Projecting a vector onto a line
Slide 1 was about exact right angles. Now we use one to answer a practical question: which point on a line is closest to a given point b? Picture the sun straight overhead at noon. A slanted stick casts a shadow on the flat ground, and the tip of the shadow sits right below the tip of the stick. That shadow is the projection p of b onto the line. The gap from the stick's tip down to the shadow, e = b − p, hangs straight down, at a right angle to the ground. That one right angle gives a short formula for the best multiple x̂ ("x-hat") of a: x̂ = aᵀb / aᵀa, so p = x̂a. It also makes p the closest point on the line to b.
Lesson 5 · slide 2Projection- The projection p lies on the line, so p = x̂a for some number x̂.
- The error e = b − p is perpendicular to a: aᵀ(b − x̂a) = 0.
- Solving for the number: x̂ = aᵀb / aᵀa.
- If b is on the line, p = b. If b is perpendicular to a, p = 0.
The closest point
Another everyday picture: you stand in a field next to a straight road. To reach the road in the fewest steps, you walk straight at it, at a right angle. The road is the line through a, you are at b, and where you arrive is p. Your path e = b − p is perpendicular to the road.
From the right angle to a formula
- p is on the line, so p = x̂a for some number x̂ ("x-hat": the best choice).
- e = b − x̂a is perpendicular to a, so its dot product with a is zero.
Two dot products and one division, in any dimension.
A worked example
Why p is the closest point
Take any other point q on the line. The path from b to q splits into two pieces: b − q = e + (p − q). First the drop e, then the step p − q along the line. Those two pieces are perpendicular, so by Pythagoras ‖b − q‖² = ‖e‖² + ‖p − q‖². That is bigger than ‖e‖² unless q = p. So q is farther than p. Try it with q = 2a = 42:
Two quick checks
If b is already on the line, the stick lies flat and is its own shadow: e = 0 and p = b. If b is perpendicular to a, the stick stands straight up and its shadow is a single point: aᵀb = 0, so p = 0.
Keep reading with the full course
Finish lesson 5 and you will be able to: fit the best line to noisy sensor data.
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Examples
Example 1 · Onto the diagonal
Project b = 31 onto the line through a = 11.
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