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Problem 314

# What is the angle between a diagonal of a cube and one of its edges?

Expert verified
The angle between a diagonal of a cube and one of its edges is 90°.
See the step by step solution

## Step 1: Visualize the cube and the diagonal

Visualize a cube with side length 'a', and choose one vertex (which we'll call A). From that vertex, draw the diagonal (let's say from A to G) passing through the interior of the cube, connecting vertex A and the vertex G which is farthest from A. Also, choose an adjacent edge from vertex A, we'll use edge AB for this example.

## Step 2: Use the Pythagorean Theorem to find the magnitude of the diagonal AG

Recall that the distance between two opposite corners of a cube can be found using the Pythagorean theorem. In our case, we will find the distance of diagonal AG. We can create a right triangle by connecting vertices A, G, and B, where AB becomes one of its legs. Denote the distance between vertices A and G by |AG|. Since we have a right triangle formed by AG, AB, and the projection along edge AD, we can use the Pythagorean theorem to find |AG|: |AG|^2 = |AB|^2 + |proj_AD(ABG)|^2 We can also use the Pythagorean theorem to find |proj_AD(ABG)|, by considering the right triangle formed by AD, DB, and the projection along AD: |proj_AD(ABG)|^2 = |AD|^2 + |DB|^2 Substituting this expression into the first equation: |AG|^2 = |AB|^2 + (|AD|^2 + |DB|^2) |AG| = $$\sqrt{|AB|^2 + |AD|^2 + |DB|^2}$$ Since AB, AD, and DB all have the same length 'a' (cube's side length), we can write: |AG| = $$\sqrt{a^2 + a^2 + a^2}$$ = $$\sqrt{3a^2}$$

## Step 3: Find the angle using the Vector Dot Product

Now that we have the magnitudes of the diagonal AG and the edge AB, we can find the angle θ between them using the vector dot product formula: $$\textbf{A}\cdot\textbf{B} = |\textbf{A}||\textbf{B}|cos(θ)$$ In our case, $$\textbf{A}$$ represents the diagonal AG and $$\textbf{B}$$ represents the edge AB. Since A and G are opposite vertices, their dot product equals zero. The magnitudes of AG and AB are |AG| and |AB|=a, so we can write: |AG||AB|cos(θ) = 0 $$\sqrt{3a^2}$$(a)cos(θ) = 0 We can solve for the angle θ: cos(θ) = $$\frac{0}{(\sqrt{3a^2})(a)}$$ cos(θ) = 0 θ = arccos(0) = $$\frac{\pi}{2}$$ or 90° So, the angle between a diagonal of a cube and one of its edges is 90 degrees.

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