10.2: Component Form
- Page ID
- 92175
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Suppose we have two vectors in rectangular form. What is the dot product of the two in terms of their components? To answer this, we begin with the definition of the dot product, Eq. (7.1.1):
\[
\mathbf{A} \cdot \mathbf{B}=A B \cos (\beta-\alpha),
\]
where \(\alpha\) is the angle vector A makes with respect to the \(x\) axis, and \(\beta\) is the angle vector \(\mathbf{B}\) makes with respect to the \(x\) axis, so that \(\beta-\alpha\) is the angle between the two vectors (Fig. \(\PageIndex{1}\)). We now use a trigonometric identity to expand the argument of the cosine:
\[
\mathbf{A} \cdot \mathbf{B}=A B(\cos \beta \cos \alpha+\sin \beta \sin \alpha)
\]
Now making use of the relations \(\cos \theta=\operatorname{adj} /\) hyp and \(\sin \theta=\mathrm{opp} / \mathrm{hyp}\), we have
\[
\cos \alpha=\frac{A_{x}}{A} ; \quad \cos \beta=\frac{B_{x}}{B} ; \quad \sin \alpha=\frac{A_{y}}{A} ; \quad \sin \beta=\frac{B_{y}}{B}
\]
Making these substitutions into Eq. \(\PageIndex{2}\), we have
\[
\begin{align}
\mathbf{A} \cdot \mathbf{B} & =A B\left(\frac{B_{x}}{B} \frac{A_{x}}{A}+\frac{B_{y}}{B} \frac{A_{y}}{A}\right) \\[6pt]
& =A_{x} B_{x}+A_{y} B_{y}
\end{align}
\]
This result can be generalized from two to three dimensions to get
\[
\mathbf{A} \cdot \mathbf{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}
\]
Suppose vectors \(\mathbf{A}=3 \mathbf{i}+4 \mathbf{j}-2 \mathbf{k}\) and \(\mathbf{B}=\mathbf{i}-5 \mathbf{j}+2 \mathbf{k}\). What is the dot product of the two vectors
Solution
Then the dot product of the two vectors is
\[
\mathbf{A} \cdot \mathbf{B}=(3)(1)+(4)(-5)+(-2)(2)=-21 \notag
\]
Notice that the final result is a scalar, not a vector.