Understanding coordinates and planes is fundamental in 3D mathematics. This topic will cover the basics of coordinate systems, the representation of points in 3D space, and the mathematical description of planes.

In a 3D Cartesian coordinate system, any point in space can be represented by three coordinates: \( (x, y, z) \). These coordinates correspond to the distances from the point to the three perpendicular axes (X, Y, and Z).

• X-axis: The horizontal axis.
• Y-axis: The vertical axis.
• Z-axis: The depth axis.

Consider a point \( P \) with coordinates \( (3, 2, 5) \):

• \( x = 3 \): The point is 3 units along the X-axis.
• \( y = 2 \): The point is 2 units along the Y-axis.
• \( z = 5 \): The point is 5 units along the Z-axis.

Exercise 1: Plot the following points in a 3D Cartesian coordinate system:

1. \( A(1, 2, 3) \)
2. \( B(-1, -2, -3) \)
3. \( C(4, 0, -2) \)

Solution:

1. Point \( A \) is 1 unit along the X-axis, 2 units along the Y-axis, and 3 units along the Z-axis.
2. Point \( B \) is -1 unit along the X-axis, -2 units along the Y-axis, and -3 units along the Z-axis.
3. Point \( C \) is 4 units along the X-axis, 0 units along the Y-axis, and -2 units along the Z-axis.

A plane in 3D space can be defined using a point and a normal vector, or by an equation in the form \( Ax + By + Cz + D = 0 \).

The general equation of a plane is: \[ Ax + By + Cz + D = 0 \] where \( A \), \( B \), \( C \), and \( D \) are constants.

Consider the plane equation \( 2x + 3y - z + 6 = 0 \):

• \( A = 2 \)
• \( B = 3 \)
• \( C = -1 \)
• \( D = 6 \)

The normal vector \( \mathbf{n} \) to the plane is \( \mathbf{n} = (A, B, C) \). For the plane \( 2x + 3y - z + 6 = 0 \), the normal vector is \( \mathbf{n} = (2, 3, -1) \).

Exercise 2: Find the normal vector for the plane given by the equation \( 4x - y + 2z - 5 = 0 \).

Solution: The normal vector \( \mathbf{n} \) is \( (4, -1, 2) \).

The distance \( d \) from a point \( P(x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \) can be calculated using the formula: \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Find the distance from the point \( (1, 2, 3) \) to the plane \( 2x + 3y - z + 6 = 0 \).

Solution:

1. Substitute \( x_1 = 1 \), \( y_1 = 2 \), \( z_1 = 3 \) into the formula: \[ d = \frac{|2(1) + 3(2) - 1(3) + 6|}{\sqrt{2^2 + 3^2 + (-1)^2}} \]
2. Calculate the numerator: \[ |2 + 6 - 3 + 6| = |11| = 11 \]
3. Calculate the denominator: \[ \sqrt{4 + 9 + 1} = \sqrt{14} \]
4. Compute the distance: \[ d = \frac{11}{\sqrt{14}} \approx 2.94 \]

Exercise 3: Calculate the distance from the point \( (2, -1, 4) \) to the plane \( 3x - 2y + z - 7 = 0 \).

Solution:

1. Substitute \( x_1 = 2 \), \( y_1 = -1 \), \( z_1 = 4 \) into the formula: \[ d = \frac{|3(2) - 2(-1) + 1(4) - 7|}{\sqrt{3^2 + (-2)^2 + 1^2}} \]
2. Calculate the numerator: \[ |6 + 2 + 4 - 7| = |5| = 5 \]
3. Calculate the denominator: \[ \sqrt{9 + 4 + 1} = \sqrt{14} \]
4. Compute the distance: \[ d = \frac{5}{\sqrt{14}} \approx 1.34 \]

In this section, we covered the basics of coordinate systems and the representation of planes in 3D space. We learned how to define a plane using its equation and how to calculate the distance from a point to a plane. These concepts are fundamental for understanding more complex topics in 3D mathematics and graphics.