Parallel and Orthogonal Projections
Senior 2 · Mathematics
Parallel and Orthogonal Projections
Lesson 2 of 4
20 min
Lean a ladder against a wall in the afternoon sun. Its shadow on the ground is not as long as the ladder, so lengths are not kept. Yet the shadow of the middle rung falls exactly in the middle of the shadow. Some things are destroyed by a projection and some are kept, and knowing which is which lets you calculate lengths you cannot measure directly.
Why? The projecting lines AA′, CC′ and BB′ are parallel, and they cut the two lines AB and A′B′. By Thales' theorem (Unit 5), parallel lines cut any two transversals in proportional segments: AC : CB = A′C′ : C′B′.
Figure 2. Along d = (1, 2), A(0, 2), M(3, 4) and B(6, 6) project to A′(−1, 0), M′(1, 0) and B′(3, 0). M is the midpoint of AB and M′ is the midpoint of A′B′. Diagram: Ganzaa.
Worked example 1
(midpoints). In Figure 2, d = 12, so a ÷ b = 0.5. A(0, 2) goes to (0 − 1, 0) = ( − 1, 0); B(6, 6) goes to (6 − 3, 0) = (3, 0); and the midpoint M(3, 4) of AB goes to (3 − 2, 0) = (1, 0). The midpoint of A′B′ is (( − 1 + 3) ÷ 2, 0) = (1, 0), which is M′. The property holds.
Worked example 2
(ratios). AB = 10 cm and C lies on AB with AC = 4 cm. Under a parallel projection, A′B′ = 15 cm. Find A′C′.
The ratio is kept: A′C′ ÷ A′B′ = AC ÷ AB = 4 ÷ 10. So A′C′ = 0.4 × 15 = 6 cm (and C′B′ = 9 cm).
To project a polygon onto a line, project every vertex. The image of the whole figure is the segment from the image furthest to one side to the image furthest to the other side.
Worked example 3
Project triangle P(1, 3), Q(4, 1), R(6, 5) onto the x-axis along d = 11.
P′ = (1 − 3, 0) = ( − 2, 0); Q′ = (4 − 1, 0) = (3, 0); R′ = (6 − 5, 0) = (1, 0). R′ lies between the others, so the image of the triangle is the segment from (−2, 0) to (3, 0), 5 units long.
Tip
Projections respect ratios, not lengths. When a question gives you one length before projection and asks for a length after, look for a ratio along the same segment (or along parallel segments) and scale it.
The images of A and B are A′(−2, 0) and B′(8, 0). Where is the image of the midpoint of AB?
Solution: at the midpoint of A′B′: (( − 2 + 8) ÷ 2, 0) = (3, 0).
AB = 12 cm and C is on AB with AC : CB = 1 : 2. The image A′B′ is 9 cm long. Find A′C′ and C′B′.
Solution: the ratio 1 : 2 is kept, so A′C′ = 9 ÷ 3 = 3 cm and C′B′ = 6 cm.
Project A(1, 2) and B(5, 4) onto the x-axis along d = 21.
Solution: a ÷ b = 2. A′ = (1 − 4, 0) = ( − 3, 0) and B′ = (5 − 8, 0) = ( − 3, 0). Both land on the same point, because AB = 42 = 2d is parallel to d.
Project the square with vertices (0, 0), (2, 0), (2, 2) and (0, 2) onto the x-axis along d = 11. How long is the image?
Solution: the images are (0, 0), (2, 0), (0, 0) and (−2, 0). The image runs from −2 to 2, so it is 4 units long, although each side of the square is only 2.
True or false: a parallel projection keeps angles.
Solution: false. In question 4 the square's right angles are flattened onto a straight line.
Remember
Sources and credits
The Rwandan examples, practice items and exercises are written by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.
