The Dot Product and Angle Between Vectors

By the end of this lesson, you will be able to compute the dot product, find the angle between two vectors, determine orthogonality, and decompose a vector into parallel and perpendicular components.

The Big Idea

The dot product produces a scalar (a number) from two vectors. Unlike vector addition, it reveals something purely about the relationship between the directions: it equals zero when vectors are perpendicular, positive when they point the same way, negative when they oppose. It's essential for work-force calculations, projections, and the definition of angle in higher dimensions.

The Dot Product

Definition (algebraic):

$$\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2$$

Geometric form:

$$\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$$

where $\theta$ is the angle between the two vectors ($0 \leq \theta \leq \pi$).

Finding the Angle Between Two Vectors

Combining both forms:

$$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$$

$$\theta = \cos^{-1}\left(\frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}\right)$$

Orthogonality

Two vectors are perpendicular (orthogonal) iff $\vec{u} \cdot \vec{v} = 0$.

Proof: $\cos 90° = 0$, so $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos 90° = 0$.

Example 1 — Computing Dot Product and Angle

$\vec{u} = \langle 3, 4 \rangle$ and $\vec{v} = \langle -4, 3 \rangle$:

$\vec{u} \cdot \vec{v} = (3)(-4) + (4)(3) = -12 + 12 = 0$

Since dot product = 0: perpendicular!

$|\vec{u}| = 5$, $|\vec{v}| = 5$. These are perpendicular unit-scaled vectors.

Example 2 — Finding the Angle

$\vec{a} = \langle 2, 1 \rangle$ and $\vec{b} = \langle -1, 3 \rangle$:

$\vec{a} \cdot \vec{b} = -2 + 3 = 1$

$|\vec{a}| = \sqrt{5}$, $|\vec{b}| = \sqrt{10}$

$\cos\theta = \dfrac{1}{\sqrt{5}\sqrt{10}} = \dfrac{1}{\sqrt{50}} = \dfrac{1}{5\sqrt{2}}$

$\theta = \cos^{-1}\left(\dfrac{1}{5\sqrt{2}}\right) \approx 81.9°$

Vector Projection

The projection of $\vec{u}$ onto $\vec{v}$ (the component of $\vec{u}$ in the direction of $\vec{v}$):