Problem 41 - Essential Calculus Early Transcendentals

Question

Use a scalar projection to show that the distance from a point \(P_{1}\left(x_{1}, y_{1}\right)\) to the line \(a x+b y+c=0\) is $$\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}$$ Use this formula to find the distance from the point \((-2,3)\) to the line \(3 x-4 y+5=0\)

Step-by-Step Solution

Answer

The distance is \(\frac{13}{5}\).

Step 1: Understanding Scalar Projection

Scalar projection is a method to find the component of one vector along the direction of another vector. For a point \(P_1(x_1, y_1)\) to a line \(ax + by + c = 0\), the line direction vector can be represented as \([a, b]\). The point's position vector relative to the line is given by \([x_1, y_1]\).

Step 2: Formula for Distance from Point to Line

The distance from a point \(P_1(x_1, y_1)\) to the line \(ax + by + c = 0\) can be found using the formula \(\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\), which is derived from the scalar projection of \([x_1, y_1]\) onto the normal vector of the line. The normal vector is \([a, b]\), which is perpendicular to the direction vector of the line.

Step 3: Substitute Point and Line to the Formula

To find the distance from the point \((-2, 3)\) to the line \(3x - 4y + 5 = 0\), we substitute \((x_1, y_1) = (-2, 3)\) and the line coefficients \(a = 3\), \(b = -4\), \(c = 5\) into the distance formula. This gives us: \(\frac{|3(-2) - 4(3) + 5|}{\sqrt{3^2 + (-4)^2}}\).

Step 4: Simplify the Expression

First calculate the numerator: \(|3(-2) - 4(3) + 5| = |-6 - 12 + 5| = |-13| = 13\). Next, calculate the denominator: \(\sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\). The distance is then \(\frac{13}{5}\).

Step 5: Conclusion

The distance from the point \((-2, 3)\) to the line \(3x - 4y + 5 = 0\) is \(\frac{13}{5}\).

Key Concepts

Distance from point to line

Calculating the distance from a point to a line involves understanding the geometric relationship between the two. A line in a two-dimensional plane can be described by the equation \(ax + by + c = 0\). The general formula to find this distance is given by

• \( \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \)

Here, \((x_1, y_1)\) is the coordinate of the point in question.
The numerator, \(|ax_1 + by_1 + c|\), is the absolute value of the line equation evaluated at the point. It represents the shortest path from the point to the line in direction perpendicular to the line itself.
The denominator, \(\sqrt{a^2 + b^2}\), normalizes this value by the length of the line's direction vector without affecting the distance component.This formula provides a straightforward way to compute distances, ensuring consistent results across different coordinates and line orientations.

Normal Vector

The normal vector is a crucial component in finding the distance from a point to a line. Unlike the line direction vector, which indicates the line's path, the normal vector is perpendicular to the line's trajectory.
For a line represented by \(ax + by + c = 0\), its normal vector is simply given by the components \([a, b]\).

• Direction Vector: \([b, -a]\)
• Normal Vector: \([a, b]\)

The normal vector is used in the scalar projection to measure how far a point is along or against this perpendicular path. It represents an essential bridge connecting the point to the specific location on the line where the minimum distance is calculated. This perpendicular nature is why it is used in the denominator of the distance formula, effectively translating the length and direction of this vector into the metric used.

Vector Component

A vector component refers to a part of a vector that aligns with a particular direction. When calculating the distance from a point to a line, vector components play a central role.
Imagine having a vector from the point to a point on the line. This vector can be broken down into components:

• Along the line
• Perpendicular to the line

The component perpendicular to the line is related to the normal vector and is the component of interest in the distance calculation.
This perpendicular vector component directly influences the path over which the shortest distance occurs. Formally, in terms of projection, the magnitude of this component is obtained using the scalar projection process. This highlights how tightly or loosely the point aligns away from the line.

• Scalar projection of the point's position vector on the normal vector gives the true measure of distance.

Distance Formula

The distance formula we use is a direct application of finding the shortest path between a stationary point and a linear path. It is essential in various geometric calculations, analytical geometry, and even practical fields like engineering and physics. This formula offers several practical benefits:

• Simplicity: The formula \(\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\) is easy to work with due to few components and real-number operations.
• Universality: It's applicable in any scenario with a straight line and a point in a 2D space.
• Precision: It reduces errors by relying on the algebraic properties of vector components rather than purely geometric intuition.

The square root term \(\sqrt{a^2 + b^2}\) acts as a normalizing factor,
while the absolute value \(|ax_1 + by_1 + c|\) guarantees the distance is a non-negative real number.Every geometric operation builds upon this principle, making it indispensable in analytical tasks.