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:

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