Problem 1 - Algebra and Geometry

Question

Let \(\theta\) be the acute angle formed by two diagonals of a cube. Show that \(\cos \theta=1 / 3\), and hence find \(\theta\)

Step-by-Step Solution

Answer

\( \cos \theta = \frac{1}{3} \); hence, \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \).

Step 1: Understand the Problem

To find the cosine of the angle \( \theta \) between the diagonals of a cube, we identify the vectors representing these diagonals. Two diagonals of a cube can be represented as vectors from one vertex to the opposite vertex.

Step 2: Define the Cube and Diagonal Vectors

Consider a cube with vertices at coordinates \((0,0,0)\), \((1,1,1)\), \((0,1,1)\), \((1,0,1)\), and so on. The diagonal from \((0,0,0)\) to \((1,1,1)\) can be represented by the vector \( \vec{a} = \langle 1, 1, 1 \rangle \). Another diagonal, say from \((0,1,0)\) to \((1,0,1)\), is \( \vec{b} = \langle 1, -1, 1 \rangle \).

Step 3: Compute the Dot Product

Calculate the dot product \( \vec{a} \cdot \vec{b} \): \(\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1\).

Step 4: Calculate Magnitudes of the Vectors

Find the magnitudes of \( \vec{a} \) and \( \vec{b} \). Both vectors have the same magnitude: \(|\vec{a}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}\) and \(|\vec{b}| = \sqrt{1^2+(-1)^2+1^2} = \sqrt{3}\).

Step 5: Determine Cosine of the Angle

The cosine of the angle \( \theta \) is given by the dot product formula:\(\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{\sqrt{3} \times \sqrt{3}} = \frac{1}{3}.\)

Step 6: Find the Angle \( \theta \)

Use the inverse cosine function to find \( \theta \): \( \theta = \cos^{-1} \left( \frac{1}{3} \right) \). Evaluate using a calculator to find the acute angle.

Key Concepts

Understanding Vector Geometry

Let's dive into the fascinating world of vector geometry to tackle this intriguing problem about cube diagonals. In essence, vector geometry involves representing objects and their relationships using vectors, which are mathematical objects with both a direction and a magnitude (or length). In our exercise, the objects of interest are the diagonals of a cube. A cube, as we know, has multiple diagonals, and each can be described as a vector. For instance, if we look at a unit cube with one vertex at the origin

• The diagonal running from the origin (0, 0, 0) to the opposite vertex (1, 1, 1) is represented by the vector \( \vec{a} = \langle 1, 1, 1 \rangle \).
• Another diagonal, for instance from (0, 1, 0) to (1, 0, 1), is represented by the vector \( \vec{b} = \langle 1, -1, 1 \rangle \).

Understanding vectors in this context helps us to find relationships between points and lines in three-dimensional space. If you visualize moving along these vectors from the origin point, you will see how directions and relative placements are represented. This is crucial to proceed with calculations such as finding angles between diagnostics using the dot product.

Using the Dot Product in Vectors

The dot product of two vectors is a fundamental concept in vector mathematics and plays a pivotal role in finding the angle between two vectors. The dot product can be thought of as a way of multiplying vectors to arrive at a scalar (a number without direction), which is particularly useful in this context. For vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is calculated as:
\[\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\]In our problem, the vectors representing the cube's diagonals are \( \vec{a} = \langle 1, 1, 1 \rangle \) and \( \vec{b} = \langle 1, -1, 1 \rangle \). To find the dot product, calculate:

• \( 1 \times 1 + 1 \times (-1) + 1 \times 1 = 1 - 1 + 1 = 1 \).

This scalar value can help us understand how much these vectors align with each other. It's important to note that the bigger the scalar value, the more these vectors point in the same direction.

Applying the Inverse Cosine Function

After computing the dot product and the magnitudes of the vectors in the exercise, the next step is to determine the cosine of the angle between them using the formula derived from the dot product:
\[\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot |\vec{b}|}\]Here, each vector has a magnitude of \( \sqrt{3} \), and the dot product is \( 1 \). So:

• \( \cos \theta = \frac{1}{\sqrt{3} \times \sqrt{3}} = \frac{1}{3} \).

To find the angle \( \theta \) itself, we utilize the inverse cosine function, also known as arccosine. This function helps us backtrack from the cosine value to the angle, which is what we're interested in here. To solve, you'd use a calculator to evaluate:

• \( \theta = \cos^{-1} \left( \frac{1}{3} \right) \).

This yields the acute angle formed by the diagonals in the cube, completing our quest using the magic of inverse trigonometric functions!