Question
In the quadrilateral pyramid , plane , plane , , and
(1) Prove that plane .
(2) Let be the reflection of point across plane . Find , where is the angle between line and plane .
Step-by-step solution
(1) Let be the midpoints of , , respectively, and connect . From , , and , , we get and . Hence quadrilateral is a parallelogram, so .
Also, , and . Since , we have plane , thus plane . Because plane ,
(2) Through , draw , meeting at . Then plane . Let . Set a 3D rectangular coordinate system with as origin, and lines containing as the -axes. Take From right-triangle projection relations, , so Since is the reflection of across plane , point is the midpoint of , thus A normal vector of plane can be . Therefore
Final answer
(1) By midpoint construction and perpendicular-plane criteria, we find a line in plane that is perpendicular to plane , so
(2) After locating the reflection point through midpoint , and using a normal vector of plane , we obtain This is the sine of the angle between line and the plane.
Marking scheme
1. Checkpoints (max 7 pts total)
Part (1): plane-perpendicular proof (3 pts)
- Correct midpoint construction and parallel relation derivation. (1.5 pts)
- Show a line in plane is perpendicular to plane . (1 pt)
- Conclude two-plane perpendicularity. (0.5 pt)
Part (2): line-plane angle (4 pts)
- Build coordinates and determine projection/symmetry point coordinates. (2 pts)
- Find a valid normal vector of plane . (1 pt)
- Apply vector formula and obtain . (1 pt)
Total (max 7)
2. Zero-credit items
- Assuming midpoint/reflection coordinates without derivation.
- No normal vector or equivalent angle computation in part (2).
3. Deductions
- Reflection-point error (-1): confusing projection point and midpoint relation.
- Angle formula misuse (-1): using cosine formula directly for line-plane angle.