10.3: Properties
- Page ID
- 92176
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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),
\[\cos \theta=\frac{\mathbf{A} \cdot \mathbf{B}}{A B} \]
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:
\[\frac{d(\mathbf{A} \cdot \mathbf{B})}{d t}=\mathbf{A} \cdot \frac{d \mathbf{B}}{d t}+\frac{d \mathbf{A}}{d t} \cdot \mathbf{B}\]