passing through \(P(2,-1,3)\) and perpendicular to the plane \(2 x+y=1\) c. Passing through \(P(0,0,0)\) and perpendicular $$ \begin{array}{l} \text { to the lines }\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right]+t\left[\begin{array}{r} 2 \\ 0 \\ -1 \end{array}\right] \text { and } \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 2 \\ 1 \\ -3 \end{array}\right]+t\left[\begin{array}{r} 1 \\ -1 \\ 5 \end{array}\right]} \end{array} $$ d. Passing through \(P(1,1,-1)\), and perpendicular to the lines $$ \begin{array}{l} {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 1 \end{array}\right]+t\left[\begin{array}{r} 1 \\ 1 \\ -2 \end{array}\right] \text { an }} \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 5 \\ 5 \\ -2 \end{array}\right]+t\left[\begin{array}{r} 1 \\ 2 \\ -3 \end{array}\right]} \end{array} $$ e. Passing through \(P(2,1,-1)\), intersecting the line \(\left[\begin{array}{l}x \\\ y \\ z\end{array}\right]=\left[\begin{array}{r}1 \\ 2 \\\ -1\end{array}\right]+t\left[\begin{array}{l}3 \\ 0 \\ 1\end{array}\right],\) and perpendicular to that line. f. Passing through \(P(1,1,2)\), intersecting the line \(\left[\begin{array}{l}x \\\ y \\ z\end{array}\right]=\left[\begin{array}{l}2 \\ 1 \\\ 0\end{array}\right]+t\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right],\) and perpendicular to line.

In each case, find \(\overrightarrow{P Q}\) and \(\|\overrightarrow{P Q}\| .\) a. \(P(1,-1,3), Q(3,1,0)\) b. \(P(2,0,1), Q(1,-1,6)\) c. \(P(1,0,1), Q(1,0,-3)\) d. \(P(1,-1,2), Q(1,-1,2)\) e. \(P(1,0,-3), Q(-1,0,3)\) f. \(P(3,-1,6), Q(1,1,4)\)

See all solutions

Recommended explanations on Economics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free