Problem 5 - Classical Mechanics

Question

Find the angle between a body diagonal of a cube and any one of its face diagonals. [Hint: Choose a cube with side 1 and with one corner at \(O\) and the opposite corner at the point (1,1,1) . Write down the vector that represents a body diagonal and another that represents a face diagonal, and then find the angle between them as in Problem 1.4.]

Step-by-Step Solution

Answer

The angle between a body diagonal and a face diagonal of the cube is \(\cos^{-1}\left(\frac{\sqrt{6}}{3}\right)\).

Step 1: Define Vectors

To solve the problem, we begin by defining vectors for the body diagonal and face diagonal. Consider a cube with a side length of 1. Place the cube with one corner at the origin, makes its coordinates \(0,0,0\). The opposite corner will be at \(1,1,1\). The body diagonal of the cube is a vector from \(0,0,0\) to \(1,1,1\), so this vector can be denoted as \vec{d_b} = (1,1,1)\. Next, consider a face of the cube. For example, the face along the x-y plane would extend from \(0,0,0\) to \(1,1,0\) and from \(0,0,0\) to \(0,1,0\). The face diagonal in this plane is \vec{d_f} = (1,1,0)\.

Step 2: Calculate the Dot Product

The angle between two vectors can be found using the dot product formula \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta}\). First, calculate the dot product of \vec{d_b} = (1,1,1)\ and \vec{d_f} = (1,1,0)\. The dot product is given by \( \vec{d_b} \cdot \vec{d_f} = 1 \times 1 + 1 \times 1 + 1 \times 0 = 2 \).

Step 3: Calculate Magnitudes

Find the magnitudes of both \vec{d_b}\ and \vec{d_f}\. The magnitude of \vec{d_b}\ is \( |\vec{d_b}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \). The magnitude of \vec{d_f}\ is \( |\vec{d_f}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \).

Step 4: Find the Cosine of the Angle

Using the formula \( \cos{\theta} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \; |\vec{b}|} \), calculate \cos{\theta}\. Substitute the previously found values: \( \cos{\theta} = \frac{2}{\sqrt{3} \times \sqrt{2}} = \frac{2}{\sqrt{6}} \). Simplify \( \frac{2}{\sqrt{6}} \) to get \( \frac{\sqrt{6}}{3} \).

Step 5: Calculate the Angle

Finally, find the angle \(\theta\) by computing \(\theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right)\). This will give us the angle we are looking for in either radians or degrees, depending on the function used (calculator setting).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector calculus, used to calculate the angle between two vectors. It's an algebraic operation that takes two equal-length sequences of numbers, usually coordinate vectors, and returns a single number.
Understanding the dot product is essential in vector analysis because it can show how much of one vector goes in the direction of another. It is defined as:

• \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \)
• In a physical sense, this result is a scalar representing the projection of one vector onto another.
• For example, in this exercise, the dot product of vectors \( \vec{d_b} = (1,1,1) \) and \( \vec{d_f} = (1,1,0) \) is calculated as:
• \( \vec{d_b} \cdot \vec{d_f} = 1\times1 + 1\times1 + 1\times0 = 2 \)

This scalar reflects the overlap in direction between the body diagonal and face diagonal of the cube. The greater this value, the smaller the angle between the vectors. Conversely, a dot product of zero indicates perpendicularity.

Cube Geometry

Understanding cube geometry is crucial for visualizing the problem at hand. A cube is a three-dimensional shape with six square faces, all of the same size, with edges of equal length. Here, the cube is positioned at the origin of a coordinate system, providing a straightforward way to define positions as vectors.
When we talk about the body diagonal of a cube, we refer to the line connecting opposite corners, such as from \((0,0,0)\) to \((1,1,1)\). This line cuts through the interior of the cube and is longer than any face diagonal.

• The face diagonal extends across one of the cube's square faces. It can be visualized as stretching from one corner of the square to the opposite corner.
• In this case, for the face of the cube on the x-y plane, the diagonal runs from \((0,0,0)\) to \((1,1,0)\).

Each type of diagonal provides key insights into the cube's geometry, and understanding their properties is critical for solving spatial problems like finding angles between vectors.

Magnitude of Vectors

Magnitude (or length) of a vector is another vital concept in vector calculus. It signifies the "size" or "length" of the vector and is calculated using the Pythagorean theorem in three dimensions.
For any vector \( \vec{v} = (v_1, v_2, v_3) \), its magnitude \( |\vec{v}| \) is expressed as follows:

• \( |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)

Finding the magnitude is essential when working with the dot product, especially when determining angles:

• For the body diagonal \( \vec{d_b} = (1, 1, 1) \), the magnitude is \( \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \).
• For the face diagonal \( \vec{d_f} = (1, 1, 0) \), the magnitude is \( \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \).

These magnitudes are necessary to solve the cosine of the angle formula using the dot product, showing how much of one vector aligns with another’s direction.