Distance of a Point from a Plane (Vector Form)

Understanding the distance of a point from a plane in vector form is a fundamental concept in 3D geometry, particularly in fields such as physics, engineering, and computer graphics. This concept is often used to determine how far a point is from a surface, which can be crucial for collision detection, rendering, and physical simulations.

Plane Equation in Vector Form

Before we delve into the distance calculation, let's recall the equation of a plane in vector form. A plane can be defined by a point ( \mathbf{A} ) (known as the position vector) and a normal vector ( \mathbf{n} ). The equation of the plane is given by:

[ \mathbf{n} \cdot (\mathbf{r} - \mathbf{A}) = 0 ]

where ( \mathbf{r} ) is the position vector of any point on the plane, and ( \mathbf{n} ) is the normal vector to the plane.

Distance Formula

The distance ( d ) of a point ( P ) with position vector ( \mathbf{p} ) from the plane is given by the formula:

[ d = \frac{|\mathbf{n} \cdot (\mathbf{p} - \mathbf{A})|}{|\mathbf{n}|} ]

where ( |\mathbf{n}| ) is the magnitude of the normal vector.

Steps to Calculate the Distance

To calculate the distance of a point from a plane using the vector form, follow these steps:

1. Identify the normal vector ( \mathbf{n} ) of the plane.
2. Choose a point ( \mathbf{A} ) on the plane (position vector).
3. Determine the position vector ( \mathbf{p} ) of the point ( P ) whose distance from the plane you want to find.
4. Substitute these vectors into the distance formula.

Important Points and Differences

Aspect	Description
Plane Equation	Defined by a point and a normal vector.
Normal Vector	A vector perpendicular to the plane.
Position Vector of the Plane	A vector pointing to a specific point on the plane.
Position Vector of the Point	A vector pointing to the point whose distance from the plane is calculated.
Distance Formula	Uses dot product and magnitude of the normal vector.
Absolute Value in Formula	Ensures the distance is non-negative.
Magnitude of Normal Vector	Normalizes the dot product result.

Example

Let's consider a plane with a normal vector ( \mathbf{n} = \begin{bmatrix} 2 \ 3 \ -1 \end{bmatrix} ) and passing through the point ( \mathbf{A} = \begin{bmatrix} 1 \ -1 \ 2 \end{bmatrix} ). We want to find the distance of the point ( P ) with position vector ( \mathbf{p} = \begin{bmatrix} 4 \ 2 \ -3 \end{bmatrix} ) from this plane.

Solution

1. The normal vector ( \mathbf{n} ) is given.
2. The position vector ( \mathbf{A} ) is given.
3. The position vector ( \mathbf{p} ) is given.
4. Substitute into the distance formula:

[ d = \frac{|\mathbf{n} \cdot (\mathbf{p} - \mathbf{A})|}{|\mathbf{n}|} ]

[ d = \frac{| \begin{bmatrix} 2 \ 3 \ -1 \end{bmatrix} \cdot \left( \begin{bmatrix} 4 \ 2 \ -3 \end{bmatrix} - \begin{bmatrix} 1 \ -1 \ 2 \end{bmatrix} \right) |}{\sqrt{2^2 + 3^2 + (-1)^2}} ]

[ d = \frac{| \begin{bmatrix} 2 \ 3 \ -1 \end{bmatrix} \cdot \begin{bmatrix} 3 \ 3 \ -5 \end{bmatrix} |}{\sqrt{4 + 9 + 1}} ]

[ d = \frac{| 2 \cdot 3 + 3 \cdot 3 + (-1) \cdot (-5) |}{\sqrt{14}} ]

[ d = \frac{| 6 + 9 + 5 |}{\sqrt{14}} ]

[ d = \frac{20}{\sqrt{14}} ]

[ d = \frac{20}{\sqrt{14}} \cdot \frac{\sqrt{14}}{\sqrt{14}} ]

[ d = \frac{20\sqrt{14}}{14} ]

[ d \approx 3.78 ]

Therefore, the distance of the point ( P ) from the plane is approximately 3.78 units.

This example illustrates how to apply the vector form of the distance formula to find the distance of a point from a plane. It is important to remember to use the absolute value in the formula to ensure the distance is a non-negative value.