Calculating Cosine of Angle Between Cube Diagonal and Edge

We need to find the angle $\theta$ between the longest diagonal of a cube and one of its edges.

Setting up the Cube Geometry

Let's place a cube with side length $a$ in a 3D coordinate system. We can place one vertex at the origin $(0, 0, 0)$ and the opposite vertex at $(a, a, a)$.

• The vector representing the longest diagonal ($\vec{d}$) connects $(0, 0, 0)$ to $(a, a, a)$. So, $\vec{d} = \langle a, a, a \rangle$.
• The vector representing an edge ($\vec{e}$) starting from the origin along the x-axis connects $(0, 0, 0)$ to $(a, 0, 0)$. So, $\vec{e} = \langle a, 0, 0 \rangle$.

Using Vector Dot Product

The relationship between the dot product of two vectors, their magnitudes, and the angle between them is given by:

$\vec{d} \cdot \vec{e} = |\vec{d}| |\vec{e}| \cos \theta$

We can rearrange this to solve for $\cos \theta$:

$\cos \theta = \frac{\vec{d} \cdot \vec{e}}{|\vec{d}| |\vec{e}|}$

Performing the Calculation

1. Calculate the dot product $\vec{d} \cdot \vec{e}$: $ \vec{d} \cdot \vec{e} = \langle a, a, a \rangle \cdot \langle a, 0, 0 \rangle = (a)(a) + (a)(0) + (a)(0) = a^2 $
2. Calculate the magnitude of the diagonal $|\vec{d}|$: $ |\vec{d}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} $
3. Calculate the magnitude of the edge $|\vec{e}|$: $ |\vec{e}| = \sqrt{a^2 + 0^2 + 0^2} = \sqrt{a^2} = a $
4. Substitute these values into the formula for $\cos \theta$: $ \cos \theta = \frac{a^2}{(a\sqrt{3})(a)} = \frac{a^2}{a^2\sqrt{3}} = \frac{1}{\sqrt{3}} $

Result

Therefore, the cosine of the angle between the longest diagonal of the cube and any one of its edges is $\frac{1}{\sqrt{3}}$.