Equation For A Line In 3d

The equation of a line in 3D space extends the familiar concepts from 2D, but requires a different approach to accurately represent direction and position. Instead of a simple slope-intercept form, 3D lines are typically described using vector equations or parametric equations. These methods utilize vectors to define both the direction of the line and a specific point that the line passes through.

Understanding Vectors in 3D Space

Before diving into the equations, it's crucial to understand the role of vectors in 3D space. A vector is a quantity that has both magnitude (length) and direction. In 3D space, a vector can be represented as an ordered triple <a, b, c>, where a, b, and c are the components of the vector along the x, y, and z axes, respectively.

• Position Vector: Represents the position of a point in space relative to the origin (0, 0, 0).
• Direction Vector: Defines the direction in which a line extends. It's the key to defining the orientation of a line in 3D.

Methods to Define a Line in 3D

There are two primary ways to define a line in 3D space:

1. Vector Equation: This uses a position vector to a point on the line and a direction vector to describe the line's orientation.
2. Parametric Equations: These express the x, y, and z coordinates of any point on the line as functions of a parameter (usually denoted as t).

1. Vector Equation of a Line

The vector equation of a line in 3D is given by:

r = r₀ + t*v

Where:

• r is the position vector of any point on the line (i.e., <x, y, z>).
• r₀ is the position vector of a specific point on the line (i.e., <x₀, y₀, z₀>).
• v is the direction vector of the line (i.e., <a, b, c>).
• t is a scalar parameter that can take any real value.

Explanation:

This equation essentially says that to reach any point r on the line, you start at a known point r₀ on the line and then move along the direction vector v by a scalar multiple t. As t varies over all real numbers, the equation traces out the entire line.

Example:

Suppose we want to find the vector equation of a line that passes through the point (1, 2, 3) and has a direction vector <4, 5, 6>.

• r₀ = <1, 2, 3>
• v = <4, 5, 6>

The vector equation of the line is:

r = <1, 2, 3> + t<<4, 5, 6>

This means any point (x, y, z) on the line can be represented as:

<x, y, z> = <1 + 4t, 2 + 5t, 3 + 6t>

2. Parametric Equations of a Line

The parametric equations of a line in 3D are derived directly from the vector equation. If we have the vector equation r = r₀ + t*v, where r = <x, y, z>, r₀ = <x₀, y₀, z₀>, and v = <a, b, c>, then the parametric equations are:

• x = x₀ + at
• y = y₀ + bt
• z = z₀ + ct

Where:

• x, y, and z are the coordinates of any point on the line.
• x₀, y₀, and z₀ are the coordinates of a specific point on the line.
• a, b, and c are the components of the direction vector.
• t is a scalar parameter.

Explanation:

Each equation expresses one of the coordinates (x, y, or z) as a function of the parameter t. By varying t, we can generate all the points along the line.

Example (Using the same data as above):

Given the point (1, 2, 3) and the direction vector <4, 5, 6>, the parametric equations of the line are:

• x = 1 + 4t
• y = 2 + 5t
• z = 3 + 6t

For example, if t = 0, we get the point (1, 2, 3). If t = 1, we get the point (5, 7, 9).

Finding the Equation of a Line Given Two Points

Often, you're given two points in 3D space and asked to find the equation of the line that passes through them. Here's how to do it:

1. Find the Direction Vector: Subtract the position vector of one point from the position vector of the other point. This gives you a vector that points from the first point to the second point, which can be used as the direction vector.

   v = r₂ - r₁

   Where:

   • r₁ is the position vector of the first point (i.e., <x₁, y₁, z₁>).
   • r₂ is the position vector of the second point (i.e., <x₂, y₂, z₂>).
   • v is the direction vector.

2. Choose a Point on the Line: You can choose either of the given points to be the point r₀ in the vector equation or the values x₀, y₀, and z₀ in the parametric equations.

3. Write the Equation: Plug the direction vector and the chosen point into either the vector equation or the parametric equations.

Example:

Find the equation of the line that passes through the points (2, 4, 6) and (7, 1, -3).

1. Find the Direction Vector:

   v = <7, 1, -3> - <2, 4, 6> = <5, -3, -9>

2. Choose a Point: Let's choose the point (2, 4, 6). So, r₀ = <2, 4, 6>.

3. Write the Equations:

   • Vector Equation: r = <2, 4, 6> + t<<5, -3, -9>
   • Parametric Equations:
     • x = 2 + 5t
     • y = 4 - 3t
     • z = 6 - 9t

Applications of Line Equations in 3D

The equations of lines in 3D space have numerous applications in various fields:

• Computer Graphics: Used to define edges of objects, create wireframe models, and perform ray tracing.
• Physics: Used to describe the trajectory of a particle moving in space.
• Engineering: Used in CAD (Computer-Aided Design) software for designing and modeling objects.
• Navigation: Used in GPS systems to determine the location and direction of movement.
• Robotics: Used in robot path planning and control.

Finding the Intersection of Two Lines in 3D

Determining whether two lines in 3D intersect, and finding the point of intersection if it exists, is a common problem. Here's the process:

1. Write the Parametric Equations for Both Lines: Let the lines be L1 and L2, with parametric equations:

   • L1:
     • x = x₁ + a₁t
     • y = y₁ + b₁t
     • z = z₁ + c₁t
   • L2:
     • x = x₂ + a₂s
     • y = y₂ + b₂s
     • z = z₂ + c₂s

   Note that we use different parameters t and s for each line.

2. Set the Equations Equal: If the lines intersect, then there must be values of t and s for which the x, y, and z coordinates are equal. This gives us a system of three equations:

   • x₁ + a₁t = x₂ + a₂s
   • y₁ + b₁t = y₂ + b₂s
   • z₁ + c₁t = z₂ + c₂s

3. Solve the System of Equations: Solve any two of these equations for t and s. You can use methods like substitution or elimination.

4. Check for Consistency: Plug the values of t and s you found into the third equation. If the equation holds true, then the lines intersect. If it doesn't, the lines do not intersect (they are skew lines).

5. Find the Point of Intersection: If the lines intersect, plug either the value of t into the parametric equations for L1 or the value of s into the parametric equations for L2 to find the coordinates of the point of intersection.

Example:

Determine if the following lines intersect and, if so, find the point of intersection:

• L1:
  • x = 1 + t
  • y = -2 + 3t
  • z = 4 - t
• L2:
  • x = 2 + s
  • y = 3 - 2s
  • z = 1 + 2s

1. Set the Equations Equal:

   • 1 + t = 2 + s
   • -2 + 3t = 3 - 2s
   • 4 - t = 1 + 2s

2. Solve the System: Let's solve the first two equations. From the first equation, we have t = 1 + s. Substituting this into the second equation:

   • -2 + 3(1 + s) = 3 - 2s
   • -2 + 3 + 3s = 3 - 2s
   • 1 + 3s = 3 - 2s
   • 5s = 2
   • s = 2/5

   Now, find t:

   • t = 1 + s = 1 + 2/5 = 7/5

3. Check for Consistency: Plug t = 7/5 and s = 2/5 into the third equation:

   • 4 - t = 1 + 2s
   • 4 - 7/5 = 1 + 2(2/5)
   • 20/5 - 7/5 = 5/5 + 4/5
   • 13/5 = 9/5

   This is not true. Therefore, the lines do not intersect. They are skew lines.

Special Cases and Considerations

• Parallel Lines: If two lines have direction vectors that are scalar multiples of each other, the lines are parallel. Parallel lines in 3D either do not intersect or are the same line.

• Skew Lines: Skew lines are lines that are neither parallel nor intersecting. They lie in different planes and do not meet. The example above illustrates skew lines.

• Coincident Lines: If two lines have the same direction vector and share a common point, they are the same line (coincident).

Distance from a Point to a Line in 3D

Finding the shortest distance from a point to a line in 3D is a common geometric problem. Here's one approach to solving it:

1. Define the Point and the Line: Let the point be P with position vector p, and let the line be defined by the vector equation r = r₀ + t*v, where r₀ is the position vector of a point on the line and v is the direction vector of the line.

2. Find a Vector from a Point on the Line to the Point P: Let w = p - r₀. This vector points from a known point on the line to the point P.

3. Project w onto v: The projection of w onto v (denoted as proj_v w) gives the component of w that lies along the direction of the line. The formula for the projection is:

   • proj_v w = ((w · v) / ||v||²) v

   Where:

   • w · v is the dot product of w and v.
   • ||v|| is the magnitude (length) of v.

4. Find the Vector Orthogonal to the Line: Subtract the projection from w to find the vector that is orthogonal (perpendicular) to the line and points from the line to the point P.

   • **u = w - proj_v w

5. Calculate the Distance: The distance d from the point P to the line is the magnitude of the vector u:

   • d = ||u||

Explanation:

This method essentially decomposes the vector w into two components: one that lies along the line and one that is perpendicular to the line. The perpendicular component represents the shortest distance from the point to the line.

Example:

Find the distance from the point P(1, 2, 3) to the line defined by the equation r = <4, 5, 6> + t<<7, 8, 9>.

1. Define the Point and the Line:

   • p = <1, 2, 3>
   • r₀ = <4, 5, 6>
   • v = <7, 8, 9>

2. Find w:

   • w = p - r₀ = <1, 2, 3> - <4, 5, 6> = <-3, -3, -3>

3. Project w onto v:

   • w · v = (-3)(7) + (-3)(8) + (-3)(9) = -21 - 24 - 27 = -72
   • ||v||² = 7² + 8² + 9² = 49 + 64 + 81 = 194
   • proj_v w = (-72/194) <7, 8, 9> = (-36/97) <7, 8, 9> = <-252/97, -288/97, -324/97>

4. Find u:

   • u = w - proj_v w = <-3, -3, -3> - <-252/97, -288/97, -324/97> = <-3 * 97/97 + 252/97, -3 * 97/97 + 288/97, -3 * 97/97 + 324/97> = <-291/97 + 252/97, -291/97 + 288/97, -291/97 + 324/97> = <-39/97, -3/97, 33/97>

5. Calculate the Distance:

   • d = ||u|| = √((-39/97)² + (-3/97)² + (33/97)²) = √(1521/9409 + 9/9409 + 1089/9409) = √(2619/9409) = √(27/97) * √(97/97) = 3√(3)/√97

Conclusion

Understanding the equations of lines in 3D space is fundamental to various fields. Whether using the vector equation or the parametric equations, the key is to grasp the role of direction vectors and points on the line. From computer graphics to physics simulations, these equations provide a powerful tool for representing and manipulating lines in three dimensions. By mastering these concepts, you can unlock a deeper understanding of spatial relationships and solve complex geometric problems.