Distance Formula from a Point to a Line in 3D

In three-dimensional (3D) space, calculating the perpendicular distance from a point to a line is a common task in geometry, physics, and engineering. The distance formula for a point to a line in 3D space helps determine the shortest distance between a given point and a line, which is always perpendicular to the line.

The Concept of Distance in 3D

Consider a point $P(x_1, y_1, z_1)$ and a line defined by a point $P_0(x_0, y_0, z_0)$ on the line and a direction vector $\mathbf{v} = (v_x, v_y, v_z)$ that gives the direction of the line. The goal is to find the shortest distance between the point and the line. This shortest distance is the perpendicular distance from the point to the line.

Equation of the Line

A line in 3D space can be represented parametrically. If $P_0(x_0, y_0, z_0)$ is a point on the line and $\mathbf{v} = (v_x, v_y, v_z)$ is the direction vector, then any point $P(x, y, z)$ on the line can be written as:

$$P(t) = P_0 + t \mathbf{v} = (x_0 + t v_x, y_0 + t v_y, z_0 + t v_z)$$

where $t$ is a scalar parameter.

Distance Formula from a Point to a Line

The shortest distance $d$ from a point $P(x_1, y_1, z_1)$ to a line is calculated using the following formula:

$$d = \frac{|\mathbf{AP} \times \mathbf{v}|}{|\mathbf{v}|}$$

Where:

• $\mathbf{A} = (x_0, y_0, z_0)$ is a point on the line,
• $\mathbf{P} = (x_1, y_1, z_1)$ is the given point,
• $\mathbf{v} = (v_x, v_y, v_z)$ is the direction vector of the line,
• $\mathbf{AP}$ is the vector from point $A$ to point $P$, and
• $\times$ denotes the cross product between the two vectors.

Step-by-Step Derivation

1. Define the vector $\mathbf{AP}$: The vector $\mathbf{AP}$ is the difference between the point $P(x_1, y_1, z_1)$ and the point $A(x_0, y_0, z_0)$ on the line:

   $$\mathbf{AP} = (x_1 - x_0, y_1 - y_0, z_1 - z_0)$$

2. Compute the cross product $\mathbf{AP} \times \mathbf{v}$: The cross product of two vectors gives a vector that is perpendicular to both input vectors. The magnitude of the cross product represents the area of the parallelogram formed by the two vectors. For $\mathbf{AP} = (x_1 - x_0, y_1 - y_0, z_1 - z_0)$ and $\mathbf{v} = (v_x, v_y, v_z)$, the cross product is:

   $$\mathbf{AP} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_1 - x_0 & y_1 - y_0 & z_1 - z_0 \\ v_x & v_y & v_z \end{vmatrix}$$

   The result is a vector perpendicular to both $\mathbf{AP}$ and $\mathbf{v}$.

3. Find the magnitude of the cross product: The magnitude of the cross product gives the area of the parallelogram formed by $\mathbf{AP}$ and $\mathbf{v}$, which is related to the perpendicular distance. The magnitude of a vector $\mathbf{u} = (u_x, u_y, u_z)$ is given by:

   $$|\mathbf{u}| = \sqrt{u_x^2 + u_y^2 + u_z^2}$$

4. Compute the final distance: Finally, divide the magnitude of the cross product by the magnitude of the direction vector $\mathbf{v}$ to get the perpendicular distance:

   $$d = \frac{|\mathbf{AP} \times \mathbf{v}|}{|\mathbf{v}|}$$

Example Calculation

Let’s calculate the distance from the point $P(1, 2, 3)$ to the line defined by $P_0(4, 5, 6)$ with direction vector $\mathbf{v} = (1, 1, 1)$.

1. Compute the vector $\mathbf{AP}$:

   $$\mathbf{AP} = (1 - 4, 2 - 5, 3 - 6) = (-3, -3, -3)$$

2. Compute the cross product $\mathbf{AP} \times \mathbf{v}$:

   $$\mathbf{AP} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & -3 & -3 \\ 1 & 1 & 1 \end{vmatrix} = (0, 0, 0)$$

3. Compute the magnitude of the cross product: The magnitude of the cross product is 0, which indicates that the point lies on the line.

Thus, the distance between the point and the line is 0.

The distance from a point to a line in 3D space is a crucial concept in geometry and various applied fields. By using the formula involving the cross product, one can easily compute the perpendicular distance from any given point to a line. The method is particularly useful in physics, computer graphics, and engineering for understanding spatial relationships in 3D space.