0.7 Projection onto a Line

A force pushing a sled up a slope does two things at once: part of it moves the sled along the slope, and part of it presses the sled into the slope. Splitting the force into those two parts — one along a direction, one across it — is the operation this section builds. Given a vector $\mathbf{v}$ and a nonzero direction $\mathbf{d}$, we want $\mathbf{v} = \mathbf{v}_{\parallel} + \mathbf{v}_{\perp}$ with $\mathbf{v}_{\parallel}$ a multiple of $\mathbf{d}$ and $\mathbf{v}_{\perp}$ orthogonal to $\mathbf{d}$. The part along $\mathbf{d}$ is called the projection of $\mathbf{v}$ onto $\mathbf{d}$; its tip is the foot of the perpendicular dropped from the tip of $\mathbf{v}$ onto the line through $\mathbf{d}$.

The figure lives in the plane, where a right angle is a right angle from every viewpoint. The direction $\mathbf{d} = (2, 1)$ is fixed, the line of its multiples is dashed, and $\mathbf{v} = (a, b)$ is on two sliders.

Interactive figureThe projection of v onto the line through dDrag the v, x-component and v, y-component sliders.

▶ Explore interactively

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Explore the figure

1. At the starting values $\mathbf{v} = (1, 3)$, the blue projection is exactly $\mathbf{d} = (2, 1)$, and the green perpendicular part runs from $(2, 1)$ to $(1, 3)$: head minus tail gives $(-1, 2)$. Check the right angle by the last section's test: $$(-1, 2) \cdot (2, 1) = -2 + 2 = 0$$.
2. Drag $b$ to $-0.5$. The red vector now points below the line, and the projection shrinks to $(0.6, 0.3)$ — three tenths of $\mathbf{d}$. The perpendicular part $(0.4, -0.8)$ drops from the line downward.
3. Set $a = -1$ and $b = 2$. The red vector is perpendicular to $\mathbf{d}$ and the blue arrow vanishes: the projection is the zero vector, and the perpendicular part is all of $\mathbf{v}$.
4. Set $a = 4$ and $b = 2$, so that $\mathbf{v} = 2\mathbf{d}$. Now the green arrow vanishes instead: a vector already on the line is its own projection.
5. Sweep $a$ across its range with $b$ fixed and watch the blue foot slide along the line. Whatever the red tip does, the green arrow meets the dashed line at a right angle. That right angle is the definition, and the formula below is what it forces.

Solving for the scalar

The parallel part is some multiple $k\mathbf{d}$; the whole problem is finding $k$. Dot both sides of $\mathbf{v} = k\mathbf{d} + \mathbf{v}_{\perp}$ with $\mathbf{d}$:

$$\mathbf{v} \cdot \mathbf{d} = k \, (\mathbf{d} \cdot \mathbf{d}) + \mathbf{v}_{\perp} \cdot \mathbf{d} = k \, |\mathbf{d}|^2 + 0$$

The perpendicular part contributed nothing, because it is orthogonal to $\mathbf{d}$ — that is the only thing we asked of it, and it is enough. So $k = \mathbf{v} \cdot \mathbf{d} / |\mathbf{d}|^2$, and

$$\operatorname{proj}_{\mathbf{d}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{d}}{\mathbf{d} \cdot \mathbf{d}} \, \mathbf{d}, \qquad \mathbf{v}_{\perp} = \mathbf{v} - \operatorname{proj}_{\mathbf{d}} \mathbf{v}$$

At the figure's starting values, $\mathbf{v} \cdot \mathbf{d} = 2 + 3 = 5$ and $\mathbf{d} \cdot \mathbf{d} = 5$, so $k = 1$ and the projection is $\mathbf{d}$ itself, as step 1 found. Notice that replacing $\mathbf{d}$ by $2\mathbf{d}$ multiplies the numerator by $2$ and the denominator by $4$, and the extra factor of $\mathbf{d}$ restores the balance: the projection depends on the line, not on which vector along it was chosen to name the direction.

For the fixed direction $(2, 1)$ the formula is a recipe with two inputs, $a$ and $b$, and two outputs. Writing it out, $k = (2a + b)/5$ and the projection is $$\big( 2(2a + b)/5, \; (2a + b)/5 \big) = \big( (4a + 2b)/5, \; (2a + b)/5 \big)$$ — each output a fixed combination of the inputs. That is exactly what a matrix does, and Chapter 3 will call the array of coefficients, with rows $(4/5, \; 2/5)$ and $(2/5, \; 1/5)$, the projection matrix. It appears below as a machine check on step 1.

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.