Dot Product and Angle

From the geometric definition of the dot product, $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$, we can solve for $\theta$:

The sign of $\mathbf{a} \cdot \mathbf{b}$ immediately classifies the angle:

• $\mathbf{a} \cdot \mathbf{b} > 0$: angle is acute ($0° < \theta < 90°$)
• $\mathbf{a} \cdot \mathbf{b} = 0$: vectors are perpendicular ($\theta = 90°$)
• $\mathbf{a} \cdot \mathbf{b} < 0$: angle is obtuse ($90° < \theta < 180°$)

Example: Find the angle between $\mathbf{a} = \begin{pmatrix}3\\4\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}4\\3\end{pmatrix}$.

$\mathbf{a} \cdot \mathbf{b} = 3(4) + 4(3) = 24$

$|\mathbf{a}| = \sqrt{9+16} = 5$, $\quad |\mathbf{b}| = \sqrt{16+9} = 5$

$\cos\theta = \dfrac{24}{25} \approx 0.96 \qquad \Rightarrow \qquad \theta \approx 16.3°$

So the angle is acute — and smaller than 45°. Check your estimate from the hero!

From the geometric definition of the dot product, $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$, we can solve for $\theta$:

Pause — copy the angle formula $\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$ with a worked 3D example computing $\theta=\cos^{-1}(\cdots)$ into your book.

We just saw the angle formula: $\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$, giving the angle $\theta\in[0^\circ,180^\circ]$ between two vectors. That raises a question: the formula gives $\cos\theta=0$ when $\vec{a}\cdot\vec{b}=0$ — is this the only condition for perpendicularity, and can a zero vector be perpendicular to everything? This card answers it → for NON-ZERO vectors, $\vec{a}\perp\vec{b}\iff\vec{a}\cdot\vec{b}=0$; the zero vector is excluded because the formula requires $|\vec{a}|\neq0$.

Two non-zero vectors are perpendicular (orthogonal) if and only if their dot product is zero:

This gives a fast algebraic test. If $\mathbf{a} = \begin{pmatrix}a_1\\a_2\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}b_1\\b_2\end{pmatrix}$, just compute $a_1 b_1 + a_2 b_2$ and check for zero.

Example: Are $\mathbf{u} = \begin{pmatrix}3\\-2\end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix}4\\6\end{pmatrix}$ perpendicular?

$\mathbf{u} \cdot \mathbf{v} = 3(4) + (-2)(6) = 12 - 12 = 0$ ✓

Yes — the dot product is zero, so $\mathbf{u} \perp \mathbf{v}$.

Finding an unknown component: If $\mathbf{a} = \begin{pmatrix}2\\k\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}3\\-4\end{pmatrix}$ are perpendicular, find $k$.

$\mathbf{a} \cdot \mathbf{b} = 2(3) + k(-4) = 6 - 4k = 0 \Rightarrow k = \dfrac{3}{2}$

Two non-zero vectors are perpendicular (orthogonal) if and only if their dot product is zero:

Pause — copy the perpendicularity test: $\vec{a}\perp\vec{b}\iff\vec{a}\cdot\vec{b}=0$ (for non-zero vectors) with a worked example finding an unknown component that makes two vectors perpendicular into your book.