Calculating Distance Between Point and Line in 3D: A Complete Guide
Table of Contents :
In 3D geometry, calculating the distance between a point and a line can be quite complex, but understanding the underlying concepts makes it manageable. Whether you're a student grappling with geometry or a professional working with 3D models, this guide will walk you through the process step by step. 🚀
Understanding the Components
What is a Point in 3D Space? 📍
A point in 3D space is defined by its coordinates (x, y, z). For example, the point A(3, 5, 2) represents a specific location in a three-dimensional coordinate system.
What is a Line in 3D Space? 📏
A line in 3D can be defined using a point and a direction vector. If we have point B(x₀, y₀, z₀) and a direction vector d(dx, dy, dz), the line can be expressed in parametric form as:
L(t) = B + t * **d**
Where t is a scalar parameter that varies.
The Formula for Distance
To calculate the distance D between a point P(px, py, pz) and a line defined by point B and direction vector d, we can use the formula:
[ D = \frac{|(\overrightarrow{BP}) \times d|}{|d|} ]
Where:
- (\overrightarrow{BP}) is the vector from point B to point P.
- (|\cdot|) denotes the magnitude of the vector.
- (\times) indicates the cross product.
Breakdown of the Formula
-
Vector BP: Find the vector from B to P: [ \overrightarrow{BP} = P - B = (px - x₀, py - y₀, pz - z₀) ]
-
Cross Product: Calculate the cross product of (\overrightarrowBP}) and d \times d = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ px - x₀ & py - y₀ & pz - z₀ \ dx & dy & dz \end{vmatrix} ]
-
Magnitude of Vectors: Calculate the magnitudes needed:
- (|\overrightarrow{BP} \times d|)
- (|d|)
Step-by-Step Calculation 🛠️
Let’s walk through an example to see how this works in practice.
Example Problem
Given:
- Point P(4, 6, 3)
- Point B(1, 2, 1)
- Direction vector d(3, 1, 2)
Step 1: Calculate Vector BP
[ \overrightarrow{BP} = P - B = (4 - 1, 6 - 2, 3 - 1) = (3, 4, 2) ]
Step 2: Calculate the Cross Product
Using the determinant: [ \overrightarrow{BP} \times d = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 2 \ 3 & 1 & 2 \end{vmatrix} = \hat{i} (4 \cdot 2 - 2 \cdot 1) - \hat{j} (3 \cdot 2 - 2 \cdot 3) + \hat{k} (3 \cdot 1 - 4 \cdot 3) ]
Calculating this gives: [ = \hat{i} (8 - 2) - \hat{j} (6 - 6) + \hat{k} (3 - 12) = \hat{i} (6) + \hat{j} (0) + \hat{k} (-9) = (6, 0, -9) ]
Step 3: Magnitude of the Cross Product
[ |\overrightarrow{BP} \times d| = \sqrt{6^2 + 0^2 + (-9)^2} = \sqrt{36 + 0 + 81} = \sqrt{117} = 10.82 ]
Step 4: Magnitude of Direction Vector
[ |d| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \approx 3.74 ]
Step 5: Calculate Distance
[ D = \frac{10.82}{3.74} \approx 2.90 ]
Final Result
The distance between the point P(4, 6, 3) and the line through point B(1, 2, 1) with direction vector d(3, 1, 2) is approximately 2.90 units. 📏
Visualization 📊
| Component | Calculation | Result |
|---|---|---|
| Point P | P(px, py, pz) | (4, 6, 3) |
| Point B | B(x₀, y₀, z₀) | (1, 2, 1) |
| Direction Vector d | d(dx, dy, dz) | (3, 1, 2) |
| Vector BP | (\overrightarrow{BP}) | (3, 4, 2) |
| Cross Product Result | (\overrightarrow{BP} \times d) | (6, 0, -9) |
| Magnitude of Cross Product | 10.82 | |
| Magnitude of Direction Vector | 3.74 | |
| Distance (D) | Final Calculation | 2.90 |
Important Notes
Understanding the geometric interpretation of distance in 3D space is crucial. The process of finding the distance can be visualized as the shortest line connecting the point and the line, which is always perpendicular to the line.
By following this guide, you should have a clearer understanding of how to calculate the distance between a point and a line in 3D space. With practice, this method can be applied to various problems across different fields, including engineering, physics, and computer graphics. Keep exploring the fascinating world of geometry! ✨
