Question

Prove that the perpendicular distance of a point \(\bar{p}\) to a plane \(\vec{u} \cdot \bar{x}=c\) in \(E^{n}\) is $$ \rho\left(\bar{p}, \bar{x}_{0}\right)=\frac{|\vec{u} \cdot \bar{p}-c|}{|\vec{u}|} . $$ \(\left(\bar{x}_{0}\right.\) is the orthogonal projection of \(\bar{p},\) i.e., the point on the plane such that \(\left.\overrightarrow{p x_{0}} \| \vec{u} .\right)\)

Short answer

The distance formula is verified as correct for calculating the distance from a point to a plane in \( E^n \).

Step-by-step solution

Understand the Problem Statement

We need to verify the formula for the perpendicular distance from a point \( \bar{p} \) to a plane given by \( \vec{u} \cdot \bar{x} = c \) in \( E^n \). Here, \( \bar{x}_0 \) refers to the orthogonal projection of \( \bar{p} \) onto the plane, meaning the vector \( \overrightarrow{px_0} \) is parallel to the normal vector \( \vec{u} \).

Definition of the Plane and Normal Vector

A plane in \( E^n \) can be described by the equation \( \vec{u} \cdot \bar{x} = c \), where \( \vec{u} \) is the normal vector to the plane. Any point \( \bar{x}_0 \) on the plane satisfies this equation.

Propose the Perpendicular Distance Formula

The perpendicular distance \( \rho \) from the point \( \bar{p} \) to the plane is given by the formula: \[ \rho(\bar{p}, \bar{x}_0) = \frac{|\vec{u} \cdot \bar{p} - c|}{|\vec{u}|} \] This formula suggests that the distance depends on the dot product of the normal vector \( \vec{u} \) with the point \( \bar{p} \), adjusted by the plane's constant \( c \), and normalized by the magnitude of \( \vec{u} \).

Prove the Distance Formula Using Perpendicular Projections

To find the perpendicular distance, consider the orthogonal projection of \( \bar{p} \) onto the plane. The perpendicular projection implies: \( \overrightarrow{px_0} = \lambda \vec{u} \) for some scalar \( \lambda \). Since \( \bar{x}_0 \) lies on the plane:\( \vec{u} \cdot \bar{x}_0 = c \). Hence, \( \bar{x}_0 = \bar{p} - \lambda \vec{u} \) solves the plane equation. Substituting, we get:\( \vec{u} \cdot (\bar{p} - \lambda \vec{u}) = c \Rightarrow \vec{u} \cdot \bar{p} - \lambda |\vec{u}|^2 = c \Rightarrow \lambda = \frac{\vec{u} \cdot \bar{p} - c}{|\vec{u}|^2} \).

Evaluate the Perpendicular Distance

The magnitude or length of \( \overrightarrow{px_0} \) is the perpendicular distance we are looking for: \[ ||\overrightarrow{px_0}|| = |\lambda||\vec{u}|| = \left|\frac{\vec{u} \cdot \bar{p} - c}{|\vec{u}|^2}\right||\vec{u}|| = \frac{|\vec{u} \cdot \bar{p} - c|}{|\vec{u}|} \] Hence, this confirms the formula for the perpendicular distance.

Key concepts

Projection onto a Plane

When we talk about projecting a point onto a plane, we are essentially finding the shadow or footprint of that point on the plane if a light were shining directly perpendicularly onto the plane. This concept is crucial as it helps in identifying the location on the plane that is closest to the point outside the plane.

The orthogonal projection of a point \( \bar{p} \) onto a plane described by the equation \( \vec{u} \cdot \bar{x} = c \) involves finding a point \( \bar{x}_0 \) on the plane such that the line joining \( \bar{p} \) to \( \bar{x}_0 \) is perpendicular to the plane. This means that \( \overrightarrow{px_0} \) is parallel to the normal vector \( \vec{u} \).

Understanding how to compute this projection helps in determining various geometric properties such as distance, direction, and placement. When you effectively project a point onto a plane, you can then calculate how far it now lies from a particular location within the plane. This builds the foundation for calculating distances in multidimensional spaces.

Normal Vector

A normal vector is fundamental in understanding the orientation of a plane within a given space. When you think about the plane, the normal vector is like an arrow sticking out perpendicularly from its surface. In mathematics, calculating with and about this vector helps determine planar equations and how points relate to each other across distances.

The normal vector \( \vec{u} \) acts as a pivotal component in the plane equation \( \vec{u} \cdot \bar{x} = c \). It's what describes the attitude of the plane in space. This vector is perpendicular to every line or vector that essentially lies within the plane itself.

• The length of the normal vector is important when measuring and comparing different vectors’ influence or direction.
• The direction of the normal vector is specified by its components, showing precisely how the plane is tilted or placed.

Understanding and using the normal vector effectively allow you to solve for positions and distances, illustrate spatial relationships, and comprehend the spatial interrelation of vectors, points, and planes.

Dot Product

The dot product is a foundational component in vector mathematics providing insight into how two vectors relate to each other directionally. This mathematical operation takes two equal-length sequences of numbers and returns a single number. It's particularly useful for determining angles between vectors and projecting vectors onto others.

In the context of plane and point calculations within vector spaces, the dot product plays an essential role. When solving the perpendicular distance of a point to a plane, you are essentially manipulating the dot product of the normal vector with another vector.

• The formula \( \vec{u} \cdot \bar{p} \) signifies how much of one vector is in the same direction as another or how aligned they are.
• The dot product provides a measure that can also help solve variables in planar equations, such as calculating a perpendicular distance or confirming alignments.
• If the dot product is zero, it confirms that two vectors are perpendicular, underscoring its significance in determining relationships in geometry.

Mastering the dot product equips you with tools for understanding vectorial relationships, tackling geometric problems, and solving equations efficiently by comprehending correlations between directional vectors.