0.6 The Dot Product and the Angle Between Vectors

Adding vectors and scaling them never leave the vectors' own world: the answer is another vector. The dot product is different. It takes two vectors of $\mathbb{R}^n$ and returns a number, by multiplying matching components and adding the results:

$$\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + \cdots + u_n v_n$$

For $\mathbf{u} = (1, 2, 2)$ and $\mathbf{v} = (2, -1, 2)$ that is $2 - 2 + 4 = 4$. The recipe is symmetric, so $\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$; it is built from products and sums, so it distributes, $\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}$, and scalars pull out, $(k\mathbf{u}) \cdot \mathbf{v} = k(\mathbf{u} \cdot \mathbf{v})$. And a vector dotted with itself is the sum of its squared components, which you have met already:

$$\mathbf{v} \cdot \mathbf{v} = v_1^2 + v_2^2 + \cdots + v_n^2 = |\mathbf{v}|^2$$

So $\mathbf{v} \cdot \mathbf{v}$ is never negative, and it is zero only for the zero vector. Length is a dot product in disguise, and that is the first hint that this number knows geometry.

What it knows is the angle. The figure fixes $\mathbf{u} = (2, 1, 0)$, puts $\mathbf{v} = (a, b, c)$ on three sliders, and fills the angle between them with a sector lying in the plane the two vectors span.

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Explore the figure

1. At the starting values $\mathbf{v} = (1, 3, 0)$, both vectors lie in the floor and the sector is an eighth of a turn: the angle is $45^\circ$. Compute the number: $\mathbf{u} \cdot \mathbf{v} = 2 + 3 + 0 = 5$, while $$|\mathbf{u}| \, |\mathbf{v}| = \sqrt{5} \, \sqrt{10} = \sqrt{50}$$, and $5/\sqrt{50} = 1/\sqrt{2}$, which is $\cos 45^\circ$.
2. Drag $c$ to $2$. The vector $\mathbf{v}$ lifts off the floor, the plane tilts with it, and the sector opens. The dot product is still $5$, because $\mathbf{u}$ has nothing in the third component to multiply against; but $|\mathbf{v}|$ has grown to $\sqrt{14}$, so the cosine has dropped to $5/\sqrt{70} \approx 0.598$ and the angle is about $53^\circ$.
3. Set $a = 1$ and $b = -2$, with any $c$. Now $\mathbf{u} \cdot \mathbf{v} = 2 - 2 + 0 = 0$, and the sector is a quarter turn — a right angle. Sweep $c$ through its whole range and watch the sector stay a quarter turn: every vector $(1, -2, c)$ is perpendicular to $\mathbf{u}$, a whole plane of them.
4. Set $a = -2$, $b = -1$ and $c = 0$, so that $\mathbf{v} = -\mathbf{u}$. The sector is a half disk, $180^\circ$, and the dot product is $-4 - 1 = -5$, the most negative it can be for this length. Then set $a = 2$ and $b = 1$: the two vectors coincide, the sector collapses, and the dot product $\mathbf{u} \cdot \mathbf{u} = 5 = |\mathbf{u}|^2$ is the most positive.

The law of cosines, rewritten

Draw $\mathbf{u}$ and $\mathbf{v}$ from a common tail, with the angle $\theta$ between them, $0 \le \theta \le \pi$. The vector $\mathbf{u} - \mathbf{v}$ from the tip of $\mathbf{v}$ to the tip of $\mathbf{u}$ closes a triangle whose sides have lengths $|\mathbf{u}|$, $|\mathbf{v}|$ and $|\mathbf{u} - \mathbf{v}|$, and the law of cosines from trigonometry says

$$|\mathbf{u} - \mathbf{v}|^2 = |\mathbf{u}|^2 + |\mathbf{v}|^2 - 2 \, |\mathbf{u}| \, |\mathbf{v}| \cos \theta$$

Now expand the left side with the dot product's own rules. Since a squared length is a self dot product, $|\mathbf{u} - \mathbf{v}|^2 = (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v})$, and distributing gives $$\mathbf{u} \cdot \mathbf{u} - 2 \, \mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{v} = |\mathbf{u}|^2 - 2 \, \mathbf{u} \cdot \mathbf{v} + |\mathbf{v}|^2$$. Setting the two expansions equal, the squared lengths cancel and what remains is

$$\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| \, |\mathbf{v}| \cos \theta$$

That is the geometric meaning of the dot product: the product of the two lengths, discounted by the cosine of the angle between them. Read backwards, it computes the angle from components alone, $\cos \theta = \mathbf{u} \cdot \mathbf{v} / (|\mathbf{u}| \, |\mathbf{v}|)$ — and since nothing in the derivation used a picture beyond the triangle, the same formula defines the angle between two vectors of $\mathbb{R}^n$, where no protractor could reach.

Two consequences carry the rest of the book. First, the sign of the dot product is the sign of the cosine: positive means the angle is acute, the vectors lean the same way; negative means obtuse. Second, and most important, the dot product is zero exactly when $\cos \theta = 0$, that is, when $\theta = 90^\circ$. Two vectors with $\mathbf{u} \cdot \mathbf{v} = 0$ are called orthogonal. For nonzero vectors in the plane or in space, orthogonal simply means perpendicular; the word is chosen because the test $\mathbf{u} \cdot \mathbf{v} = 0$ makes sense in $\mathbb{R}^n$, and even for the zero vector, which is orthogonal to everything without having an angle with anything.

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.