Body diagonals of a cube. What is the angle between two intersecting body diagonals of a cube? (A body diagonal connects two corners and passes through the interior of the cube. A face diagonal connects two corners and runs on one face of the cube.)
The correct answer and explanation is:
The angle between two intersecting body diagonals of a cube is 109.47° (approximately).
Explanation:
A cube is a three-dimensional shape with six square faces, eight vertices, and twelve edges. A body diagonal is a line segment connecting two opposite corners of the cube, passing through its interior. Each cube has four body diagonals that intersect at the cube’s center.
To determine the angle between two intersecting body diagonals, consider a cube with vertices labeled in a coordinate system. Assume a cube with side length s, and let its vertices be:
- A(0,0,0)A(0,0,0)
- B(s,0,0)B(s,0,0)
- C(s,s,0)C(s,s,0)
- D(0,s,0)D(0,s,0)
- E(0,0,s)E(0,0,s)
- F(s,0,s)F(s,0,s)
- G(s,s,s)G(s,s,s)
- H(0,s,s)H(0,s,s)
Two intersecting body diagonals are AEAE and CGCG:
- The vector for AE (from AA to EE) is: AE→=(0,0,s)−(0,0,0)=(0,0,s)\overrightarrow{AE} = (0,0,s) – (0,0,0) = (0,0,s)
- The vector for CG (from CC to GG) is: CG→=(s,s,s)−(s,s,0)=(0,0,s)\overrightarrow{CG} = (s,s,s) – (s,s,0) = (0,0,s)
Using the dot product formula: cosθ=u⋅v∣u∣∣v∣\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}
where u and v are body diagonal vectors: AC→=(s,s,s)\overrightarrow{AC} = (s,s,s) BD→=(−s,−s,s)\overrightarrow{BD} = (-s,-s,s)
Dot product: s2+s2+s2=3s2s^2 + s^2 + s^2 = 3s^2
Magnitude: 3s2=3s\sqrt{3s^2} = \sqrt{3}s cosθ=3s23s2=−13\cos \theta = \frac{3s^2}{3s^2} = -\frac{1}{3} θ=cos−1(−13)≈109.47∘\theta = \cos^{-1}(-\frac{1}{3}) \approx 109.47^\circ
Thus, the angle between two intersecting body diagonals of a cube is 109.47°.
