To find the angle formed by the diagonal of a cube and a diagonal of one of its faces, follow these steps:

Step 1: Understanding the Geometry of the Cube

Consider a cube with side length $a$.

• The diagonal of the cube is a line connecting two opposite vertices of the cube, passing through its interior.
• A diagonal of a face is a line connecting two opposite vertices of one of the cube’s square faces.

Step 2: Diagonal of the Cube

The length of the diagonal of the cube can be found using the Pythagorean theorem in three dimensions. If the side of the cube is $a$, the diagonal $D_{\text{cube}}$ is given by:

$D_{\text{cube}} = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$

Step 3: Diagonal of the Face

The diagonal of a square face of the cube is simply the diagonal of a square with side length $a$. Using the Pythagorean theorem for a square:

$D_{\text{face}} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$

Step 4: Finding the Angle Between the Two Diagonals

The diagonal of the cube and the diagonal of the face form an angle. To find this angle, we use the dot product formula for the angle $\theta$ between two vectors:

$\cos \theta = \frac{\vec{v}_1 \cdot \vec{v}_2}{|\vec{v}_1| |\vec{v}_2|}$

Where $\vec{v}_1$ and $\vec{v}_2$ are the vectors representing the two diagonals, and $|\vec{v}_1|$ and $|\vec{v}_2|$ are their magnitudes.

Step 4a: Define the Vectors

Let’s assume the cube is aligned such that one corner is at the origin.

• The vector for the diagonal of the cube from $(0,0,0)$ to $(a,a,a)$ is: $\vec{v}_1 = (a, a, a)$

• The vector for the diagonal of a face, say the face on the $xy$-plane, is from $(0,0,0)$ to $(a,a,0)$: $\vec{v}_2 = (a, a, 0)$

Step 4b: Dot Product

The dot product of $\vec{v}_1 = (a, a, a)$ and $\vec{v}_2 = (a, a, 0)$ is: $\vec{v}_1 \cdot \vec{v}_2 = a \cdot a + a \cdot a + a \cdot 0 = 2a^2$

Step 4c: Magnitudes of the Vectors

The magnitude of $\vec{v}_1$ is: $|\vec{v}_1| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$

The magnitude of $\vec{v}_2$ is: $|\vec{v}_2| = \sqrt{a^2 + a^2} = a\sqrt{2}$

Step 4d: Cosine of the Angle

Now, applying the dot product formula: $\cos \theta = \frac{2a^2}{(a\sqrt{3})(a\sqrt{2})} = \frac{2a^2}{a^2\sqrt{6}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}$

Thus, the angle $\theta$ is: $\theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right)$

Using a calculator to find the inverse cosine: $\theta \approx 35.2644^\circ$

Final Answer:

The angle formed by the diagonal of a cube and a diagonal of one of its faces is approximately 35.26°.

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Would you like a deeper explanation or further steps on a related topic? Here are some related questions:

1. How do you find the angle between two vectors in 3D space?
2. What is the general formula for the diagonal of a cube in terms of side length?
3. How can the angle between two vectors in space be applied to real-life problems?
4. What other geometric properties of a cube can we explore using vectors?
5. How would the angle change if the cube had a different shape (e.g., a rectangular prism)?

Tip: When working with 3D geometry problems involving angles between vectors, always start by defining the vectors clearly and use the dot product formula to simplify the process.