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\begin{center}
  {\bf University of Kentucky, Physics 306 \\
    Homework \#3, Rev. A, due Wednesday, 2025-02-05 }
\end{center}

\vspace{.3 in}

{\bf 1. Orthogonal Projections}: Parallel and perpendicular projections can be expressed as $P_{\parallel}=\bh n \bh n \cdot$ and $P_{\perp}=-\bh n \times \bh n \times$, in terms of the dot and cross products respectively, which themselves are projected products $\bv  a\cdot \bv v=av_{\parallel}$ and $\bv a\times \bv v=\bh n a v_{\perp}$.  We will build them up geometrically piece-by-piece, and explore their matrix versions in this problem.

{\bf \quad a)} Starting with the vector $\bv r=\pmatrix{x \cr y}$, i) draw $\bv r$ for some value of $x$ and $y$, ii) illustrate the length $r_{x}=\bh x\cdot \bv r$, and the vector $\bv r_{x}=\bh x \bh x\cdot \bv r$ with the same magnitude, in the direction $\bh x$.  iii) Calculate $\bv r_{x}=\bh x (\bh x\cdot \bv r)$, where $\bh x=\pmatrix{1\cr 0}$ and $\bh x\cdot = \bh x^{T}=\pmatrix{1 & 0}$. iv) Calculate the matrix $P_{x}=\bh x \bh x\cdot$ and then $\bv r_{x}=P_{x}\bv r$, which should be the same by associativity. v) Show that $P_{x}^{2}=P_{x}$ algebraically and by multiplying matrices. vi) Calculate $P_y=\bh y \bh y\cdot$ and show that $P_{x}+P_{y}=I$, so that $\bv r_{x}+\bv r_{y}=\bv r$.

{\bf \quad b)} Starting with some vector $\bv r=\pmatrix{x \cr y \cr z}$, i) draw the vectors $\bv r$, $\bh nr_{\perp}=\bh z\times \bv r$ (projected and rotated by $90^{\circ}$),  $-\bv r_{\perp}=\bh z \times (\bh z\times \bv r)$ (another rotation), and $\bv r_{\perp}$, the projection in the $xy$-plane perpendicular to $\bh z$.  ii) Calculate the components of $\bv r_{\perp}=-\bh z \times (\bh z\times \bv r)$ iii) to obtain the matrix $P_{\perp}=-\bh z \times \bh z\times$ by its action on $\bv r$. iv) Show  $P_{\perp}^{2}=P_{\perp}$ by matrix multiplication and v) $P_{\parallel} + P_{\perp}=I$ by adding the matrices, and by using the rule $\bv a \times (\bv  b\times \bv c)=\bv b (\bv a \cdot \bv c)-\bv c(\bv a\cdot \bv b)$.

{\bf 2. General Projections}: relational versus parametric linear/planar geometry.

{\bf \quad a)} Show graphically that the following equations define the set of points $\{\bs x\}$ \mbox{on a {\bf line} or {\bf plane},}
\begin{center}
  \begin{tabular}{r|@{\qquad}l@{\qquad}l}
    & relational & parametric \\ \hline
    line & $\{\bv x ~|~ \bv a \times \bv x = \bv d\}$  &  $\{\bv x = \bv x_1 + \bv a \alpha \phantom{+ \bv c \gamma} ~~|~ \quad \alpha \in \mathbb{R}\}$ \\
    plane& $\{\bv x ~|~ \bv A \cdot \bv x = D\}$  &  $\{\bv x = \bv x_2 + \bv b \beta + \bv c \gamma ~|~ \beta,\gamma \in \mathbb{R}\}$ \\
  \end{tabular}
\end{center}
where $\bv a, \bv d, \bv A, D$ are constants which define the geometry, $\bv x_{1,2}$ are fixed points on the line and plane respectively, and $\alpha, \beta, \gamma$ are parameters that vary along the line/plane to uniquely parametrize points in the line/plane.  [{\it bonus}: Show the 5th relation $\{\bv x = \bv x_2 + \bv A \times \bv \delta ~|~ \bv \delta \in \mathbb{R}^{3}\}$.]

{\bf \quad b)} What constraint between $\bv a$ and $\bv d$ is implicit in the formula $\bv a \times \bv x = \bv d$?  What is the relation between $\bv b$, $\bv c$, and $\bv A$?  Substitute $\bv x$ of each parameterization into its relational equation to show the consistency between the two forms and to derive $\bv d$ and $D$ in terms of $\bv a, \bv x_1$ and $\bv A, \bv x_2$.

{\bf \quad c)} Define $\bt a \equiv \bv A /( \bv a{\cdot} \bv A)$, which is parallel to $\bv A$, and {\it normalized} in the sense that $\bv a \cdot \bt a = 1$.  Using the BAC-CAB rule, show that $\bv x = \bv a (\bt a\cdot \bv x) - \bt a \times (\bv a \times \bv x)$ for all $\bv x$.  Illustrate this non-orthogonal projection of $\bv x$ onto vectors $\bv x_{1}$ parallel to the line plus $\bv x_{2}$ parallel to the plane.  [{\it bonus}: use this to calculate the point $\bv x_0$ at the intersection of the line and plane in terms of $\bv a, \bv d, \bv A, D$.  Verify this by showing that $\bv x_0$ satisfies the relational equation for both the line and plane.]

{\bf \quad d)} Let ~$\bt a \equiv \frac{\bv b\times \bv c}{\bv a \cdot \bv b \times \bv c}$, ~$\bt b \equiv \frac{\bv c\times \bv a}{\bv a \cdot \bv b \times \bv c}$,~and ~$\bt c \equiv \frac{\bv a\times \bv b}{\bv a \cdot \bv b \times \bv c}$, where the arrows on $\bv a, \bv b, \bv c$ distinguish them from their {\it covectors} $\bt a, \bt b, \bt c$.  The definition of $\bt a$ is the same as in c) with $\bv A = \bv b \times \bv c$.  Calculate the nine combinations of $(\bv a, \bv b, \bv c)^T \cdot (\bt a, \bt b, \bt c)=I$, i.e.  $\bv a \cdot\bt a=1$, $\bv a \cdot\bt b=0$, \ldots to show they are {\it mutually orthonormal}.  In this sense, $(\bt a, \bt b, \bt c)$ is the \href{https://en.wikipedia.org/wiki/Dual_basis}{\it dual} basis of $(\bv a, \bv b, \bv c)$. [{\it bonus}: Show using Cramer's rule that $[\bt a|\bt b|\bt c]$ = $[\bv a|\bv b|\bv c]^{-1}$, ie. that it is a {\it reciprocal basis}.]

{\bf \quad e)} The {\it contravariant components} of $\bv x$ are defined as the components $(\alpha,\beta,\gamma)$ that satisfy the equation $\bv x = \bv a \alpha + \bv b \beta + \bv c \gamma$; i.e., $(\bv a, \bv b, \bv c)$ is the contravariant basis.  Using $\bv x \cdot \bt a$, etc., calculate the three contravariant components of $\bv x$ in terms of dot products.  Likewise, find the {\it covariant components} $(\tilde\alpha,\tilde\beta,\tilde\gamma)$ of $\bv x$, defined as the components $(\tilde\alpha,\tilde\beta,\tilde\gamma)$ which satisfy $\bv x = \bt a \tilde\alpha + \bt b \tilde\beta + \bt c \tilde\gamma$; i.e.  $(\bt a, \bt b, \bt c)$ is the covariant basis.




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