In this section, we will explore the mathematical representations of planes and lines in three-dimensional space. Understanding these concepts is crucial for various applications in 3D graphics, computer vision, and geometric modeling.

A plane in 3D space can be defined using a point and a normal vector. The general equation of a plane is given by:

\[ ax + by + cz = d \]

where:

• \( (a, b, c) \) is the normal vector to the plane.
• \( (x, y, z) \) are the coordinates of any point on the plane.
• \( d \) is a constant.

Given a point \( P_0(x_0, y_0, z_0) \) on the plane and a normal vector \( \mathbf{n} = (a, b, c) \), the plane equation can be derived as follows:

1. Any point \( P(x, y, z) \) on the plane satisfies the condition that the vector \( \overrightarrow{P_0P} \) is perpendicular to the normal vector \( \mathbf{n} \).
2. The vector \( \overrightarrow{P_0P} \) is given by \( (x - x_0, y - y_0, z - z_0) \).
3. The dot product of \( \overrightarrow{P_0P} \) and \( \mathbf{n} \) must be zero:

\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]

4. Expanding and rearranging terms, we get the plane equation:

\[ ax + by + cz = ax_0 + by_0 + cz_0 \]

Let \( d = ax_0 + by_0 + cz_0 \), then:

\[ ax + by + cz = d \]

Consider a plane passing through the point \( (1, 2, 3) \) with a normal vector \( \mathbf{n} = (2, -1, 4) \).

1. The equation of the plane is:

\[ 2(x - 1) - 1(y - 2) + 4(z - 3) = 0 \]

2. Simplifying, we get:

\[ 2x - 2 - y + 2 + 4z - 12 = 0 \]

\[ 2x - y + 4z - 12 = 0 \]

\[ 2x - y + 4z = 12 \]

A line in 3D space can be represented parametrically using a point and a direction vector. The parametric equations of a line are given by:

\[ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{d} \]

where:

• \( \mathbf{r}(t) = (x(t), y(t), z(t)) \) are the coordinates of any point on the line.
• \( \mathbf{r}_0 = (x_0, y_0, z_0) \) is a point on the line.
• \( \mathbf{d} = (d_x, d_y, d_z) \) is the direction vector of the line.
• \( t \) is a parameter.

The symmetric form of the line equation is derived from the parametric form by eliminating the parameter \( t \):

\[ \frac{x - x_0}{d_x} = \frac{y - y_0}{d_y} = \frac{z - z_0}{d_z} \]

Consider a line passing through the point \( (1, 2, 3) \) with a direction vector \( \mathbf{d} = (4, -2, 5) \).

1. The parametric equations of the line are:

\[ x(t) = 1 + 4t \] \[ y(t) = 2 - 2t \] \[ z(t) = 3 + 5t \]

2. The symmetric form of the line equation is:

\[ \frac{x - 1}{4} = \frac{y - 2}{-2} = \frac{z - 3}{5} \]

To find the intersection of a line and a plane, substitute the parametric equations of the line into the plane equation and solve for the parameter \( t \).

Find the intersection of the line \( \mathbf{r}(t) = (1 + 4t, 2 - 2t, 3 + 5t) \) with the plane \( 2x - y + 4z = 12 \).

1. Substitute the parametric equations into the plane equation:

\[ 2(1 + 4t) - (2 - 2t) + 4(3 + 5t) = 12 \]

2. Simplify and solve for \( t \):

\[ 2 + 8t - 2 + 2t + 12 + 20t = 12 \]

\[ 30t + 12 = 12 \]

\[ 30t = 0 \]

\[ t = 0 \]

3. Substitute \( t = 0 \) back into the parametric equations to find the intersection point:

\[ x(0) = 1 + 4(0) = 1 \] \[ y(0) = 2 - 2(0) = 2 \] \[ z(0) = 3 + 5(0) = 3 \]

The intersection point is \( (1, 2, 3) \).

Find the equation of the plane passing through the point \( (2, -1, 3) \) with a normal vector \( (1, 4, -2) \).

Solution:

\[ 1(x - 2) + 4(y + 1) - 2(z - 3) = 0 \]

\[ x - 2 + 4y + 4 - 2z + 6 = 0 \]

\[ x + 4y - 2z + 8 = 0 \]

Find the parametric and symmetric forms of the line passing through the point \( (0, 1, -1) \) with a direction vector \( (3, -2, 4) \).

Solution:

Parametric Form:

\[ x(t) = 0 + 3t \] \[ y(t) = 1 - 2t \] \[ z(t) = -1 + 4t \]

Symmetric Form:

\[ \frac{x}{3} = \frac{y - 1}{-2} = \frac{z + 1}{4} \]

Determine the intersection point of the line \( \mathbf{r}(t) = (2 + t, -1 + 2t, 3 - t) \) with the plane \( x + y + z = 6 \).

Solution:

1. Substitute the parametric equations into the plane equation:

\[ (2 + t) + (-1 + 2t) + (3 - t) = 6 \]

2. Simplify and solve for \( t \):

\[ 2 + t - 1 + 2t + 3 - t = 6 \]

\[ 4 + 2t = 6 \]

\[ 2t = 2 \]

\[ t = 1 \]

3. Substitute \( t = 1 \) back into the parametric equations to find the intersection point:

\[ x(1) = 2 + 1 = 3 \] \[ y(1) = -1 + 2(1) = 1 \] \[ z(1) = 3 - 1 = 2 \]

The intersection point is \( (3, 1, 2) \).

In this section, we have covered the equations of planes and lines in 3D space, including their parametric and symmetric forms. We also explored how to find the intersection of a line and a plane. These concepts are fundamental for understanding and manipulating 3D geometric objects. In the next module, we will delve into the representation of 3D objects and their transformations.