To find the angle between two body diagonals in a unit cube, we first need to understand what body diagonals are and how they are positioned within a cube.
A unit cube has vertices at the following coordinates: (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), and (1, 1, 1). The body diagonals connect opposite corners of the cube. In this case, we can take two body diagonals: from (0, 0, 0) to (1, 1, 1) and from (1, 0, 0) to (0, 1, 1).
The vectors corresponding to these diagonals can be written as follows:
- Vector A (from (0, 0, 0) to (1, 1, 1)): A = (1, 1, 1)
- Vector B (from (1, 0, 0) to (0, 1, 1)): B = (-1, 1, 1)
To find the angle between the two vectors, we can use the dot product formula:
A · B = |A| |B| cos(θ)
Where:
- A · B is the dot product of vectors A and B.
- |A| is the magnitude of vector A.
- |B| is the magnitude of vector B.
- θ is the angle between the two vectors.
Calculating the dot product:
A · B = (1)(-1) + (1)(1) + (1)(1) = -1 + 1 + 1 = 1
Next, we calculate the magnitudes of vectors A and B:
|A| = √(1² + 1² + 1²) = √3
|B| = √((-1)² + 1² + 1²) = √3
Now we can substitute these values back into the dot product formula:
1 = (√3)(√3) cos(θ)
1 = 3 cos(θ)
cos(θ) = 1/3
This gives us:
θ = cos-1(1/3)
Thus, the angle between the two body diagonals in a unit cube is the arccosine of 1/3.