Math Core

You can add vectors and multiply them by scalars, but there is also a useful way to multiply two vectors together. The dot product turns two vectors into a single number, and that number tells you how closely the vectors point in the same direction. It gives you the angle between vectors, a quick test for perpendicular lines, and the physics formula for work.

Definition

Dot product

The dot product of $\mathbf{u} = \langle u_1, u_2 \rangle$ and $\mathbf{v} = \langle v_1, v_2 \rangle$ is the scalar

\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2.

Multiply matching components, then add.

The result is a number, not a vector. That is why the dot product is sometimes called the scalar product.

Worked example: Computing a dot product

Find $\langle 3, 4 \rangle \cdot \langle -2, 5 \rangle$ .

(3)(-2) + (4)(5) = -6 + 20 = 14.

The dot product follows familiar-looking rules. For vectors $\mathbf{u}$ , $\mathbf{v}$ , $\mathbf{w}$ and a scalar $k$ :

$\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$
$\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}$
$(k\mathbf{u}) \cdot \mathbf{v} = k(\mathbf{u} \cdot \mathbf{v})$
$\mathbf{v} \cdot \mathbf{v} = v_1^2 + v_2^2 = \|\mathbf{v}\|^2$

The last rule links the dot product to length, and it is the key to everything that follows.

The angle between two vectors

Place $\mathbf{u}$ and $\mathbf{v}$ tail to tail. They form two sides of a triangle whose third side is $\mathbf{u} - \mathbf{v}$ . The law of cosines from the first lesson in this unit says

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

Expanding the left side with the dot product rules gives $\|\mathbf{u}\|^2 - 2\,\mathbf{u} \cdot \mathbf{v} + \|\mathbf{v}\|^2$ . Compare the two sides and cancel, and you are left with a beautiful formula.

Dot product and angle

If $\theta$ is the angle between nonzero vectors $\mathbf{u}$ and $\mathbf{v}$ (with $0^\circ \le \theta \le 180^\circ$ ), then

\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta, \qquad\text{so}\qquad \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\,\|\mathbf{v}\|}.

Placed tail to tail, two vectors make an angle θ between 0° and 180°.

Because the lengths are always positive, the sign of the dot product matches the sign of $\cos\theta$ :

$\mathbf{u} \cdot \mathbf{v}$	Angle $\theta$
positive	acute (vectors point roughly the same way)
zero	$90^\circ$ (perpendicular)
negative	obtuse (vectors point roughly opposite ways)

Worked example: Finding the angle

Find the angle between $\mathbf{u} = \langle 2, 1 \rangle$ and $\mathbf{v} = \langle -1, 3 \rangle$ , to the nearest tenth of a degree.

\mathbf{u} \cdot \mathbf{v} = -2 + 3 = 1, \qquad \|\mathbf{u}\| = \sqrt{5}, \qquad \|\mathbf{v}\| = \sqrt{10}.

\cos\theta = \frac{1}{\sqrt{5}\sqrt{10}} = \frac{1}{\sqrt{50}} \approx 0.1414, \qquad \theta \approx 81.9^\circ.

The dot product is small and positive, so the angle is acute but close to $90^\circ$ , which matches.

Orthogonal vectors

Two vectors are orthogonal when the angle between them is $90^\circ$ . Since $\cos 90^\circ = 0$ , this happens exactly when the dot product is zero.

Orthogonality test

Nonzero vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if and only if $\mathbf{u} \cdot \mathbf{v} = 0$ .

This test needs no angles and no square roots, which makes it a fast way to check for right angles in coordinate geometry.

Worked example: Making vectors orthogonal

Find $k$ so that $\langle k, 6 \rangle$ is orthogonal to $\langle 3, -2 \rangle$ .

Set the dot product to zero: $3k + 6(-2) = 0$ , so $3k = 12$ and $k = 4$ . Check: $\langle 4, 6 \rangle \cdot \langle 3, -2 \rangle = 12 - 12 = 0$ .

Common mistake

The dot product of two vectors is a number, so you cannot "dot" three vectors in a row, and $\mathbf{u} \cdot \mathbf{v} = 0$ does not mean either vector is zero. Also, don't confuse the dot product with component-by-component multiplication: $\langle 3, 4 \rangle \cdot \langle -2, 5 \rangle$ is $14$ , not $\langle -6, 20 \rangle$ .

Projections

Often you want to know how much of one vector points along another, for example how much of gravity pulls a cart down a ramp. Shine a light perpendicular to $\mathbf{v}$ and look at the shadow that $\mathbf{u}$ casts on the line of $\mathbf{v}$ . That shadow is the projection of $\mathbf{u}$ onto $\mathbf{v}$ .

The projection of u onto v is the shadow of u on the line through v. The dashed segment meets that line at a right angle.

Projection formulas

The scalar projection (the signed length of the shadow) and the vector projection of $\mathbf{u}$ onto $\mathbf{v}$ are

\text{comp}_{\mathbf{v}}\,\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|}, \qquad \text{proj}_{\mathbf{v}}\,\mathbf{u} = \left(\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2}\right)\mathbf{v}.

The scalar projection is $\|\mathbf{u}\|\cos\theta$ , which is the adjacent side of a right triangle with hypotenuse $\|\mathbf{u}\|$ . The vector projection is that length times the unit vector $\dfrac{\mathbf{v}}{\|\mathbf{v}\|}$ .

Worked example: Projecting one vector onto another

Find the vector projection of $\mathbf{u} = \langle 6, 2 \rangle$ onto $\mathbf{v} = \langle 3, 4 \rangle$ .

\mathbf{u} \cdot \mathbf{v} = 18 + 8 = 26, \qquad \|\mathbf{v}\|^2 = 9 + 16 = 25.

\text{proj}_{\mathbf{v}}\,\mathbf{u} = \frac{26}{25}\langle 3, 4 \rangle = \langle 3.12,\ 4.16 \rangle.

Check: the leftover piece $\mathbf{u} - \text{proj}_{\mathbf{v}}\,\mathbf{u} = \langle 2.88, -2.16 \rangle$ should be orthogonal to $\mathbf{v}$ , and indeed $3(2.88) + 4(-2.16) = 8.64 - 8.64 = 0$ .

Work

In physics, the work done by a constant force $\mathbf{F}$ that moves an object along a displacement $\mathbf{d}$ counts only the part of the force in the direction of motion:

W = \mathbf{F} \cdot \mathbf{d} = \|\mathbf{F}\|\,\|\mathbf{d}\|\cos\theta.

Only the horizontal part of the pull, ‖F‖ cos 30°, moves the box along the floor.

Worked example: Pulling a box

You drag a box 20 m across a floor by pulling on a rope with a force of 50 N at $30^\circ$ above the horizontal. How much work do you do?

W = (50)(20)\cos 30^\circ = 1000 \cdot \frac{\sqrt{3}}{2} \approx 866.0 \text{ joules}.

If you pulled horizontally, the work would be the full $1000$ joules. The angled pull wastes some effort lifting instead of moving.

Tip

If you already have components, use $W = F_1 d_1 + F_2 d_2$ . If you have magnitudes and an angle, use $W = \|\mathbf{F}\|\|\mathbf{d}\|\cos\theta$ . Both compute the same dot product.