35.3: Inverse
- Page ID
- 92268
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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 vectors \(\mathbf{A}, \mathbf{B}\), and \(\mathbf{C}\) such that \(\mathbf{A} \times \mathbf{B}=\mathbf{C}\). If vectors \(\mathbf{B}\) and \(\mathbf{C}\) are known, can we solve for vector \(\mathbf{A}\) ?
There is no such thing as a "cross division" operation, so we can't do anything similar to \(A=C / B\). In fact, there is no unique solution for vector \(\mathbf{A}\). There are an infinite number of vectors that can be crossed with \(\mathbf{B}\) to yield vector \(\mathbf{C}\); the smaller the angle between \(\mathbf{A}\) and \(\mathbf{B}\), the larger the magnitude \(\mathbf{A}\) must have to yield a given vector \(\mathbf{C}\).
To solve \(\mathbf{A} \times \mathbf{B}=\mathbf{C}\) for vector \(\mathbf{A}\), we will need to know vectors \(\mathbf{B}\) and \(\mathbf{C}\), along with one other piece of information, such as the magnitude of vector \(\mathbf{A}\) or the angle \(\theta\) between \(\mathbf{A}\) and \(\mathbf{B}\). Suppose the magnitude \(A\) of vector \(\mathbf{A}\) is known; then since \(|\mathbf{A} \times \mathbf{B}|=A B \sin \theta=C\), we have
\[\sin \theta=\frac{C}{A B}\]
On the other hand, if \(\theta\) is known, then
\[A=\frac{C}{B \sin \theta} .\]
In either case, we now know both the magnitude \(A\) and the angle \(\theta\). Then since \(\mathbf{A} \cdot \mathbf{B}=A B \cos \theta\), we can now find the dot product \(\mathbf{A} \cdot \mathbf{B}\). Now let's take
\[\mathbf{A} \times \mathbf{B}=\mathbf{C} .\]
Crossing both sides on the right with vector \(\mathbf{B}\), we get
\[(\mathbf{A} \times \mathbf{B}) \times \mathbf{B}=\mathbf{C} \times \mathbf{B} .\]
The left-hand side is a vector triple product; applying Eq. (32.3.8), we get
\[\mathbf{B}(\mathbf{A} \cdot \mathbf{B})-B^{2} \mathbf{A}=\mathbf{C} \times \mathbf{B}\]
Solving for vector \(\mathbf{A}\), we find
\[\mathbf{A}=\frac{1}{B^{2}}[(\mathbf{B} \times \mathbf{C})+(\mathbf{A} \cdot \mathbf{B}) \mathbf{B}]\]
So if either \(A\) or \(\theta\) is known, then we can find \(\mathbf{A} \cdot \mathbf{B}=A B \cos \theta\); knowing this and \(\mathbf{B}\) and \(\mathbf{C}\), Eq. (32.25) lets us solve for vector \(\mathbf{A}\) (provided \(\mathbf{B}, \mathbf{C} \neq \mathbf{0}\) ).