In a unit cube with side length , find the dihedral angle along edge between faces and .

The dihedral angle along is (i.e., ).

Solid Geometry Solution – Problem 13: find the dihedral angle along edge between faces and

Question

In a unit cube $ABCD-A_1B_1C_1D_1$ with side length $1$ , find the dihedral angle along edge $AB$ between faces $ABCD$ and $ABB_1A_1$ .

Step-by-step solution

Set up coordinates: $A(0,0,0),\ B(1,0,0),\ C(1,1,0),\ D(0,1,0),\ A_1(0,0,1),\ B_1(1,0,1).$ Face $ABCD$ lies in plane $z=0$ , so a normal vector is $\vec n_1=(0,0,1)$ . Face $ABB_1A_1$ lies in plane $y=0$ , so a normal vector is $\vec n_2=(0,1,0)$ . Compute $\vec n_1\cdot\vec n_2=0.$ Therefore the angle $\theta$ between the two planes satisfies $\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}=0,$ so $\theta=90^\circ.$

Final answer

The dihedral angle along $AB$ is $90^\circ$ (i.e., $\frac\pi2$ ).

Marking scheme

1. Checkpoints (max 7 pts total)

Coordinate setup (2 pts): Assign consistent cube coordinates.
Plane normals (3 pts): Identify correct normals for $z=0$ and $y=0$ .
Angle computation (2 pts): Use $\cos\theta=\frac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}$ to get $90^\circ$ .

2. Zero-credit items

Stating “adjacent faces of a cube are perpendicular” without any vector/coordinate computation.
Using a wrong plane equation for a face.

3. Deductions

Normal direction error (-1): swapping normals (e.g., using $(1,0,0)$ for plane $y=0$ ).
Angle formula misuse (-1): omitting absolute value or norms in the cosine formula.

Solid Geometry – Problem 13: find the dihedral angle along edge between faces and