Why can the area projected onto a plane be calculated with the dot product of the vector area and the unit vector of the plane?

Question

I was trying to find the projection of a flat surface onto an arbitrary plane and I came across this Wikipedia article,

The projected area onto a plane is given by the dot product of the vector area $\mathbf S$ and the target plane unit normal ${\hat {\mathbf {m} }}$: $${\displaystyle A_{\parallel }=\mathbf {S} \cdot {\hat {\mathbf {m} }}}$$

While this result is intuitive, since one can think that the vector projection of $\mathbf {S}$ over ${\hat {\mathbf {m} }}$ would yield a new vector area $\mathbf S_{\hat {\mathbf {m} }}= (\mathbf{S} \cdot \hat{\mathbf{m}}) \hat{\mathbf{m}}$ proportional to the projected surface, I wonder what would the proper way to justify this result be. Unfortunately, the article doesn't give more details about this.

Answer 1

I think maybe it is not worth giving a rigorous proof, if you can understand this property well.

The key of it is $\operatorname{Area}(S_{m})=\operatorname{Area}(S)\cdot\cos\theta$. Consider the simplest case. If the original rectangle and the projected rectangle share a same side, as you mentioned, then the result is obvious by simple geometry. Note that the translation and rotation of the original rectangle don't change the area of it, nor the area of the projected one. We know the relation remains true no matter where the rectangle it is on the plane. For the general surface, you can devide it into so many small rectangles, and conclude by taking the limit. If you would like a formal proof, it would essentially say the same thing except it is expressed with integrals.

Answer 2

$ \newcommand\R{\mathbb R} \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\MVects[1]{\mathop{\textstyle\bigwedge^{\mkern-0.5mu#1}}} \newcommand\form[1]{\langle#1\rangle} $The determinant of a linear transformation $T : \R^n \to \R^n$ is the factor by which $T$ scales $n$-volumes. This is shown concretely by the change-of-variables formula for integrals: if $S$ is some volume then $$ \int_{T(S)}\mathrm dV = \int_S\det(\mathrm DT)\,\mathrm dV = \det(T)\int_s\mathrm dV \tag{Int} $$ where $\mathrm DT$ is the total differential of $T$ (making $\det(\mathrm DT)$ the Jacobian determinant) and the second equality follows since the differential at any point of a linear transformation is the transformation itself.

This fact takes a more abstract form within the exterior algebra $\Ext\R^n$. The space of $n$-vectors $\MVects n\R^n$ is one-dimensional, corresponding to the fact that $\R^n$ has exactly one $n$-dimensional subspace; it follows that there is a unique scalar $\delta$ such that $T(I) = \delta I$ for all $I \in \MVects n\R^n$ (where here we are conflating $T$ with its outermorphism: its natural extension to a homomorphism over $\Ext\R^n$, which is much like its action over subspaces of $\R^n$). This scalar is precisely $\delta = \det T$.

Now suppose $n = 3$, we have planes-through-the-origin $P_a$ and $P_b$ with unit normals $a, b$, and $\mathcal P : \R^3 \to P_b$ is the orthogonal projection onto $P_b$. More properly we can represent $P_a$ and $P_b$ with bivectors $p_a, p_b \in \MVects 2\R^3$, related to the normals through the Hodge star: $$ a = \star p_a,\quad b = \star p_b. $$ Though much more could be said, we will say that $p_a, p_b$ are unit whenever $a, b$ are. Requiring $p_a$ and $p_b$ to be units makes each unique up to a sign. Because $\mathcal P$ maps $P_a$ to $P_b$, the outermorphism $\Ext\R^ 3\to \Ext P_b$ of $\mathcal P$ maps $p_a$ to a multiple of $p_b$, i.e. there is some scalar $\alpha$ such that $$ \mathcal P(p_a) = \alpha p_b. \tag{$*$} $$ Because $p_a, p_b$ are unique up to a sign, so is $\alpha$.

The analogy of this $\alpha$ with a determinant should be clear. Let's make this more precise. Intuitively, there are exactly two rotations about $P_a\cap P_b$ taking $P_b$ to $P_a$; WLOG choose one and call it $R$. Then $R\circ\mathcal P : P_a \to P_a$ has a well-defined determinant. Because $R$ is an isometry, the equation (Int) gives this determinant a clear interpretation as the factor by which $\mathcal P$ scales areas within $P_a$ when they are projected onto $P_b$. Again because $R$ is an isometry we can say $R(p_b) = \pm p_a$; then we see $$ R(\mathcal P(p_a)) = \alpha R(p_b) = \pm\alpha p_a $$ and so $\det(R\circ\mathcal P) = \pm\alpha$ as desired.

Finally we concern ourselves with $a\cdot b$, which is the factor in the projection of $a$ onto $b$; our last task is to show that this is $\pm\alpha$ since scaling $a$ by an area factor $A$ then gives $b$ a magnitude of $\alpha A$. This is simplest in the formalism of geometric algebra. Let $I$ be our right-handed unit pseudoscalar (i.e. 3-vector). Then $\star B = -BI$ for any bivector $B$, and we will also take it as fact that $\mathcal P(B) = -\form{Bp_b}p_b$ where $\form{\cdot}$ is the scalar part operator. From ($*$) we see that $$ -\form{p_ap_b}p_b = \alpha p_b \implies \alpha = -\form{p_ap_b} = \form{p_aI^2p_b} = \form{(-p_aI)(-p_bI)} = (\star p_a)\cdot(\star p_b) = a\cdot b $$ using the fact that $I^2 = -1$, that $I$ is in the center of the algebra, and that for two vector $x, y$ we have $xy = x\cdot y + x\wedge y$.