Skill: Determine the angle between two vectors

For two nonzero vectors $\mathbf a$ and $\mathbf b$, the angle $\theta$ between them is found using the dot product formula:

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

The angle is always chosen between $0^\circ$ and $180^\circ$.

Worked example

Find the angle between

$$\mathbf a=\begin{pmatrix}2\\1\end{pmatrix} \quad\text{and}\quad \mathbf b=\begin{pmatrix}1\\3\end{pmatrix}.$$

Step 1: Find the dot product.

Multiply corresponding components and add:

$$\mathbf a\cdot\mathbf b=(2)(1)+(1)(3)=2+3=5.$$

Step 2: Find the length of each vector.

For $\mathbf a$:

$$\ \vert \mathbf a\ \vert =\sqrt{2^2+1^2}=\sqrt5.$$

For $\mathbf b$:

$$\ \vert \mathbf b\ \vert =\sqrt{1^2+3^2}=\sqrt{10}.$$

Step 3: Substitute into the formula.

$$\cos\theta =\frac{5}{(\sqrt5)(\sqrt{10})} =\frac{5}{\sqrt{50}} =\frac{1}{\sqrt2}.$$

Step 4: Use the inverse cosine.

$$\theta=\cos^{-1}\left(\frac{1}{\sqrt2}\right)=45^\circ.$$

Therefore, the angle between the vectors is

$$\boxed{45^\circ}.$$

Because the dot product was positive, the angle is acute. In general:

• If $\mathbf a\cdot\mathbf b>0$, the angle is acute.
• If $\mathbf a\cdot\mathbf b=0$, the vectors are perpendicular, so the angle is $90^\circ$.
• If $\mathbf a\cdot\mathbf b<0$, the angle is obtuse.

The zero vector cannot be used because it has no direction, so its angle with another vector is undefined.