Question

Use projections to find a general formula for the (smallest) distance between the point \(\left.P\left(x_{0}, y_{0}\right) \text { and the line } a x+b y=c . \text { (See Exercises } 62-65 .\right)\)

Short answer

Answer: The general formula for the smallest distance between point P and line ax + by = c is given by: $ d(P,Q) = \frac{|ax_0+by_0-c|}{\sqrt{a^{2}+b^{2}}} $

Step-by-step solution

Find the normal vector of the line

The normal vector of a line ax + by = c is given by the coefficients of x and y. So the normal vector is N = (a, b).

Find the unit normal vector of the line

To find the unit normal vector, we need to divide the normal vector by its magnitude. The magnitude of N is given by \(\sqrt{a^{2}+b^{2}}\). Therefore, the unit normal vector \(\hat{N}\) is given by: \(\hat{N}=\frac{1}{\sqrt{a^{2}+b^{2}}}(a, b)\)

Project the point P onto the line

To project the point P onto the line, we subtract its coordinates from the coordinates of the point (x0, y0) and dot the result with the unit normal vector. We will call the projection point Q: $ Q = \text{proj}_{\hat{N}}(P) = \left( x_0 - a\frac{\left\cdot\left}{a^2+b^2}, y_0 - b\frac{\left\cdot\left}{a^2+b^2}\right) $

Calculate the distance between point P and its projection Q

To find the distance between point P and its projection Q, we need to find the difference between their coordinates and take the magnitude of the resulting vector: $ d(P,Q) = \sqrt{\left(x_0 - a\frac{\left\cdot\left}{a^2+b^2}\right)^{2}+\left(y_0 - b\frac{\left\cdot\left}{a^2+b^2}\right)^{2}} $

Simplify the expression and obtain the general formula for the distance

We know that the required distance between point P and the line is given by d(P, Q). Now, let's simplify the expression: $ d(P,Q) = \frac{1}{\sqrt{a^{2}+b^{2}}} \sqrt{(ax_0+by_0-c)^{2}} $ Since we are calculating the smallest distance between the point and the line, we can drop the square root and apply the absolute value for the given expression. So the general formula to find the smallest distance between point P and line ax + by = c is: $ d(P,Q) = \frac{|ax_0+by_0-c|}{\sqrt{a^{2}+b^{2}}} $

Key concepts

Projection

Projection in the context of geometry, particularly when discussing lines and points, essentially measures how one vector relates to a line as it "projects" or "drops down" onto it.
The projection of a point onto a line helps us determine the perpendicular distance to the line from that point.

• To project a point onto a line means finding the closest point on that line.
• This involves utilizing vectors and dot products for accuracy.
• It describes how far along the line a particular point is when viewed directly from a perpendicular perspective.

Imagine shining a light directly above point P that casts a shadow straight onto the line—this shadow is the projection. Through calculations, notably by employing the normal vector, you'll locate this shadow point mathematically.

Normal Vector

A normal vector in terms of a line in 2D is a vector that is perpendicular to the line.
In the equation of the line given by \[ ax + by = c \], the normal vector is represented by the coefficients of \( a \) and \( b \).

• For a line \( ax + by = c \), the normal vector is \( N = (a,b) \).
• This vector points perpendicularly from the line, determining direction and slope characteristics of the line.
• The normal vector is crucial as it becomes a baseline for more complex vector calculations like finding distances and projections.

Understanding how the coefficients shape this vector helps visualize how lines are oriented in plane geometry. Grasping the role of the normal vector is essential when using it to carry out projections.

Distance Formula

The Distance Formula is a mathematical equation used to compute the distance between two points in some coordinate space.
For our specific task of finding the distance between a point and a line, this formula is slightly adapted.

• For two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the Euclidean distance is given by \[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \].
• In determining the distance from a point to a line, we utilize the formula: \[ \frac{|ax_0 + by_0 - c|}{\sqrt{a^2 + b^2}} \].
• This results from projecting the point onto the line and calculating the magnitude of that point's vector perpendicular to the line.

By using this streamlined approach, the formula aids in efficiently calculating the shortest distance, crucial in optimizing many geometric problems.

Vector Magnitude

Vector magnitude is akin to the "length" or "size" of a vector.
It represents how long or large a vector is in a given space, irrespective of its direction.

• The magnitude of a vector \( (a, b) \) is computed via \[ \sqrt{a^2 + b^2} \].
• This forms the basis for computing a unit normal vector, which scales the normal vector to just a unit length.
• When finding distances, this magnitude factors into normalizing vectors, leading to simpler calculations.

Comprehending vector magnitude is fundamental to grasping other vector operations. It not only quantifies vectors but also normalizes them—making calculations neat and practical in different mathematical and physics applications.