The Three Perpendiculars Theorem
For a line l on plane α:
Line PH is perpendicular to plane α（H is the foot）
Line HQ is perpendicular to l（Q is a point on l）
⟹ Line PQ ⊥ l
（The converse also holds: PQ⊥l, PH⊥α ⟹ HQ⊥l）
Example ① (Application of the Three Perpendiculars Theorem)
In cuboid ABCD−EFGH（AB=3, BC=4, AE=5）, find the angle between AG and BC.
The vector from A to G is (3,4,5) (a diagonal). The vector of BC is (0,4,0) (the y-direction).
cos θ = (3×0+4×4+5×0)/(√50×4) = 16/(4√50) = 4/√50 = 4/(5√2) = 2√2/5
Answer:θ = arccos(2√2/5)
In cuboid ABCD−EFGH（AB=3, BC=4, AE=5）, find the angle between AG and BC.
The vector from A to G is (3,4,5) (a diagonal). The vector of BC is (0,4,0) (the y-direction).
cos θ = (3×0+4×4+5×0)/(√50×4) = 16/(4√50) = 4/√50 = 4/(5√2) = 2√2/5
Answer:θ = arccos(2√2/5)
Example ② (Distance to a plane)
Find the distance (height) h from a vertex of a regular tetrahedron (edge length a) to its base.
From the centroid G of the base to the midpoint M of a base edge = a/√3 (centroid of an equilateral triangle)
From the vertex to the midpoint of each base edge = a×√3/2 (height of an equilateral triangle)
h² + (a/√3)² = (a√3/2)² → h² = 3a²/4 − a²/3 = 5a²/12 → h = a√(2/3) = a√6/3
Answer:h = a√6/3（height of a regular tetrahedron）
Find the distance (height) h from a vertex of a regular tetrahedron (edge length a) to its base.
From the centroid G of the base to the midpoint M of a base edge = a/√3 (centroid of an equilateral triangle)
From the vertex to the midpoint of each base edge = a×√3/2 (height of an equilateral triangle)
h² + (a/√3)² = (a√3/2)² → h² = 3a²/4 − a²/3 = 5a²/12 → h = a√(2/3) = a√6/3
Answer:h = a√6/3（height of a regular tetrahedron）
Key points
- Three Perpendiculars Theorem: reduce perpendicularity in space to a plane
- Angle between lines: calculate cosθ using the dot product and magnitudes of direction vectors
- Height of a regular tetrahedron: h = a√6/3 (important formula)
- For problems in space, set up coordinates or use geometric properties
Practice Problems
- Find the height and volume of a regular tetrahedron with edge length 6.
- In cuboid ABCD−EFGH（AB=2, BC=3, AE=6）, find the length of diagonal AG.
- Using the Three Perpendiculars Theorem: find the distance from the vertex to the base of a regular square pyramid whose base is a square with edge length 4 and whose height is 3 (the distance from the vertex to the centroid of the base).
Answers and Explanations
- Height h=6×√6/3=2√6。Volume=(1/3)×(6²×√3/4)×2√6=(1/3)×9√3×2√6=6√18=18√2
- AG=√(AB²+BC²+AE²)=√(4+9+36)=√49=7
- The centroid of the base is its center. Distance from the vertex to the centroid = height = 3（in a regular square pyramid, the vertex is directly above the centroid of the square base）