Question

By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle between any two diagonals of a cube, where e?ch diagonal connects diametrically opposite corners and passes through the center of the cube.

Short answer

Answer: The smaller angle between any two diagonals of a cube is approximately \(70.53^\circ\).

Step-by-step solution

Express the diagonals as vectors

Let's first label the vertices of the cube as A, B, C, D, E, F, G, and H, where A is the origin (0,0,0). We can represent the diagonals of the cube with vectors connecting pairs of diametrically opposite points. Let AB = a, BC = b, DA = c. Connecting (0,0,0) to the diagonally opposite vertex E, we obtain the following diagonals: - For diagonal AE: \(\vec{d_1} = a+b+c\) - For diagonal BG: \(\vec{d_2} = -a+b+c\) - For diagonal DH: \(\vec{d_3} = a-b+c\) - For diagonal CF: \(\vec{d_4} = -a-b+c\)

Find the dot product of the diagonals

Using the dot product definition, the dot product of \(\vec{d_1}\) and \(\vec{d_2}\) is given by: \(\vec{d_1} \cdot \vec{d_2} = (a+b+c) \cdot (-a+b+c)\) By calculating the products: \(\vec{d_1} \cdot \vec{d_2} = -a^2+b^2+c^2+ab-ac+bc\) We can choose any pair of vectors since all the diagonal pairs will make the same angle with each other.

Find the magnitudes of the vectors

Now, we need to find the magnitudes of both vectors. This is given by: \(|\vec{d_1}| = \sqrt{a^2+b^2+c^2}\) \(|\vec{d_2}| = \sqrt{(-a)^2+b^2+c^2}\) As both vectors have the same magnitudes, we can denote them as M: \(M = \sqrt{a^2+b^2+c^2}\)

Apply the dot product formula

Applying the dot product formula to find the angle θ between \(\vec{d_1}\) and \(\vec{d_2}\): \(\cos(\theta) = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} = \frac{-a^2+b^2+c^2+ab-ac+bc}{M^2}\)

Solve for the angle θ

Now let's solve for the angle θ: \(\cos(\theta) = \frac{-a^2+b^2+c^2+ab-ac+bc}{(a^2+b^2+c^2)}\) As we know that in a cube, all edges are equal in length, a = b = c. Let a = b = c = s. Substitute s into the equation: \(\cos(\theta) = \frac{-s^2+s^2+s^2+s^2-s^2+s^2}{3s^2} = \frac{s^2}{3s^2}\) Finally, take the inverse cosine: \(\theta = \arccos(\frac{s^2}{3s^2}) = \arccos(\frac{1}{3})\) The smaller angle between any two diagonals of the cube is approximately \(70.53^\circ\).

Key concepts

Diagonals of a Cube

In understanding the geometry of a cube, it's essential to grasp the concept of diagonals. Unlike edges, which connect two adjacent corners or vertices, a diagonal inside a cube links two corners that are completely opposite each other. Imagine a line slicing through the cube, passing through its center and connecting these two distant points.

In a cube, there are several such diagonals. Each diagonal can be visualized by selecting a pair of diametrically opposite vertices. These lines are three-dimensional, meaning they extend through all three axes (x, y, z) of the cube. This creates a fascinating internal symmetry, as every diagonal is equal in length, reflecting the uniformity of the cube's structure.

Dot Product

The dot product is a fundamental operation in vector calculus. It provides a way to multiply two vectors, resulting in a scalar, rather than another vector. The dot product is calculated as:
\[\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3\]
where \(\vec{u}\) and \(\vec{v}\) are vectors in a three-dimensional space.

In the context of diagonals in a cube, the dot product helps us calculate the cosine of the angle between two vectors. By finding \(\vec{d_1} \cdot \vec{d_2}\), we can determine how closely aligned the two diagonal vectors are. If the result is positive, they point in somewhat similar directions; if it's zero, they are perpendicular; if negative, they point in opposite directions. This property makes the dot product incredibly useful for solving geometrical problems involving angles.

Angle Between Vectors

Determining the angle between two vectors is a common problem in vector calculus. To find this angle, \(\theta\), between vectors \(\vec{u}\) and \(\vec{v}\), we use the formula:
\[\cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}\]
where \(|\vec{u}|\) and \(|\vec{v}|\) are the magnitudes of the vectors.

In cubes, using this formula allows us to understand the spatial arrangement of the diagonals. Since the lengths (magnitudes) of the diagonals are equal, we substitute the values into the formula to find the cosine of the angle \(\theta\). This procedure reveals the structure and angles formed by the intersections of the cube's diagonals, providing a clearer picture of its three-dimensional nature.

Geometry of a Cube

The geometry of a cube is characterized by its six equal, square faces and twelve equal edges. Its symmetry makes it a simple yet rich subject in geometry. A cube has eight vertices, and at each vertex, three edges meet.

All corners—or vertices—are at the same distance from the center of the cube. This symmetrical arrangement is one reason the cube's diagonals are all the same length. Because of the right angles between the edges, the cube doesn’t just present straightforward two-dimensional surface interactions but also complex three-dimensional structures, like intersecting diagonals.

This inherent symmetry makes cubes extremely suitable for exercises involving spatial visualization and geometric calculations, such as finding the angles between diagonals. Understanding these properties assists in applying vector calculus effectively to real-world problems dealing with cubic shapes.