Area of projection of cube in $\mathbb{Z}^3$ onto a hyperplane

Question

A cube with vertices $(\pm 1, \pm 1, \pm 1)$ gets projected into the plane perpendicular to vector $\mathbf{n}\in S^2$. The projection is a hexagon, how do I find the area?

I think I can just compute $\mathbf{n}\cdot e_{\mathbf{x}},\mathbf{n}\cdot e_{\mathbf{y}}$ and $\mathbf{n}\cdot e_{\mathbf{z}}$ and the area would just be the some of the 3 parallelograms arising from projection:

$$A = (\mathbf{n}\cdot e_{\mathbf{x}})(\mathbf{n}\cdot e_{\mathbf{y}}) + (\mathbf{n}\cdot e_{\mathbf{y}})(\mathbf{n}\cdot e_{\mathbf{z}}) + (\mathbf{n}\cdot e_{\mathbf{z}})(\mathbf{n}\cdot e_{\mathbf{x}}) $$

Something doesń't feel quite right. I feel like these should be related to Euler angles somehow.

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EDIT The answer seems to be $A = |\cos \alpha| + |\cos \beta| + |\cos \gamma|$ using the direction cosines of the vector $\mathbf{n}$ although it seems to be difficult to check that the two responses below are equivalent.

Possibly related: Volume of the projection of the unit cube on a hyperplane

Answer 1

Let $\mathbf{n}$ be a unit vector in $\mathbb{R}^3$ be normal to the plane $P$. The projection $p$ of a vector $v\in\mathbb{R}^3$ onto the plane $P$ is given by $p(v)=v-(v\cdot\mathbf{n})\mathbf{n}$.

The projection of the unit cube onto $P$ along the vector $\mathbf{n}$ is a hexagon whose interior has preimage which intersects the boundary of the unit cube in exactly two points and so $p$ restricted to the boundary is almost everywhere two-to-one. It follows that the area of the hexagon is equal to half the sum of the area of each projected face. Seeing as projections of opposite faces have the same area, this means that the area of the hexagon is equal to the sum of the area of the projected coordinate faces (the three faces which have one coordinate identically zero). The area of a parallelogram whose vertices are $0$, $\mathbf{a}$, $\mathbf{b}$, $\mathbf{a}+\mathbf{b}$ is given by $|\det(\mathbf{a},\mathbf{b})|$ We're working with vectors in three-space though so we'll need to add a unit vector which is orthogonal to the parallelogram to get the volume of this new parallelepiped which will be equal to the area of the parallelogram face. Such a unit vector is $\mathbf{n}$.

So, the area $A$ of the hexagon is given by $$\begin{array}{rcl}A & = & \sum_{1\leq i<j\leq 3}|\det(p(e_i),p(e_j))|\\ & = & \sum_{1\leq i<j\leq 3} |\det((e_i-\mathbf{n}_i\mathbf{n}),(e_j-\mathbf{n}_j\mathbf{n}),\mathbf{n})|\\ &=&|\det((e_1-\mathbf{n}_1\mathbf{n}),(e_2-\mathbf{n}_2\mathbf{n}),\mathbf{n})|+|\det((e_1-\mathbf{n}_1\mathbf{n}),(e_3-\mathbf{n}_3\mathbf{n})\mathbf{n},\mathbf{n})|+\\&&|\det((e_2-\mathbf{n}_2\mathbf{n}),(e_3-\mathbf{n}_3\mathbf{n}),\mathbf{n})|\end{array}$$ where $\mathbf{n}_i$ is the $i$th component of the vector $\mathbf{n}=(\mathbf{n}_1,\mathbf{n}_2,\mathbf{n}_3)^T$.

After some algebraic manipulation I believe this comes out to be $A=|\mathbf{n}_1|+|\mathbf{n}_2|+|\mathbf{n}_3|$ but please check (I ended up plugging into wolfram).

[edit]It was brought to my attention that the question is looking at a cube whose edge lengths are each $2$ units instead of the usual $1$, so the above answer needs to be scaled linearly by a factor of $2$ everywhere to compensate, which means the area should be scaled by a factor of $4$.

Answer 2

As remarked by Daniel Rust the area of the hexagon in question is the sum of the projected areas of the three "kinds" of cube facets. When a piece of a plane $\Sigma$ is orthogonally projected onto another plane $\Pi$ then the area is multiplied by $|\cos\phi|$, where $\phi$ is the angle between the planes, or equivalently: the angle between the corresponding normals. Therefore it immediately follows that the area of our hexagon is given by $$A=4\bigl(|n_1|+|n_2|+|n_3|\bigr)\ .$$