Orthogonal Projections of the Edges of a Cube
Orthogonal Projections of the Edges of a Cube
This Demonstration shows that the sum of the squares of the lengths of the orthogonal projections of the edges of a cube with edge length to a plane equals .
a
8
2
a
Let the bottom corner of the cube (a trihedron) have bottom vertex and three sides , , of length . Let be perpendicular to with . Let the angles of to the three sides be , , . Take the trihedron as the axes of a coordinate system with . Then , , , and so α+β+γ=++=OP=1 [1, p.27].
O
OA
OB
OC
a
OP
OP=1
OP
∠AOP=α
∠BOP=β
∠COP=γ
P=(x,y,z)
x=±cos(α)
y=±cos(β)
z=±cos(γ)
2
cos
2
cos
2
cos
2
x
2
y
2
z
The length of the projection of to is , and similarly for and . So the lengths of the projections of the three edges are , , . Now
OA
O'A'=OA''=sinα
OB
OC
asinα
asinβ
asinγ
2
sin
2
sin
2
sin
2
cos
2
cos
2
cos
2
cos
2
cos
2
cos
The 12 edges of the cube are parallel in sets of four, so the sum of the squares of all the edge lengths equals .
8
2
a