Problem 56 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the angle between a main diagonal of a cube and one of its faces.

Step-by-Step Solution

Answer

The angle between a main diagonal of a cube and one of its faces is \( \cos^{-1}\left(\frac{\sqrt{6}}{3}\right) \).

Step 1: Understand the Problem

We need to find the angle between a main diagonal of a cube and one of its faces. A cube has six faces and four main diagonals. Each main diagonal passes through the cube from one vertex to the opposite vertex.

Step 2: Set Up a Coordinate System

Let's place the cube in a 3D coordinate system such that one vertex is at the origin, (0, 0, 0), and the cube extends along the positive x, y, and z axes. If the side length of the cube is \(a\), the vertices will be at coordinates (0,0,0), (a,0,0), (0,a,0), (0,0,a), etc.

Step 3: Determine the Main Diagonal Vector

Choose a main diagonal that goes from (0,0,0) to (a,a,a). The vector representing this diagonal is \(\vec{d} = \langle a, a, a \rangle\).

Step 4: Determine a Face Diagonal Vector

Choose a face of the cube, for example, the face in the xy-plane (z=0). A diagonal of this face runs from (0,0,0) to (a,a,0). The vector for this diagonal is \(\vec{f} = \langle a, a, 0 \rangle\).

Step 5: Calculate the Dot Product

The dot product of \(\vec{d}\) and \(\vec{f}\) is \(\vec{d} \cdot \vec{f} = a \times a + a \times a + a \times 0 = 2a^2\).

Step 6: Calculate the Magnitudes

The magnitude of \(\vec{d}\) is \(|\vec{d}| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}\). The magnitude of \(\vec{f}\) is \(|\vec{f}| = \sqrt{a^2 + a^2} = a\sqrt{2}\).

Step 7: Find the Cosine of the Angle

The cosine of the angle \(\theta\) between the vectors is given by:\[\cos(\theta) = \frac{\vec{d} \cdot \vec{f}}{|\vec{d}| \times |\vec{f}|} = \frac{2a^2}{a\sqrt{3} \cdot a\sqrt{2}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}\]

Step 8: Calculate the Angle

The angle \(\theta\) is found by taking the inverse cosine of the result:\[ \theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right) \]

Key Concepts

3D Coordinate System

A 3D coordinate system is a method of representing points in three-dimensional space. It uses three perpendicular axes: usually labeled as x, y, and z. This system helps to precisely define the position of any point with a set of three numbers known as coordinates. For example, the point (2, 3, 4) is located 2 units along the x-axis, 3 units along the y-axis, and 4 units along the z-axis.

• The x-axis typically represents horizontal movement.
• The y-axis indicates vertical movement.
• The z-axis shows depth.

In exercises like finding the angle between vectors, placing an object like a cube in a 3D coordinate system allows us to describe its edges, faces, and diagonals through vector coordinates.

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar or simply a number. It is very useful in calculating angles between vectors because it relates directly to the angle. Here's the formula:\[ \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3 \]where \( \vec{u} = \langle u_1, u_2, u_3 \rangle \) and \( \vec{v} = \langle v_1, v_2, v_3 \rangle \).

• A high dot product indicates smaller angles between vectors.
• A zero dot product suggests a 90-degree angle or perpendicular vectors.

To find the angle between vectors, you combine the dot product with the magnitudes of the vectors.

Vector Magnitude

The magnitude of a vector, often represented as \(|\vec{v}|\), tells us the vector's length in space. It's calculated by taking the square root of the sum of the squares of its components. Here’s the formula:\[|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}\]For example, if \( \vec{v} = \langle a, b, c \rangle \), then its magnitude is \( \sqrt{a^2 + b^2 + c^2} \).

• A vector with all zero components has a magnitude of zero.
• Magnitude helps to normalize vectors.

When finding angles between vectors, having their magnitudes helps to normalize them, ensuring calculations are accurate.

Trigonometric Functions

Trigonometric functions are mathematical functions that relate angles to side lengths in right-angled triangles. They are crucial when working with angles in vector analysis, such as computing the angle between two vectors. The cosine function, in particular, is fundamental here. It represents the ratio of the adjacent side to the hypotenuse in a right triangle:\[\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\]In vector calculations, the cosine of the angle \(\theta\) between vectors \(\vec{u}\) and \(\vec{v}\) is given by:\[\cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| \times |\vec{v}|}\]To find the angle itself, the inverse cosine function (\( \cos^{-1} \)) is used, which gives the angle when you know its cosine value.