To determine the angle between the diagonal $B_1D$ and the face $DD_1C_1C$ in a rectangular parallelepiped:

Explanation:

1. Identify vectors:

   • The diagonal $B_1D$ can be represented as a vector $\vec{B_1D}$.
   • The face $DD_1C_1C$ lies in a plane, so we need a normal vector to this plane. A suitable vector perpendicular to the plane $DD_1C_1C$ can be derived from its vertices or edges.

2. Calculate direction:

   • For $B_1D$, the coordinates of the endpoints can be used to compute the direction vector.
   • For the plane $DD_1C_1C$, use the cross product of two non-parallel vectors in the plane to find a normal vector.

3. Cosine formula: The angle $\theta$ between $\vec{B_1D}$ and the plane is given by: $\cos \theta = \frac{|\vec{B_1D} \cdot \vec{n}|}{\|\vec{B_1D}\| \cdot \|\vec{n}\|}$ Here, $\vec{n}$ is the normal vector to the plane $DD_1C_1C$.

4. Options: Among the given choices, $B_1DC_1$ seems relevant because it involves points from both the diagonal and the face.

Answer:

The correct choice is $B_1DC_1$.

Would you like me to provide a step-by-step computation for this or clarify any concept?

Related Questions:

1. How can you compute the normal vector to a plane given three points?
2. What is the role of the dot product in finding angles between vectors?
3. How do you calculate the length of a vector in 3D space?
4. How can the cross product be used to determine perpendicular vectors?
5. What are some applications of angle computations in geometry?

Tip:

Always visualize the geometry of a parallelepiped to understand vector directions and planes better.