Parallel and Orthogonal Projections
Senior 2 · Mathematics
Parallel and Orthogonal Projections
Lesson 1 of 4
20 min
At three o'clock on a sunny afternoon in Nyagatare, every pole, tree and person casts a shadow, and all the shadows point the same way. The sun is so far away that its rays arrive parallel to each other. The shadow of the top of a pole is found by following a ray from the top of the pole until it hits the ground.
That is exactly what a parallel projection does: it sends every point, along one fixed direction, onto a line (or, in three dimensions, onto a surface). Architects use projections to draw plans, map-makers use them to flatten hills onto paper, and computer games use them to draw 3D worlds on a flat screen.
A parallel projection in the plane needs:
To project a point P onto l parallel to d: draw the line through P parallel to d. It meets l at a point P′ (read "P prime"). P′ is the image of P, or the parallel projection of P onto l along d.
With a ruler and a set square: lay one edge of the set square along d, hold a ruler firmly against another edge, then slide the set square along the ruler until its edge passes through P. Draw along that edge to l.
If P already lies on l, then P′ = P: it does not move.
Project both end points. The image of segment AB is the segment A′B′.
Figure 1. Parallel projection onto the x-axis along d = (1, 1): A(1, 2) goes to A′(−1, 0) and B(6, 4) goes to B′(2, 0). Diagram: Ganzaa.
Two special cases:
Take l to be the x-axis and d = ab with b ≠ 0. Moving from P(x, y) in the direction of d, the y-coordinate drops to 0 after t = −y ÷ b steps of d. So the image of P(x, y) is
P′ = (x − (a ÷ b) × y, 0)
Worked example 1
d = 11 and P is (5, 3). Here a ÷ b = 1, so P′ = (5 − 3, 0) = (2, 0). Check: from (5, 3), three steps of −1−1 reach (2, 0).
Worked example 2
d = 2−1 and Q is (1, 4). Here a ÷ b = 2 ÷ ( − 1) = −2, so Q′ = (1 − ( − 2) × 4, 0) = (1 + 8, 0) = (9, 0).
Worked example 3
With d = 11, project the segment from A(1, 2) to B(6, 4). A′ = (1 − 2, 0) = ( − 1, 0) and B′ = (6 − 4, 0) = (2, 0), so A′B′ is 3 units long. But AB = √(5² + 2²) = √29 ≈ 5.39 units. A parallel projection can change lengths.
Tip
Every projecting line is parallel to d, and every image lies ON l. If your image point is not on the line of projection, or your dashed lines are not parallel, something has gone wrong. Use the sliding set square; never draw the parallels by eye.
In questions 1 to 5 the line of projection is the x-axis.
Project P(4, 6) along d = 12.
Solution: a ÷ b = 1 ÷ 2 = 0.5, so P′ = (4 − 0.5 × 6, 0) = (1, 0).
Project P(2, 5) along d = −11.
Solution: a ÷ b = −1, so P′ = (2 − ( − 1) × 5, 0) = (7, 0).
Project P(3, 7) along d = 01.
Solution: a ÷ b = 0, so P′ = (3, 0). The projecting line is vertical: this special case is the next-but-one lesson.
Project the segment from A(1, 1) to B(3, 3) along d = 11.
Solution: A′ = (0, 0) and B′ = (3 − 3, 0) = (0, 0). The image is a single point, because AB = 22 is parallel to d.
Find the length of the image of the segment from A(0, 4) to B(5, 2) along d = 11.
Solution: A′ = (−4, 0) and B′ = (3, 0), so A′B′ = 3 − ( − 4) = 7 units.
Remember
Sources and credits
Written by Ganzaa.
