Question

Distance between a point and a line in the plane Use projections to find a general formula for the (least) distance between the point $P(x_0,y_0)\text{ and the line }ax+by=c.\text{ (See Exercises }62-65.)$.

Short answer

The general formula for the least distance between a point (x₀, y₀) and a line with equation ax + by = c is: $$d_{min}=\frac{|c-a{x}_0-b{y}_0|}{\sqrt{a^2+b^2}}$$

Step-by-step solution

Determine the line's direction vector.

Find a direction vector for the given line by using the coefficients $a$ and $b$ from the equation $ax+by=c$. Since any two points on the line can give us its direction vector, we can use the coefficients to get the direction vector. The direction vector, $\overrightarrow{d}$, of the given line is: $$\overrightarrow{d}=⟨a,b⟩$$

Determine the vector from the point P to any point on the line.

To determine the shortest distance, find the projection of a vector from point P to any point on the line. We can use the point $(0,\frac{c}{b})$ as a point on the line (assuming b is non-zero). The vector from P to the line, $\overrightarrow{v}$, would be: $$\overrightarrow{v}=⟨-x_0,\frac{c}{b}-y_0⟩$$

Calculate the scalar projection of $\overrightarrow{v}$ onto $\overrightarrow{d}$.

We will now calculate the scalar projection of the vector $v$ onto the direction vector $d$. The scalar projection, $pro{j}_{\overrightarrow{d}}\overrightarrow{v}$, is given by $$pro{j}_{\overrightarrow{d}}\overrightarrow{v}=\frac{\overrightarrow{v}\cdot \overrightarrow{d}}{\Vert \overrightarrow{d}\Vert }$$ Where $\overrightarrow{v}\cdot \overrightarrow{d}$ denotes the dot product of the two vectors and $\Vert \overrightarrow{d}\Vert$ denotes the magnitude of the direction vector. Perform the operations: $$pro{j}_{\overrightarrow{d}}\overrightarrow{v}=\frac{(-x_0\cdot a)+(\frac{c}{b}-y_0)\cdot b}{\sqrt{a^2+b^2}}$$

Calculate the least distance.

Now that we have the scalar projection, we can calculate the least distance between the point and the line. The least distance, $d_{min}$, is given by the magnitude of the vector from P to the line, subtracted by the scalar projection of this vector onto the direction vector $d$: $$d_{min}=\frac{(c-a{x}_0-b{y}_0)}{\sqrt{a^2+b^2}}$$ Here is the general formula for the least distance between a point $(x_0,y_0)$ and a line with equation $ax+by=c$.

Key concepts

Scalar Projection

The scalar projection of a vector offers a way to determine how much one vector extends in the direction of another. Imagine you have a vector $\overrightarrow{v}$ and you want to determine how much of this vector points in the same direction as another vector $\overrightarrow{d}$. This is where scalar projection comes into play. It's like holding a flashlight above $\overrightarrow{v}$, and seeing the shadow it casts onto $\overrightarrow{d}$. This shadow's length is the scalar projection.
To calculate the scalar projection of $\overrightarrow{v}$ onto $\overrightarrow{d}$, you use:

• Find the dot product of $\overrightarrow{v}$ and $\overrightarrow{d}$.
• Divide the result by the magnitude (or length) of $\overrightarrow{d}$.

This gives you a scalar, a single number that tells you the magnitude of this extension. Mathematically, it is represented as:$$pro{j}_{\overrightarrow{d}}\overrightarrow{v}=\frac{\overrightarrow{v}\cdot \overrightarrow{d}}{\Vert \overrightarrow{d}\Vert }$$
Scalar projections provide insight into how vectors relate in terms of direction and distance, making them invaluable for geometric and physical analysis.

Direction Vector

A direction vector is a critical concept in understanding the trajectory or alignment of a line. Given a line equation in the form $ax+by=c$, the coefficients $a$ and $b$ play a vital role. These coefficients can directly be used to determine the direction vector $\overrightarrow{d}$, which is $⟨a,b⟩$. This vector encapsulates the slope and orientation of the line.
The beauty of a direction vector lies in its simplicity. It transforms complex line equations into a straightforward vector representation that can be visually and mathematically manipulated. With a direction vector:

• You gain insights into the line's slope and movement.
• It makes it easier to perform vector operations like scalar projections and cross products.

Using the direction vector, you can quickly assess any line's characteristics and apply vector algebra techniques, such as finding distances or angles relative to other geometric entities.

Dot Product

The dot product, also known as the scalar product, tells you how two vectors interact when they point at one another. Imagine it as a measure of similarity in direction. For two vectors $\overrightarrow{u}=⟨u_1,u_2⟩$ and $\overrightarrow{v}=⟨v_1,v_2⟩$, the dot product $\overrightarrow{u}\cdot \overrightarrow{v}$ is given by:$$\overrightarrow{u}\cdot \overrightarrow{v}=u_1{v}_1+u_2{v}_2$$
This operation results in a scalar. If the dot product is:

• Positive, the vectors point in a generally similar direction.
• Negative, the vectors point in more opposite directions.
• Zero, the vectors are perpendicular, or orthogonal.

In practical terms, the dot product is crucial for calculating angles between vectors and projecting one vector onto another. Its application extends beyond geometric notions, enriching various fields such as physics, computer graphics, and machine learning by offering foundational building blocks for vector operations.

Magnitude of a Vector

The magnitude of a vector, often known as its length, gives you an idea of how far the vector extends in space. Think of it as the actual distance from the start to the end tip of the vector. If you have a vector $\overrightarrow{v}=⟨x,y⟩$, its magnitude is calculated using the Pythagorean theorem, as follows:$$\Vert \overrightarrow{v}\Vert =\sqrt{x^2+y^2}$$
Magnitude is a fundamental concept in vector analysis. It allows you to determine how large or small a vector is without concerning its direction. This concept is particularly useful because:

• It helps in normalizing vectors, which means scaling them to a unit length.
• It aids in calculating distances and is handy in the application of the distance formula.

The importance of vector magnitude lies in its ability to translate complex multi-dimensional quantities into easily understandable and comparable single numbers, simplifying computations and analyses in physics, engineering, and computer science.