Distance of a point from a plane (vector form)
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:
- Identify the normal vector ( \mathbf{n} ) of the plane.
- Choose a point ( \mathbf{A} ) on the plane (position vector).
- Determine the position vector ( \mathbf{p} ) of the point ( P ) whose distance from the plane you want to find.
- 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
- The normal vector ( \mathbf{n} ) is given.
- The position vector ( \mathbf{A} ) is given.
- The position vector ( \mathbf{p} ) is given.
- 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.