Question

Given the cube with vertices \(P(x, y, z)\) where each of \(x, y,\) and \(z\) is either 0 or \(2,\) consider the plane perpendicular to the diagonal through \(P(0,0,0)\) and \(P(2,2,2)\) and bisecting it. a. Show that the plane meets six of the edges of the cube and bisects them. b. Show that the six points in (a) are the vertices of a regular hexagon.

Short answer

Plane bisects six edges of the cube, vertices forming a regular hexagon.

Step-by-step solution

Understand the Cube and Diagonal

The cube has vertices at all points where each of the coordinates \(x, y, z\) can be either 0 or 2. Thus, the vertices are \((0,0,0), (2,0,0), (0,2,0), (0,0,2), (2,2,0), (2,0,2), (0,2,2), (2,2,2)\). The diagonal is through \((0,0,0)\) and \((2,2,2)\).

Equation of the Plane

The line through \((0,0,0)\) and \((2,2,2)\) is given by \(x = y = z\). A plane perpendicular to this line will have a normal vector along \(x = y = z\), like \((1,1,1)\). The plane bisects the diagonal, so it intersects halfway, at \((1,1,1)\). The equation for the plane is \(x + y + z = 3\).

Find Intersection Points with Cube Edges

For the plane \(x + y + z = 3\), solve it for each edge of the cube. Edges parallel to axis intersect the plane at points where two coordinates are fixed and the equation determines the third. Examples: on edge from \((0,0,0)\) to \((2,0,0)\), we find \((1,0,2)\) since \(x = 1\) and \(x + y + z = 3\). Repeat for each edge.

Verify Each Edge Bisected

Verify every intersection point equally divides its edge. For instance, the midpoint of \((0,0,0)\) and \((2,0,0)\) is \((1,0,0)\). Similarly calculate and check for all edges, confirming the plane gives the midpoint for each.

Identify the Regular Hexagon

Identify six intersection points: \((1,0,2), (0,1,2), (2,1,0), (1,2,0), (0,2,1), (2,0,1)\). To form a regular hexagon, verify equal distances between consecutive points. Use distance formula \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\) to confirm each side's length equals \(\sqrt{2}\), proving hexagon regularity.

Key concepts

Diagonal of a Cube

When understanding the properties of a cube, one key aspect is the diagonal. A cube is a three-dimensional shape, having 12 edges, 8 vertices, and 6 faces.
The diagonal we typically refer to runs from one corner of the cube to the opposite corner, through the center of the cube.
Given a cube with vertices where coordinates are either 0 or 2 (such as at \((0,0,0)\) and \((2,2,2)\)), this specific diagonal would connect these two points.The length of the diagonal in a cube can be calculated using the distance formula in three dimensions, \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \).
For our example, plugging in gives \[\sqrt{(2-0)^2 + (2-0)^2 + (2-0)^2} = \sqrt{12} = 2\sqrt{3}.\]
Understanding this diagonal is important because in some problems, like the one at hand, we deal with planes or lines that are perpendicular or parallel to a cube's diagonal.
This allows us to explore symmetries and other geometric properties related to the cube.

Equation of a Plane

Finding the equation of a plane can seem tricky, but it's crucial in understanding how it relates to other geometric figures like the diagonal of a cube.
A plane in a three-dimensional space is often defined by its normal vector, which is perpendicular to every point on the plane.Consider the diagonal in a cube going from \((0, 0, 0)\) to \((2, 2, 2)\). The line connecting these points has the direction where \(x = y = z\). To find a plane perpendicular to this diagonal, we use a normal vector like \((1, 1, 1)\).
Since the plane also bisects the diagonal, it must pass through the midpoint \((1, 1, 1)\).To derive the equation for this plane, the strategy is to use the normal vector and point:
\[x + y + z = 3.\]
This equation is crucial as it helps to determine the interactions between the plane and the various edges of the cube.
By solving \(x + y + z = 3\) along different edges, we identify various points of intersection, crucial for other aspects of geometry, such as forming new shapes within the cube.
These calculations are powerful tools for further exploration in geometry and computer graphics.

Regular Hexagon

A regular hexagon is a polygon with six equal sides and angles. In the context of a cube, identifying a regular hexagon requires strategic exploration of intersections between planes and cube edges.
In the problem scenario, we form a hexagon from six points of intersection between the plane \(x + y + z = 3\) and the cube's edges.To locate these points, evaluate the equation along edges intersecting the plane:

• Consider the edge from \((0, 0, 0)\) to \((2, 0, 0)\); it yields \((1, 0, 2)\).


• Similar calculations for other edges produce: \((0, 1, 2), (2, 1, 0), (1, 2, 0), (0, 2, 1), (2, 0, 1)\).

To confirm a regular hexagon, we need equal distances between consecutive points, ensuring symmetry and equal side lengths.
Using the distance formula between these points should yield identical results, specifically \(\sqrt{2}\).
Confirming these distances ensures that the hexagon is regular, demonstrating another exciting geometric result derived from intersecting regular shapes like cubes and planes.
This is not only central to understanding geometry in theory but very applicable in areas like computer graphics, where constructions within three-dimensional spaces are fundamental.