# The dot product: projection, angle and why the formula works

> Multiply matching coordinates and add. The result measures how far one arrow reaches along another, which gives angles, projections and Cauchy-Schwarz.

Series: Linear Algebra (https://www.monocurl.com/monocurl-essays/linear-algebra/)
Canonical: https://www.monocurl.com/monocurl-essays/linear-algebra/dot-product/
Keywords: dot product, dot product meaning, dot product and angle, a dot b = |a||b| cos theta, vector projection formula, Cauchy-Schwarz inequality, perpendicular vectors, scalar product

[Figure 1, animation: The arrows \mathbf{a} = (3, 1) (blue) and \mathbf{b} = (1, 2) (orange) are turned together through a quarter turn; the faint arrows mark where they started. All four coordinates change, and so do the two products a_1b_1 and a_2b_2, but their sum stays at 5 the whole way.]

Take two arrows from the origin, $\mathbf{a} = (3, 1)$ and $\mathbf{b} = (1, 2)$. Multiply their first coordinates, multiply their second coordinates, and add: $3 \cdot 1 + 1 \cdot 2 = 5$. Now turn the whole picture a quarter turn counterclockwise. Both arrows get new coordinates, $\mathbf{a}$ becomes $(-1, 3)$ and $\mathbf{b}$ becomes $(-2, 1)$, and the recipe gives $(-1)(-2) + 3 \cdot 1 = 5$ again. The animation makes the turn gradually. The two products change the entire time, and their sum does not move.

That needs an explanation. Coordinates depend on where we happened to put the axes, so a recipe built out of coordinates has no obvious right to ignore them. This one behaves as if it responded only to the arrows themselves: how long they are and how they sit relative to each other. Our job is to find out what the 5 is measuring, and then to prove it.

I assume you know that a [vector](@linear-algebra/vectors) $\mathbf{v} = (v_1, v_2)$ is an arrow from the origin to the point $(v_1, v_2)$, that vectors add coordinate by coordinate and scale by multiplying each coordinate, and that the length of $\mathbf{v}$ is $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2}$ by the [Pythagorean theorem](@geometry/pythagorean-theorem). The basis arrows are $\mathbf{i} = (1, 0)$ and $\mathbf{j} = (0, 1)$, and $(v_1, v_2) = v_1\mathbf{i} + v_2\mathbf{j}$.

## The definition and its algebra

**Definition.** The **dot product** of $\mathbf{a} = (a_1, a_2)$ and $\mathbf{b} = (b_1, b_2)$ is the number

$$
\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2.
$$

For vectors with $n$ coordinates, $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + \dots + a_nb_n$.

The inputs are two vectors and the output is a single number. Before asking what the number means, it pays to learn how it behaves.

**Pause and try.** With $\mathbf{a} = (3, 1)$ and $\mathbf{b} = (1, 2)$: (a) A quarter turn clockwise sends $(x, y)$ to $(y, -x)$. Turn both arrows this way and compute the dot product. (b) Compute $(2\mathbf{a}) \cdot \mathbf{b}$. (c) Compute $\mathbf{a} \cdot \mathbf{a}$ and compare it with the length of $\mathbf{a}$.

**Answer.** (a) The arrows become $(1, -3)$ and $(2, -1)$, and $1 \cdot 2 + (-3)(-1) = 5$, unchanged again. (b) $2\mathbf{a} = (6, 2)$, so $(2\mathbf{a}) \cdot \mathbf{b} = 6 + 4 = 10$: doubling one arrow doubles the result. (c) $\mathbf{a} \cdot \mathbf{a} = 9 + 1 = 10$, and $|\mathbf{a}| = \sqrt{10}$. The dot product of a vector with itself is its length squared.

Parts (b) and (c) are instances of general rules, and all of them come straight from the definition.

**Theorem.** **(Rules of the dot product.)** For all vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ with the same number of coordinates and every number $k$:

- $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$ (symmetry);
- $(\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}$;
- $(k\mathbf{a}) \cdot \mathbf{b} = k\,(\mathbf{a} \cdot \mathbf{b})$;
- $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$, which is positive unless $\mathbf{a} = \mathbf{0}$.

**Proof.** I write the proof for two coordinates; with $n$ coordinates every line has $n$ terms of the same form. (1) $a_1b_1 + a_2b_2 = b_1a_1 + b_2a_2$ because numbers multiply in either order. (2) $(a_1 + b_1)c_1 + (a_2 + b_2)c_2 = (a_1c_1 + a_2c_2) + (b_1c_1 + b_2c_2)$ by the distributive law for numbers. (3) $(ka_1)b_1 + (ka_2)b_2 = k(a_1b_1 + a_2b_2)$. (4) $\mathbf{a} \cdot \mathbf{a} = a_1^2 + a_2^2$, which is $|\mathbf{a}|^2$ by the definition of length. A sum of squares is zero only when every term is zero.

Rules 2 and 3 are stated for the first vector, and rule 1 lets us swap the two, so they hold for the second vector as well: $\mathbf{c} \cdot (\mathbf{a} + \mathbf{b}) = \mathbf{c} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{b}$ and $\mathbf{a} \cdot (k\mathbf{b}) = k\,(\mathbf{a} \cdot \mathbf{b})$. A product that is linear in each of its two inputs separately is called **bilinear**. In practice it means that brackets expand the way they do for numbers. One expansion will carry the main proof of this lesson:

$$
(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} - \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} = |\mathbf{a}|^2 - 2\,\mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2.
$$

## The easy case: one arrow lies along an axis

The meaning of the number is easiest to see when $\mathbf{b}$ points along the $x$ axis. Let $\mathbf{b} = (4, 0)$. Then $\mathbf{a} \cdot \mathbf{b} = 4a_1 + 0 \cdot a_2 = 4a_1$: the second coordinate of $\mathbf{a}$ is ignored, and the first is multiplied by the length of $\mathbf{b}$.

Physics has a use for exactly this. Suppose a cart rolls 4 meters east along a straight track while you pull it with a force $\mathbf{a}$, measured in newtons, aimed off to one side. Only the eastward part of your pull, $a_1$, moves the cart along the track, and the work you do is that part times the distance: $4a_1$ joules. Pull straight north and you do no work on the cart. Pull partly west and your work is negative, because you are holding it back.

Geometrically, $a_1$ is the shadow of $\mathbf{a}$ on the line of $\mathbf{b}$, if the light shines straight down onto that line. It is a signed length: positive when the shadow falls on the side $\mathbf{b}$ points to, negative on the other side.

**Pause and try.** Let $\mathbf{b} = (4, 0)$ and let $\mathbf{a}$ be an arrow of length 3 that you may aim anywhere. For which directions of $\mathbf{a}$ is $\mathbf{a} \cdot \mathbf{b}$ largest, zero, and most negative, and what are those three values?

**Answer.** The dot product is 4 times the $x$ coordinate of $\mathbf{a}$, which ranges from $3$ to $-3$. It is largest, $12$, when $\mathbf{a} = (3, 0)$ points along $\mathbf{b}$. It is $0$ when $\mathbf{a} = (0, 3)$ or $(0, -3)$, at a right angle to $\mathbf{b}$. It is most negative, $-12$, when $\mathbf{a} = (-3, 0)$ points against $\mathbf{b}$.

[Figure 2, animation: The orange arrow \mathbf{b} = (4, 0) is fixed and the blue arrow \mathbf{a}, of length 3, swings from 25^\circ to 155^\circ. The thick band on the axis is the shadow of \mathbf{a} on the line of \mathbf{b}. The dot product is the signed length of the shadow times 4: positive, zero at the right angle, then negative.]

If $\theta$ is the angle from $\mathbf{b}$ up to $\mathbf{a}$, the shadow has signed length $|\mathbf{a}|\cos\theta$. That is what the cosine is: an arrow of length $L$ at angle $\theta$ from the positive $x$ axis has $x$ coordinate $L\cos\theta$. So in this special position

$$
\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\cos\theta \cdot |\mathbf{b}|.
$$

Now recall the opening experiment, where turning both arrows together left the dot product alone. If that is true in general, we could turn any pair of arrows until $\mathbf{b}$ lies along the positive $x$ axis, and the formula above would hold for every pair. So we have a conjecture: $\mathbf{a} \cdot \mathbf{b}$ is always the signed shadow of $\mathbf{a}$ on $\mathbf{b}$, times the length of $\mathbf{b}$. So far it rests on one numerical experiment. I want to do two things with it: see directly why a sum of coordinate products should compute a shadow on a tilted line, and then prove the statement.

## Why the formula sees a shadow on a tilted line

Take a line through the origin that is not an axis, the one along $\mathbf{u} = (0.8, 0.6)$. I chose $\mathbf{u}$ to have length 1, since $0.8^2 + 0.6^2 = 1$, so that $\mathbf{u}$ only names a direction. How long is the shadow of $\mathbf{a} = (3, 1)$ on this line? The conjecture says $\mathbf{a} \cdot \mathbf{u} = 3 \cdot 0.8 + 1 \cdot 0.6 = 3$. Step through the figure to see where each term comes from.

[Figure 3, steps: The shadow of \mathbf{a} = (3, 1) on the line along the unit vector \mathbf{u} = (0.8, 0.6), found by walking to the tip of \mathbf{a} in two legs. The leg 3\mathbf{i} casts a shadow of length 3 \times 0.8, the leg 1\mathbf{j} continues it by 1 \times 0.6, and the total is 3.]

  - step: Drop a perpendicular from the tip of a (blue) to the line along u (orange). The shadow runs from the origin to its foot.
  - step: Walk to the tip of a in two legs: 3 steps of i, then 1 step of j.
  - step: One step of i casts a shadow of length 0.8 on the line, so the first leg casts 3 x 0.8 = 2.4.
  - step: One step of j casts a shadow of length 0.6, and the second leg's shadow starts where the first one ended.
  - step: The two shadows together are the shadow of a: 2.4 + 0.6 = 3.0, which is 3 x 0.8 + 1 x 0.6.
Two facts about shadows are doing the work. The first is that shadows add. Walk from the origin to the tip of $\mathbf{a}$ along $3\mathbf{i}$ and then along $1\mathbf{j}$, and your shadow on the line moves from the origin to the foot of the perpendicular from $\mathbf{a}$, however you got there. So the shadow of $\mathbf{a}$ is the shadow of the first leg plus the shadow of the second, and a leg three times as long casts a shadow three times as long. Casting a shadow on a fixed line is a linear operation.

The second fact is about the basis arrows. How long is the shadow of $\mathbf{i}$ on the line along $\mathbf{u}$? Both are arrows of length 1, and the picture of the two of them is symmetric under the reflection that swaps them. So the shadow of $\mathbf{i}$ on $\mathbf{u}$ is as long as the shadow of $\mathbf{u}$ on $\mathbf{i}$. But the shadow of $\mathbf{u}$ on the $x$ axis is the first coordinate of $\mathbf{u}$, which is $0.8$. In the same way the shadow of $\mathbf{j}$ on $\mathbf{u}$ is $u_2 = 0.6$.

Put the two together: the shadow of $a_1\mathbf{i} + a_2\mathbf{j}$ on the line along a unit vector $\mathbf{u}$ is $a_1u_1 + a_2u_2$. The coordinates of $\mathbf{u}$ appear in the formula because they are the shadows of the basis arrows. For a vector $\mathbf{b}$ of any length in the direction of $\mathbf{u}$, we have $\mathbf{b} = |\mathbf{b}|\,\mathbf{u}$, and rule 3 gives $\mathbf{a} \cdot \mathbf{b} = |\mathbf{b}|\,(\mathbf{a} \cdot \mathbf{u})$: the shadow times the length of $\mathbf{b}$, as conjectured.

This is an explanation, and I would not yet call it a proof, because "shadows add" was read off a picture in which everything had convenient signs. The proof below takes a different road and leans on one theorem from trigonometry.

## The theorem: lengths and the angle

First we need to say what the angle between two vectors is. For nonzero $\mathbf{a}$ and $\mathbf{b}$, the **angle between them** is the angle $\theta$ at the origin in the triangle with corners $\mathbf{0}$, $\mathbf{a}$ and $\mathbf{b}$, so $0^\circ \le \theta \le 180^\circ$. It is $0^\circ$ when $\mathbf{b}$ is a positive multiple of $\mathbf{a}$ and $180^\circ$ when it is a negative multiple; in those two cases the triangle is flat.

[Figure 4, image: The triangle with corners at the origin, \mathbf{a} = (3, 1) and \mathbf{b} = (1, 2). Its third side is the vector \mathbf{a} - \mathbf{b} = (2, -1), drawn from the tip of \mathbf{b} to the tip of \mathbf{a}, and it lies opposite the angle \theta between the two arrows.]

The theorem we borrow is the [law of cosines](@trigonometry/law-of-sines-and-cosines): in a triangle with sides of lengths $p$, $q$, $r$, where the angle between the sides $p$ and $q$ is $\theta$,

$$
r^2 = p^2 + q^2 - 2pq\cos\theta.
$$

It is the Pythagorean theorem with a correction term for triangles whose angle is not $90^\circ$, and the linked lesson proves it by dropping an altitude and applying Pythagoras twice.

**Theorem.** Let $\mathbf{a}$ and $\mathbf{b}$ be nonzero vectors in the plane and let $\theta$ be the angle between them. Then

$$
\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta.
$$

**Proof.** In the triangle with corners $\mathbf{0}$, $\mathbf{a}$, $\mathbf{b}$, the sides at the origin have lengths $|\mathbf{a}|$ and $|\mathbf{b}|$ and enclose the angle $\theta$. The third side joins the tips, so its length is $|\mathbf{a} - \mathbf{b}|$. The law of cosines says

$$
|\mathbf{a} - \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2\,|\mathbf{a}|\,|\mathbf{b}|\cos\theta.
$$

On the other hand, rule 4 and the expansion from the first section give

$$
|\mathbf{a} - \mathbf{b}|^2 = (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2\,\mathbf{a} \cdot \mathbf{b}.
$$

The two right-hand sides are equal. Cancel $|\mathbf{a}|^2 + |\mathbf{b}|^2$ and divide by $-2$ to get $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta$.

If the triangle is flat, the law of cosines is still a true equation, which we check directly. For $\theta = 0^\circ$ the tips are $\bigl||\mathbf{a}| - |\mathbf{b}|\bigr|$ apart, and $(|\mathbf{a}| - |\mathbf{b}|)^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|$. For $\theta = 180^\circ$ they are $|\mathbf{a}| + |\mathbf{b}|$ apart, and $(|\mathbf{a}| + |\mathbf{b}|)^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2|\mathbf{a}||\mathbf{b}|$.

This is a complete proof, given the law of cosines. Its content is one comparison: the squared distance between the tips can be computed from coordinates, where bilinearity produces the term $-2\,\mathbf{a} \cdot \mathbf{b}$, or from the triangle, where trigonometry produces $-2|\mathbf{a}||\mathbf{b}|\cos\theta$.

**Key idea.** The dot product has two faces that are always equal: $a_1b_1 + a_2b_2 = |\mathbf{a}|\,|\mathbf{b}|\cos\theta$. The left side is easy to compute. The right side says what was computed: how far $\mathbf{a}$ reaches along $\mathbf{b}$, times the length of $\mathbf{b}$.

The opening puzzle is now settled. Turning both arrows together changes neither length and does not change the angle between them, so the right side stays put, and the left side has to follow it. The shadow conjecture is settled too: $|\mathbf{a}|\cos\theta$ is the signed shadow of $\mathbf{a}$ on the line of $\mathbf{b}$, and it gets multiplied by $|\mathbf{b}|$.

**Worked example.** **The angle between $\mathbf{a} = (3, 1)$ and $\mathbf{b} = (1, 2)$.** We know three of the four quantities in the theorem and solve for the fourth. The dot product is $5$, and the lengths are $|\mathbf{a}| = \sqrt{10}$ and $|\mathbf{b}| = \sqrt{5}$. So

$$
\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|} = \frac{5}{\sqrt{10}\sqrt{5}} = \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}},
$$

and the angle between $0^\circ$ and $180^\circ$ with that cosine is $\theta = 45^\circ$.

*Check.* The third side of the triangle is $\mathbf{a} - \mathbf{b} = (2, -1)$, of length $\sqrt{5}$. The sides $\sqrt{5}, \sqrt{5}, \sqrt{10}$ satisfy $5 + 5 = 10$, so the triangle has a right angle at the tip of $\mathbf{b}$ and two equal legs. Its other two angles are $45^\circ$ each.

## Perpendicular, acute, obtuse

The sign of $\cos\theta$ sorts angles into three kinds, and lengths are positive, so the sign of the dot product does the same sorting without any trigonometry.

**Theorem.** For nonzero vectors $\mathbf{a}$ and $\mathbf{b}$ in the plane: $\mathbf{a} \cdot \mathbf{b} > 0$ exactly when the angle between them is less than $90^\circ$, $\mathbf{a} \cdot \mathbf{b} < 0$ exactly when it is more than $90^\circ$, and $\mathbf{a} \cdot \mathbf{b} = 0$ exactly when they are perpendicular.

**Proof.** In $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ the factors $|\mathbf{a}|$ and $|\mathbf{b}|$ are positive, so the dot product has the sign of $\cos\theta$. For $0^\circ \le \theta \le 180^\circ$, the cosine is positive below $90^\circ$, zero at $90^\circ$ and negative above.

The zero vector has no direction and makes no angle with anything, but $\mathbf{0} \cdot \mathbf{b} = 0$ for every $\mathbf{b}$. The usual convention is to call two vectors **orthogonal** when their dot product is 0, which counts $\mathbf{0}$ as orthogonal to everything and agrees with "perpendicular" otherwise.

**Pause and try.** Describe all vectors $\mathbf{v} = (x, y)$ that are orthogonal to $\mathbf{a} = (3, 1)$. Where do they lie?

**Answer.** The condition is $3x + y = 0$, so $y = -3x$ and $\mathbf{v} = (x, -3x) = -x\,(-1, 3)$. These are the multiples of $(-1, 3)$: a line through the origin. In general $(-a_2, a_1)$ is orthogonal to $(a_1, a_2)$, since $-a_1a_2 + a_2a_1 = 0$. It is $\mathbf{a}$ turned a quarter turn.

[Figure 5, image: Each whole-number point \mathbf{v} is colored by the sign of \mathbf{a} \cdot \mathbf{v}, where \mathbf{a} = (3, 1) is the blue arrow. The points with dot product 0 lie on the dashed line through the origin perpendicular to \mathbf{a}; the green points are on the side \mathbf{a} points to and the red points are on the other side.]

The dashed line is the set of solutions of $3x + y = 0$. An equation like $3x + y = c$ says "the dot product of $(x, y)$ with $(3, 1)$ equals $c$", and its solutions form a line perpendicular to $(3, 1)$. The coefficients of a linear equation are a vector, and that vector points straight across the line.

## Projection: the shadow as a vector

So far the shadow has been a length. Often we want the shadow itself, as a vector, together with what is left over.

Let us find the shadow of $\mathbf{b} = (1, 2)$ on the line of $\mathbf{a} = (3, 1)$ without any angles. The shadow is some multiple $t\,\mathbf{a}$. What singles out the right $t$ is the perpendicular: the leftover $\mathbf{b} - t\,\mathbf{a}$ must be orthogonal to $\mathbf{a}$. The video slides $t$ and watches the dot product of the leftover with $\mathbf{a}$.

[Figure 6, animation: Looking for the multiple t\,\mathbf{a} of \mathbf{a} = (3, 1) closest to the tip of \mathbf{b} = (1, 2). The purple arrow is the leftover \mathbf{b} - t\,\mathbf{a}. Its dot product with \mathbf{a} is 5 - 10t, which reaches 0 at t = 0.5, and there the leftover is perpendicular to \mathbf{a}.]

The condition can be solved for $t$ with the rules: $(\mathbf{b} - t\,\mathbf{a}) \cdot \mathbf{a} = \mathbf{b} \cdot \mathbf{a} - t\,(\mathbf{a} \cdot \mathbf{a}) = 5 - 10t$, which is zero for $t = \frac12$. The shadow is $\frac12(3, 1) = (1.5, 0.5)$ and the leftover is $(1, 2) - (1.5, 0.5) = (-0.5, 1.5)$. Nothing in that calculation was special to these numbers.

**Theorem.** **(Projection.)** Let $\mathbf{v}$ and $\mathbf{w}$ be vectors with $\mathbf{w} \neq \mathbf{0}$. There is exactly one way to write $\mathbf{v} = \mathbf{p} + \mathbf{r}$ with $\mathbf{p}$ a multiple of $\mathbf{w}$ and $\mathbf{r}$ orthogonal to $\mathbf{w}$, namely

$$
\mathbf{p} = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}}\,\mathbf{w}, \qquad \mathbf{r} = \mathbf{v} - \mathbf{p}.
$$

The vector $\mathbf{p}$ is called the **projection of $\mathbf{v}$ onto $\mathbf{w}$**.

**Proof.** Write $\mathbf{p} = t\,\mathbf{w}$ for an unknown number $t$, so that $\mathbf{r} = \mathbf{v} - t\,\mathbf{w}$. By rules 2 and 3, $\mathbf{r} \cdot \mathbf{w} = \mathbf{v} \cdot \mathbf{w} - t\,(\mathbf{w} \cdot \mathbf{w})$. Since $\mathbf{w} \neq \mathbf{0}$, rule 4 gives $\mathbf{w} \cdot \mathbf{w} > 0$, so this expression is zero for exactly one value of $t$, which is $t = \dfrac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}}$.

The proof used only the four rules, so it holds for vectors with any number of coordinates. The length of the projection agrees with the shadow from before: $|\mathbf{p}| = \dfrac{|\mathbf{v} \cdot \mathbf{w}|}{|\mathbf{w}|^2}\,|\mathbf{w}| = \dfrac{|\mathbf{v} \cdot \mathbf{w}|}{|\mathbf{w}|} = |\mathbf{v}|\,|\cos\theta|$.

**Worked example.** **A crate on a ramp.** A ramp rises 3 m over a horizontal run of 4 m. A crate of mass 10 kg sits on it, so gravity pulls it straight down with a force of about 98 N. How much of that force acts along the ramp, and how much work does gravity do if the crate is pushed from the bottom to the top?

Put the $x$ axis along the ground and the $y$ axis up. The ramp runs along $\mathbf{d} = (4, 3)$, in meters, and gravity is $\mathbf{F} = (0, -98)$, in newtons. We want the projection of $\mathbf{F}$ onto $\mathbf{d}$, so we need two dot products: $\mathbf{F} \cdot \mathbf{d} = 0 \cdot 4 + (-98) \cdot 3 = -294$ and $\mathbf{d} \cdot \mathbf{d} = 16 + 9 = 25$. Then

$$
\mathbf{p} = \frac{-294}{25}(4, 3) = (-47.04, -35.28).
$$

Its length is $\frac{294}{25} \cdot |\mathbf{d}| = 11.76 \times 5 = 58.8$ N, and the minus sign says it points down the ramp. So holding the crate still takes 58.8 N along the ramp, well short of the 98 N needed to lift it. The leftover $\mathbf{r} = \mathbf{F} - \mathbf{p} = (47.04, -62.72)$ presses the crate into the surface.

The work done by gravity over the displacement $\mathbf{d}$ is $\mathbf{F} \cdot \mathbf{d} = -294$ J, negative because the crate moves against the pull.

*Check.* $\mathbf{r} \cdot \mathbf{d} = 47.04 \cdot 4 - 62.72 \cdot 3 = 188.16 - 188.16 = 0$, so the leftover is perpendicular to the ramp. And lifting 10 kg straight up by 3 m costs $98 \times 3 = 294$ J: the ramp lowers the force and leaves the work unchanged.

[Figure 7, image: A crate on a ramp that rises 3 m over 4 m. Gravity (black) pulls straight down with 98 N. Its projection onto the ramp direction (red) has length 58.8 N and points down the slope, and the dashed leftover is perpendicular to the ramp.]

## The Cauchy-Schwarz inequality

A shadow is never longer than the arrow that casts it. Written out with dot products, that plain remark becomes one of the most used inequalities in mathematics.

[Figure 8, animation: The blue arrow \mathbf{a} of length 3 makes a full turn while \mathbf{b} = (2, 1) stays fixed. The green band is the projection of \mathbf{a} onto \mathbf{b}. The left bar is |\mathbf{a} \cdot \mathbf{b}| and the right bar is |\mathbf{a}||\mathbf{b}|: the left never exceeds the right, and they are equal only when the arrows lie on one line.]

**Theorem.** **(Cauchy-Schwarz inequality.)** For all vectors $\mathbf{a}$ and $\mathbf{b}$ with the same number of coordinates,

$$
|\mathbf{a} \cdot \mathbf{b}| \le |\mathbf{a}|\,|\mathbf{b}|,
$$

with equality exactly when one of the vectors is a multiple of the other.

**Proof.** If $\mathbf{b} = \mathbf{0}$, both sides are 0 and $\mathbf{b} = 0\,\mathbf{a}$ is a multiple of $\mathbf{a}$. Otherwise, split $\mathbf{a} = \mathbf{p} + \mathbf{r}$ as in the projection theorem, with $\mathbf{p} = t\,\mathbf{b}$, $t = \dfrac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}}$, and $\mathbf{r} \cdot \mathbf{b} = 0$. Then $\mathbf{p} \cdot \mathbf{r} = t\,(\mathbf{b} \cdot \mathbf{r}) = 0$, and expanding,

$$
|\mathbf{a}|^2 = (\mathbf{p} + \mathbf{r}) \cdot (\mathbf{p} + \mathbf{r}) = |\mathbf{p}|^2 + 2\,\mathbf{p} \cdot \mathbf{r} + |\mathbf{r}|^2 = |\mathbf{p}|^2 + |\mathbf{r}|^2 \ge |\mathbf{p}|^2.
$$

Now $|\mathbf{p}|^2 = t^2\,|\mathbf{b}|^2 = \dfrac{(\mathbf{a} \cdot \mathbf{b})^2}{|\mathbf{b}|^4}\,|\mathbf{b}|^2 = \dfrac{(\mathbf{a} \cdot \mathbf{b})^2}{|\mathbf{b}|^2}$. So $(\mathbf{a} \cdot \mathbf{b})^2 \le |\mathbf{a}|^2|\mathbf{b}|^2$, and taking square roots of these nonnegative numbers gives the inequality. Equality holds exactly when $|\mathbf{r}|^2 = 0$, that is, when $\mathbf{r} = \mathbf{0}$ and $\mathbf{a} = t\,\mathbf{b}$.

The middle line of the proof is the Pythagorean theorem for the right triangle with legs $\mathbf{p}$ and $\mathbf{r}$ and hypotenuse $\mathbf{a}$, derived here from bilinearity alone. In the plane the inequality also follows from $|\cos\theta| \le 1$, but the proof above never mentions an angle, and that matters. For vectors with three or more coordinates we have not defined an angle at all. Cauchy-Schwarz says that for nonzero vectors the number $\dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ lies between $-1$ and $1$, so it is the cosine of exactly one angle between $0^\circ$ and $180^\circ$, and that angle is taken as the *definition* of the angle between $\mathbf{a}$ and $\mathbf{b}$ in any dimension.

**Worked example.** **An angle in three dimensions.** Find the angle between $\mathbf{a} = (1, 2, 2)$ and $\mathbf{b} = (2, -1, 2)$.

$\mathbf{a} \cdot \mathbf{b} = 2 - 2 + 4 = 4$, and $|\mathbf{a}| = \sqrt{1 + 4 + 4} = 3$, $|\mathbf{b}| = \sqrt{4 + 1 + 4} = 3$. So $\cos\theta = \frac{4}{9}$ and $\theta \approx 63.6^\circ$. Cauchy-Schwarz holds with room to spare: $4 \le 9$.

## Where the analogy with multiplication breaks

The dot product expands like a product of numbers, which tempts people to treat it like one in other ways.

**Common mistake.** **Cancelling a vector.** From $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c}$ with $\mathbf{a} \neq \mathbf{0}$ it does not follow that $\mathbf{b} = \mathbf{c}$. With $\mathbf{a} = (3, 1)$, both $\mathbf{b} = (1, 2)$ and $\mathbf{c} = (2, -1)$ give 5. What does follow is $\mathbf{a} \cdot (\mathbf{b} - \mathbf{c}) = 0$: the two vectors differ by something orthogonal to $\mathbf{a}$, here $(-1, 3)$. They have the same shadow on $\mathbf{a}$ and may differ sideways as much as they like.

**Common mistake.** **Dividing by the length once in a projection.** A common slip is to write the projection of $\mathbf{v}$ onto $\mathbf{w}$ as $\dfrac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{w}|}\,\mathbf{w}$. Try it on $\mathbf{v} = (1, 2)$, $\mathbf{w} = (3, 1)$: it gives $\frac{5}{\sqrt{10}}(3, 1) \approx (4.74, 1.58)$, an arrow of length 5, longer than $\mathbf{v}$ itself, whose length is $\sqrt5 \approx 2.24$. A shadow cannot be longer than the arrow. The number $\frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{w}|}$ is the *length* of the shadow; to turn it into a vector it must multiply the unit vector $\frac{\mathbf{w}}{|\mathbf{w}|}$, and that brings in the second factor of $|\mathbf{w}|$ in the denominator: $|\mathbf{w}|^2 = \mathbf{w} \cdot \mathbf{w}$.

What was hard here was that the definition and the meaning live in different worlds. The definition is arithmetic on coordinates, the meaning is a shadow, and the bridge between them is a single computation of $|\mathbf{a} - \mathbf{b}|^2$ done two ways. The reusable idea is bilinearity: once a product expands like ordinary multiplication, lengths, angles, projections and the Pythagorean theorem can all be handled by algebra, in any number of dimensions. The dot product returns whenever a [matrix](@linear-algebra/matrices-as-transformations) meets a vector, because each coordinate of $A\mathbf{x}$ is a row of $A$ dotted with $\mathbf{x}$; the lesson on [linear systems](@linear-algebra/linear-systems-and-inverses) reads each equation of $A\mathbf{x} = \mathbf{b}$ that way.

**Check yourself.** Two nonzero vectors satisfy $\mathbf{a} \cdot \mathbf{b} = -6$. What can you conclude?
Answer: The angle between them is more than $90^\circ$ The lengths are positive, so $\cos\theta < 0$, which happens exactly for angles between $90^\circ$ and $180^\circ$.

**Recap.** - $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2$ is a number. It is symmetric, linear in each vector, and $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$.
- $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$: the signed shadow of one vector on the other, times the other's length. The proof compares two computations of $|\mathbf{a} - \mathbf{b}|^2$, one by bilinearity and one by the law of cosines.
- The dot product is positive for acute angles, negative for obtuse ones, and zero exactly for orthogonal vectors.
- The projection of $\mathbf{v}$ onto $\mathbf{w} \neq \mathbf{0}$ is $\dfrac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}}\,\mathbf{w}$, and what is left of $\mathbf{v}$ is orthogonal to $\mathbf{w}$.
- Cauchy-Schwarz: $|\mathbf{a} \cdot \mathbf{b}| \le |\mathbf{a}||\mathbf{b}|$, with equality only when one vector is a multiple of the other. For nonzero vectors it makes $\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ a valid definition of angle in any dimension.

## Problems

The first three are computations, the next three use the rules to explain something, and the last two are proofs.

**Problem 1.** Compute: (a) $(2, -1) \cdot (3, 4)$; (b) $(1, 2, 3) \cdot (4, -5, 6)$; (c) the length of $(2, -1)$. (d) Which pairs among $(2, -1)$, $(1, 2)$ and $(-4, 2)$ are orthogonal?

**Solution.** (a) $6 - 4 = 2$. (b) $4 - 10 + 18 = 12$. (c) $\sqrt{4 + 1} = \sqrt5$. (d) $(2, -1) \cdot (1, 2) = 2 - 2 = 0$ and $(1, 2) \cdot (-4, 2) = -4 + 4 = 0$, so those two pairs are orthogonal. $(2, -1) \cdot (-4, 2) = -8 - 2 = -10$, which is not zero. In fact $(-4, 2) = -2\,(2, -1)$, so these two point in opposite directions.

**Problem 2.** Find the angle between $(3, 4)$ and $(5, 12)$ to the nearest tenth of a degree.

**Solution.** The dot product is $15 + 48 = 63$. The lengths are $\sqrt{9 + 16} = 5$ and $\sqrt{25 + 144} = 13$. So $\cos\theta = \frac{63}{65} \approx 0.9692$ and $\theta \approx 14.3^\circ$. Check with directions: $(3, 4)$ is at $53.13^\circ$ from the $x$ axis and $(5, 12)$ at $67.38^\circ$, which differ by $14.25^\circ$.

**Problem 3.** Find the projection of $\mathbf{v} = (4, 7)$ onto $\mathbf{w} = (3, 4)$ and the leftover $\mathbf{r}$. Use them to find the distance from the point $(4, 7)$ to the line through the origin along $(3, 4)$.

**Solution.** $\mathbf{v} \cdot \mathbf{w} = 12 + 28 = 40$ and $\mathbf{w} \cdot \mathbf{w} = 25$, so $\mathbf{p} = \frac{40}{25}(3, 4) = (4.8, 6.4)$ and $\mathbf{r} = (4, 7) - (4.8, 6.4) = (-0.8, 0.6)$. Check: $\mathbf{r} \cdot \mathbf{w} = -2.4 + 2.4 = 0$. The point $\mathbf{p}$ is the foot of the perpendicular from $(4, 7)$ to the line, so the distance is $|\mathbf{r}| = \sqrt{0.64 + 0.36} = 1$.

**Problem 4.** For which $k$ are $(k, 2)$ and $(3, k - 5)$ orthogonal?

**Solution.** The dot product is $3k + 2(k - 5) = 5k - 10$, which is zero exactly for $k = 2$. Check: $(2, 2) \cdot (3, -3) = 6 - 6 = 0$.

**Problem 5.** A sled is pulled 20 m across level snow by a rope that makes an angle of $30^\circ$ with the ground. The tension in the rope is 50 N. How much work does the rope do on the sled? Answer once with the angle formula and once with coordinates.

**Solution.** With the angle: $W = |\mathbf{F}||\mathbf{d}|\cos\theta = 50 \times 20 \times \cos 30^\circ = 1000 \times 0.8660 \approx 866$ J. With coordinates: put the $x$ axis along the motion, so $\mathbf{d} = (20, 0)$ and $\mathbf{F} = (50\cos 30^\circ, 50\sin 30^\circ) \approx (43.30, 25)$. Then $\mathbf{F} \cdot \mathbf{d} \approx 43.30 \times 20 + 25 \times 0 = 866$ J. The upward 25 N of the pull does no work, because the sled does not move upward.

**Problem 6.** Let $\mathbf{a} = (3, 1)$. Describe the set of all $\mathbf{v} = (x, y)$ with $\mathbf{a} \cdot \mathbf{v} = 10$. Which point of that set is closest to the origin, and how far away is it?

**Solution.** The condition is $3x + y = 10$, a line. One point on it is $\mathbf{a}$ itself, since $\mathbf{a} \cdot \mathbf{a} = 10$. If $\mathbf{v}$ is any point on it, then $\mathbf{a} \cdot (\mathbf{v} - \mathbf{a}) = 10 - 10 = 0$, so $\mathbf{v} = \mathbf{a} + \mathbf{r}$ with $\mathbf{r}$ orthogonal to $\mathbf{a}$: the line passes through the tip of $\mathbf{a}$ and is perpendicular to $\mathbf{a}$. For such a point, $|\mathbf{v}|^2 = |\mathbf{a}|^2 + 2\,\mathbf{a} \cdot \mathbf{r} + |\mathbf{r}|^2 = 10 + |\mathbf{r}|^2$, which is smallest when $\mathbf{r} = \mathbf{0}$. So the closest point is $(3, 1)$, at distance $\sqrt{10} \approx 3.16$.

**Problem 7.** Prove that the diagonals of a parallelogram are perpendicular if and only if all four sides have the same length (the parallelogram is a rhombus).

**Solution.** Let the sides from one corner be the vectors $\mathbf{a}$ and $\mathbf{b}$. The diagonals are $\mathbf{a} + \mathbf{b}$ and $\mathbf{a} - \mathbf{b}$. Expanding by bilinearity and using symmetry,

$$
(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} = |\mathbf{a}|^2 - |\mathbf{b}|^2.
$$

The diagonals are perpendicular exactly when this is 0, that is, when $|\mathbf{a}| = |\mathbf{b}|$. Opposite sides of a parallelogram are equal, so $|\mathbf{a}| = |\mathbf{b}|$ says all four sides are equal. (A parallelogram with nonzero area has nonzero diagonals, so "dot product zero" does mean perpendicular here.)

**Problem 8.** Prove the triangle inequality $|\mathbf{a} + \mathbf{b}| \le |\mathbf{a}| + |\mathbf{b}|$ for vectors with any number of coordinates. When does equality hold?

**Solution.** Both sides are nonnegative, so it is enough to compare their squares. By bilinearity, $|\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + 2\,\mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2$. By Cauchy-Schwarz, $\mathbf{a} \cdot \mathbf{b} \le |\mathbf{a} \cdot \mathbf{b}| \le |\mathbf{a}||\mathbf{b}|$, so

$$
|\mathbf{a} + \mathbf{b}|^2 \le |\mathbf{a}|^2 + 2|\mathbf{a}||\mathbf{b}| + |\mathbf{b}|^2 = (|\mathbf{a}| + |\mathbf{b}|)^2.
$$

Equality needs $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|$. By the equality case of Cauchy-Schwarz one vector is a multiple of the other, say $\mathbf{a} = t\,\mathbf{b}$ (or $\mathbf{b} = \mathbf{0}$), and then $\mathbf{a} \cdot \mathbf{b} = t|\mathbf{b}|^2$ must be nonnegative, so $t \ge 0$. Equality holds exactly when one vector is a nonnegative multiple of the other: the two arrows point the same way and walking along both is a straight path.