6.1 Projection onto a Plane

Here is a new kind of question. The plane $z = 0.4x + 0.2y$ is a subspace of $\mathbb{R}^3$ — flat, through the origin, closed under the usual operations. The vector $\mathbf{b} = (1, 1, 3)$ is not on it: the plane above the point $(1, 1)$ sits at height $0.4 + 0.2 = 0.6$ , far below $3$ . Chapter 2 would stop there — not reachable — but the useful question is quantitative: which point of the plane is closest to $\mathbf{b}$ ?

The answer is found by dropping a perpendicular from $\mathbf{b}$ onto the plane. The figure shows the whole construction: the dark point $\mathbf{b}$ floating above the plane, the dashed red segment falling from it along the plane's normal direction, and the blue point $(1.8, 1.4, 1)$ where it lands — the orthogonal projection of $\mathbf{b}$ . Two solid segments from the origin trace $\mathbf{b}$ itself (gray) and its projection (blue).

A semi-transparent tilted plane through the origin with a dark point floating above it. A dashed red segment drops from the point straight down onto the plane along the plane's normal direction, ending at a blue point on the plane. Two solid segments from the origin show the original vector and its shadow in the plane. There are no sliders. — The point $\mathbf{b} = (1, 1, 3)$ above the plane $z = 0.4x + 0.2y$ , with the dashed red perpendicular dropping to its projection $(1.8, 1.4, 1)$ , marked in blue. The gray and blue segments from the origin are $\mathbf{b}$ and its projection as vectors.

Explore the figure

Orbit until you view the plane edge-on. The dashed segment meets it at a right angle — not a right angle on your screen from every viewpoint, but a true one in space, visible when the plane degenerates to a line in your view.
A natural first guess for the closest point is "straight down": drop vertically from $\mathbf{b}$ to $(1, 1, 0.6)$ . That vertical trip has length $3 - 0.6 = 2.4$ . The dashed route is visibly shorter and lands at a different point — closest is measured perpendicular to the plane, not perpendicular to the floor.
Confirm by eye that the blue point actually lies in the plane, and that its coordinates are plausible: $0.4(1.8) + 0.2(1.4) = 0.72 + 0.28 = 1$ , exactly its height.
Follow the triangle made by the gray segment, the blue segment, and the dashed drop: $\mathbf{b}$ splits into a piece inside the plane plus a piece perpendicular to it. Nothing overlaps and nothing is missing.

Computing the drop

Rewrite the plane as $0.4x + 0.2y - z = 0$ ; its normal vector is $\mathbf{n} = (0.4, 0.2, -1)$ , and the perpendicular from $\mathbf{b}$ must travel along $\mathbf{n}$ . So the projection is $\mathbf{p} = \mathbf{b} - t\,\mathbf{n}$ for the value of $t$ that lands on the plane, which works out to $t = (\mathbf{b} \cdot \mathbf{n})/(\mathbf{n} \cdot \mathbf{n})$ . Here $\mathbf{b} \cdot \mathbf{n} = 0.4 + 0.2 - 3 = -2.4$ and $\mathbf{n} \cdot \mathbf{n} = 0.16 + 0.04 + 1 = 1.2$ , so $t = -2$ and

$\mathbf{p} = \mathbf{b} + 2\mathbf{n} = (1 + 0.8, \; 1 + 0.4, \; 3 - 2) = (1.8, \; 1.4, \; 1)$

The error vector is $\mathbf{b} - \mathbf{p} = (-0.8, -0.4, 2) = -2\mathbf{n}$ : a pure multiple of the normal, with length $\sqrt{0.64 + 0.16 + 4} = \sqrt{4.8} \approx 2.19$ — beating the vertical guess of $2.4$ , as the figure promised. And perpendicularity is what makes $\mathbf{p}$ provably the winner: for any other point $\mathbf{q}$ of the plane, the segment from $\mathbf{q}$ to $\mathbf{p}$ lies in the plane, at right angles to $\mathbf{b} - \mathbf{p}$ , so Pythagoras gives $|\mathbf{b} - \mathbf{q}|^2 = |\mathbf{b} - \mathbf{p}|^2 + |\mathbf{p} - \mathbf{q}|^2$ , strictly larger. A perpendicular error is not one good option — it is a certificate that no better option exists.

-2

The drop parameter t, solved ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The landing condition $(\mathbf{b} - t\,\mathbf{n}) \cdot \mathbf{n} = 0$ written out with the numbers of this example — $\mathbf{b} \cdot \mathbf{n} = -2.4$ and $\mathbf{n} \cdot \mathbf{n} = 1.2$ — and solved by a computer algebra system: $t = -2$ , the value that drops $\mathbf{b}$ exactly onto $(1.8, 1.4, 1)$ .

Example 1 Certify the projection

Verify that $\mathbf{p} = (1.8, 1.4, 1)$ lies in the plane and that the error $\mathbf{b} - \mathbf{p}$ is perpendicular to it.

In the plane: $0.4(1.8) + 0.2(1.4) = 0.72 + 0.28 = 1$ , which matches the third coordinate. Perpendicular: the plane is spanned by $(1, 0, 0.4)$ and $(0, 1, 0.2)$ — its points with $x = 1, y = 0$ and $x = 0, y = 1$ — so it is enough to check the error against both. With $\mathbf{b} - \mathbf{p} = (-0.8, -0.4, 2)$ : first, $(-0.8)(1) + (-0.4)(0) + (2)(0.4) = -0.8 + 0.8 = 0$ ; second, $(-0.8)(0) + (-0.4)(1) + (2)(0.2) = -0.4 + 0.4 = 0$ . Both vanish, so the error is orthogonal to every combination of the spanning vectors — the entire plane. The decomposition $(1, 1, 3) = (1.8, 1.4, 1) + (-0.8, -0.4, 2)$ splits $\mathbf{b}$ into plane part plus normal part exactly.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.