Understanding the Equation of a Plane

The equation of a plane in three-dimensional space can take different forms depending on the information given. In this problem, we are given two conditions about the plane:

1. It cuts an intercept of 5 units on the z-axis.
2. It is parallel to the xy-plane.

Plane Parallel to the XY-Plane

Let's first consider the condition that the plane is parallel to the xy-plane. The equation of the xy-plane is \(z = 0\). Any plane parallel to the xy-plane will have a constant z-coordinate for all its points. This is because the normal vector to such a plane will be parallel to the z-axis (which is the normal vector of the xy-plane, \(\vec{k}\) or \(<0, 0, 1>\)).

Therefore, the general equation of a plane parallel to the xy-plane is of the form:

\(z = c\)

where \(c\) is a constant.

Using the Z-axis Intercept

The second condition is that the plane cuts an intercept of 5 units on the z-axis. An intercept on the z-axis means the point where the plane crosses the z-axis. Any point on the z-axis has coordinates of the form \((0, 0, z)\). The z-intercept is the value of \(z\) at the point where the plane intersects the z-axis. An intercept of 5 units on the z-axis means the plane passes through the point \((0, 0, 5)\).

Combining the Conditions to Find the Equation

We know the plane has the form \(z = c\) and it passes through the point \((0, 0, 5)\).

To find the value of \(c\), we substitute the coordinates of the point \((0, 0, 5)\) into the equation \(z = c\).

Substituting \(z = 5\), we get:

\(5 = c\)

So, the constant \(c\) is 5.

Therefore, the equation of the plane that is parallel to the xy-plane and cuts an intercept of 5 units on the z-axis is:

\(z = 5\)

Verifying the Options

Let's quickly look at the given options:

• Option 1: \(x + y = 5\). This plane is parallel to the z-axis. It does not cut an intercept on the z-axis (unless it passes through the origin, which it doesn't).
• Option 2: \(z = 5\). This is a plane parallel to the xy-plane (constant z value) and it passes through \((0, 0, 5)\) (setting x=0, y=0 gives z=5), meaning it cuts a 5-unit intercept on the z-axis. This matches our derived equation.
• Option 3: \(z = 0\). This is the equation of the xy-plane itself. It cuts a 0-unit intercept on the z-axis.
• Option 4: \(x + y + z = 5\). This plane cuts intercepts on the x, y, and z axes. To find the z-intercept, set x=0 and y=0, which gives \(z = 5\). While it has a z-intercept of 5, it is not parallel to the xy-plane (it has x and y terms).

Based on the analysis, the equation \(z = 5\) correctly represents the plane that is parallel to the xy-plane and has a z-intercept of 5 units.