10.3: Properties

Commutativity

Let's look at a few properties of the dot product. First of all, the dot product is commutative:

$$\mathbf{A} \cdot \mathbf{B}=\mathbf{B} \cdot \mathbf{A}$$

The proof of this property should be obvious from Eqs. (7.1.1) and (7.2.6). This isn't a trivial property; in fact, the other two types of vector multiplication are non-commutative.

Projections

The dot product is defined as it is because it gives the projection of one vector onto the direction of another. For example, dotting a vector $\mathbf{A}$ with any of the cartesian unit vectors gives the projection of $\mathbf{A}$ in that direction:

$$\begin{align} \mathbf{A} \cdot \mathbf{i} & =A_{x} \\[6pt] \mathbf{A} \cdot \mathbf{j} & =A_{y} \\[6pt] \mathbf{A} \cdot \mathbf{k} & =A_{z} \end{align}$$

In general, the projection of vector $\mathbf{A}$ in the direction of unit vector $\hat{\mathbf{u}}$ is $\mathbf{A} \cdot \hat{\mathbf{u}}$.

Magnitude

From Eq. (7.2.6), it follows that $\mathbf{A} \cdot \mathbf{A}=A_{x}^{2}+A_{y}^{2}+A_{z}^{2}=A^{2}$; so the magnitude of a vector $\mathbf{A}$ is given in terms of the dot product by

$$\begin{align} A^{2} & =\mathbf{A} \cdot \mathbf{A} \\[6pt] A & =\sqrt{\mathbf{A} \cdot \mathbf{A}} \end{align}$$

Angle between Two Vectors

The dot product is also useful for computing the separation angle between two vectors. From Eq. (7.1.1),

$$\cos \theta=\frac{\mathbf{A} \cdot \mathbf{B}}{A B}$$

Example $\PageIndex{1}$

We wish to find the angle between the two vectors $\mathbf{A}=3 \mathbf{i}+4 \mathbf{j}-2 \mathbf{k}$ and $\mathbf{B}=\mathbf{i}-5 \mathbf{j}+2 \mathbf{k}$.

Solution

We first find the dot product of the two vectors:

$$\mathbf{A} \cdot \mathbf{B}=(3)(1)+(4)(-5)+(-2)(2)=-21 \notag$$

The magnitudes of the two vectors are

$$\begin{align} & A=\sqrt{3^{2}+4^{2}+(-2)^{2}}=\sqrt{29} \notag \\[6pt] & B=\sqrt{1^{2}+(-5)^{2}+2^{2}}=\sqrt{30} \notag \end{align} \notag$$

Therefore

$$\cos \theta=\frac{-21}{\sqrt{29} \sqrt{30}}=-0.711967 \notag$$

and so the angle between $\mathbf{A}$ and $\mathbf{B}$ is

$$\theta=135.40^{\circ} \notag$$

You do not need to worry about getting the angle $\theta$ in the correct quadrant, because $\theta$ will necessarily always be between $0^{\circ}$ and $180^{\circ}$, and the inverse cosine function will always return its result in this range.

Orthogonality

Another useful property of the dot product is: if two vectors are orthogonal, then their dot product is zero. For example, for the cartesian unit vectors:

$$\mathbf{i} \cdot \mathbf{j}=\mathbf{j} \cdot \mathbf{k}=\mathbf{i} \cdot \mathbf{k}=0$$

The converse is also true: if the dot product is zero, then the two vectors are orthogonal.

The cartesian unit vectors $\mathbf{i}, \mathbf{j}$, and $\mathbf{k}$ are orthonormal, so that

$$\mathbf{i} \cdot \mathbf{i}=\mathbf{j} \cdot \mathbf{j}=\mathbf{k} \cdot \mathbf{k}=1$$

Derivative

The derivative of the dot product is similar to the familiar product rule for scalars:

$$\frac{d(\mathbf{A} \cdot \mathbf{B})}{d t}=\mathbf{A} \cdot \frac{d \mathbf{B}}{d t}+\frac{d \mathbf{A}}{d t} \cdot \mathbf{B}$$