Website owner: James Miller
ORTHOGONAL PROJECTIONS OF PLANE AREA REGIONS
Theorem. Let a region Q in a plane A be projected orthogonally onto some other plane B to produce a region P. Then
Area of P = (area of Q) cos θ
where θ is the acute angle between the two planes. See Fig. 1.
If n and m are unit vectors perpendicular to planes A and B respectively, then
Area of P = | n·m | (area of Q)
where | n·m | is the absolute value of the dot product of n and m.
Proof. Prove that the area of the projection P of region Q is given by (area of Q) cos θ .
To prove this result consider Fig.
2 where we have planes A and B
intersecting along line RS and
making an angle of θ with each
other. A rectangular region Q of
dimensions a
b in plane A is
projected orthogonally onto
region P of plane B. The area of
Q is ab. The area of P is ac = ab
cos θ. Thus the area of P is given
by
Area of P = (area of Q) cos θ .
We can use this result to find the area of the orthogonal projection of any arbitrary region Q in plane A onto the plane B by partitioning Q into narrow rectangular slices that run perpendicular to the line of intersection RS of planes A and B as shown in Fig. 3. Each slice projects into an image whose area is given by
(area of slice)
cos θ .
Summing all slices gives the final result.
Prove. cos θ = | n·m |
Proof. The acute angle between two planes is equal to the acute angle between two lines perpendicular to the planes.
Angle between two lines with direction numbers l1, m1, n1 and l2, m2, n2 . The angle θ between two lines with direction numbers l1, m1, n1 and l2, m2, n2 is given by
If n and m are unit vectors perpendicular to planes A and B, the components of n and m can be taken as direction numbers and this formula becomes
cos θ = | n·m | .
Using the absolute value insures that the formula gives the acute angle between the planes and not the obtuse angle. Depending on the direction of the vectors n and m, the dot product could be negative and the angle be the obtuse angle.
