Parallel and Orthogonal Projections
Senior 2 · Mathematics
Parallel and Orthogonal Projections
Lesson 3 of 4
20 min
Rwanda lies just south of the equator, so twice a year the midday sun passes almost directly overhead and shadows fall straight down. A projection whose rays meet the line (or the ground) at a right angle is called an orthogonal projection. A floor plan of a house is the orthogonal projection of the building onto the ground, and the orthogonal projection of a point onto a line also gives the shortest distance from the point to the line.
An orthogonal projection onto a line l is a parallel projection whose direction is perpendicular to l. The image P′ of a point P is the foot of the perpendicular from P to l.
Because it is a parallel projection, it keeps everything from the last lesson: points on l, order, midpoints and ratios along a segment.
Worked example 1
The orthogonal projections of A(3, 5) onto the x-axis and the y-axis are (3, 0) and (0, 5).
Project A(x1, y1) and B(x2, y2) onto the x-axis. The image A′B′ has length |x2 − x1|, the horizontal component of AB. The vertical component is the height the segment rises. By Pythagoras' theorem:
AB² = A′B′² + (rise)²
So the image is never longer than the segment: A′B′ ≤ AB. The two are equal only when AB is parallel to l, and A′B′ = 0 (a single point) when AB is perpendicular to l.
Figure 3. Orthogonal projection onto the x-axis: the segment from A(1, 1) to B(5, 4) has length 5, and its image A′B′ has length 4. Diagram: Ganzaa.
Worked example 2
A(1, 2) and B(7, 10). The image of AB on the x-axis runs from 1 to 7: length 6. The image on the y-axis runs from 2 to 10: length 8. And AB = √(6² + 8²) = 10, longer than both images.
Worked example 3
One face of a classroom roof slopes 2.6 m from the ridge down to the edge, and the edge is 1 m lower than the ridge. How wide is this face on the building's floor plan?
The plan shows the orthogonal projection onto the ground. Width² = 2.6² − 1² = 6.76 − 1 = 5.76, so the plan width is √5.76 = 2.4 m.
Of all the segments from P to points of l, the perpendicular PP′ is the shortest. So the distance from P to l is PP′.
Worked example 4
P is (4, −3). Its distance from the x-axis is 3 (to P′(4, 0)); from the line x = 1 it is 4 − 1 = 3 (to (1, −3)); from the line y = 2 it is 2 − ( − 3) = 5 (to (4, 2)).
Tip
"Keep x, make y zero." That is the whole rule for projecting onto the x-axis. Then sanity-check lengths: an orthogonal image can never be longer than the original segment. If yours is, you have mixed up the segment and its projection.
Find the orthogonal projections of P(−4, 7) onto the x-axis and onto the y-axis.
Solution: (−4, 0) and (0, 7).
A(−2, 5) and B(4, −3). Find the lengths of the orthogonal projections of AB onto each axis, and the length AB.
Solution: onto the x-axis: 4 − ( − 2) = 6; onto the y-axis: 5 − ( − 3) = 8; AB = √(36 + 64) = 10.
A ladder 13 m long leans against a wall. Its orthogonal projection on the ground is 5 m long. How high up the wall does it reach?
Solution: height² = 13² − 5² = 169 − 25 = 144, so the height is 12 m.
Project P(2, 9) orthogonally onto the line y = 4, and find the distance from P to that line.
Solution: P′ = (2, 4); the distance is 9 − 4 = 5 units.
A segment is perpendicular to the line of projection. What is its orthogonal projection?
Solution: a single point, because both ends slide along the same perpendicular.
Remember
Sources and credits
Written by Ganzaa.
