Planes and Lines in Three-Dimensional Space

Planes in Three-Dimensional Space

Definition and Equation of a Plane

A plane in three-dimensional space is a flat, two-dimensional surface that extends infinitely in all directions. To uniquely determine a plane, we require:

• A point in the plane, P0 = (x0, y0, z0)

• A normal vector to the plane, \( \vec{n} = A \hat{i} + B \hat{j} + C \hat{k} \)

If a point P = (x, y, z) lies on the plane, then the vector from P0 to P (denoted \( \overrightarrow{P_0P} = (x-x_0, y-y_0, z-z_0) \)) must be perpendicular to the normal vector \( \vec{n} \). This leads to the condition:

• \( \overrightarrow{P_0P} \cdot \vec{n} = 0 \)

• Or, equivalently, \( (\vec{r} - \vec{r}_0) \cdot \vec{n} = 0 \)

The general equation for a plane is:

\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]

Alternatively, this can be written as:

\[ Ax + By + Cz = D \]

where \( D = Ax_0 + By_0 + Cz_0 \).

Examples of Plane Equations

• Example 1: Find the equation of the plane containing the point (1, 0, 5) and having a normal vector \( \vec{n} = 3\hat{i} + \hat{j} - \hat{k} \).

Solution:

• \( A = 3, B = 1, C = -1 \)

• Equation: \( 0 = 3(x-1) + 1(y-0) + (-1)(z-5) \)

• Simplifies to: \( 3x + y - z = -2 \)

• Example 2: Find the equation of the plane containing the points P1 = (1, 1, 0), P2 = (2, 0, 2), and P3 = (0, 3, 3).

Solution:

• Find vectors in the plane: \( \overrightarrow{P_1P_2} = (1, -1, 2) \), \( \overrightarrow{P_1P_3} = (-1, 2, 3) \)

• Normal vector: \( \vec{n} = \overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3} = (-7, -5, 1) \)

• Equation: \( -7(x-1) -5(y-1) + 1(z-0) = 0 \)

• Simplifies to: \( -7x -5y + z = -12 \)

Lines in Three-Dimensional Space

Definition and Equations of a Line

A line in three-dimensional space is determined by:

• A point on the line, P0 = (x0, y0, z0)

• A direction vector, \( \vec{v} = a\hat{i} + b\hat{j} + c\hat{k} \)

If a point P = (x, y, z) lies on the line, then the vector \( \overrightarrow{P_0P} = (x-x_0, y-y_0, z-z_0) \) must be parallel to \( \vec{v} \). Thus,

• \( \overrightarrow{P_0P} = t \vec{v} \) for some scalar parameter t

• Or, \( \vec{r} = \vec{r}_0 + t\vec{v} \)

In component form, the equations are:

\[ \begin{cases} x = x_0 + at \\ y = y_0 + bt \\ z = z_0 + ct \end{cases} \]

Alternatively, eliminating t (assuming a, b, c ≠ 0):

\[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \]

If a = 0, but b and c are not zero:

\[ \frac{y - y_0}{b} = \frac{z - z_0}{c}, \quad x = x_0 \]

Example: Intersection of Two Planes

• Example 3: Find the standard form equation for the line of intersection of the planes x + y − z = 0 and y + 2z = 6.

Solution:

• Normals: \( \vec{n}_1 = (1, 1, -1) \), \( \vec{n}_2 = (0, 1, 2) \)

• Direction vector: \( \vec{v} = \vec{n}_1 \times \vec{n}_2 = (3, -2, 1) \)

• Find a point on the line by solving the system:

  • x0 + y0 − z0 = 0

  • y0 + 2z0 = 6

• One solution: (3, 0, 3)

• Standard form equation:

  • \( \frac{x-3}{3} = \frac{y}{-2} = \frac{z-3}{1} \)

• Alternative solution: (0, 2, 2) yields \( \frac{x}{3} = \frac{y-2}{-2} = \frac{z-2}{1} \)

Additional info: The intersection of two non-parallel planes is always a line, and the direction vector of this line is given by the cross product of the planes' normal vectors.