Prove that the area of a unit cube’s projection on any plane equals the length of the cube’s projection on the perpendicular of this plane.
Source: Found this problem in solving mathematical problems by terence tao but I couldn't find any solutions online.
Have a look at the Volume of a parallelepiped obtained by scalar triple product of the three edge vectors and to the fact that it is invariant under a rotation.
Also have a look at the Volume of a parallelepiped obtained by wedge product of the edge vectors $$ {\bf V} = {\bf u}_1 \wedge {\bf u}_2 \wedge {\bf u}_3 $$ which can be generalized to higher dimension.
As a matter of fact, given $n$ vectors in ${\mathbb R}^m$, their wedge product $$ {\bf V}_n^{\left( m \right)} = {\bf x}_1 \wedge {\bf x}_2 \wedge \cdots \wedge {\bf x}_n = \sum\limits_{\left\{ {\begin{array}{*{20}c} {\left| {\left\{ r \right\}_{\,k} } \right| = n} \\ {\left\{ r \right\}_{\,k} \in \left\{ {1,2, \cdots ,m} \right\}} \\ \end{array}} \right.} {V_k \;{\bf e}_{\left\{ r \right\}_{\,k} } } $$ where $$ \left\{ \begin{array}{l} {\bf e}_{\left\{ r \right\}_{\,k} } = {\bf e}_{j_{\,1} } {\bf e}_{j_{\,2} } \cdots {\bf e}_{j_{\,n} } \\ j_{\,1} < \, j_{\,2} < \cdots < j_{\,n} \\ \left\{ r \right\}_{\,k} = \left\{ {j_{\,1} , j_{\,2} , \cdots ,j_{\,n} } \right\} \in \left\{ {1,2, \cdots ,m} \right\} \\ k = \left[ {1,\binom{m}{n}} \right] \\ \end{array} \right. $$
so that the sum is over all the $n$-subsets of $ \left\{ {1,2, \cdots ,m} \right\}$ and $\bf V$ is the $n$-vector associated with the $n$-parallelotope whose edges are parallel to the $\bf x$ vectors.
The magnitude of $\bf V$ represents the ($n$-) volume (absolute value) of that parallelotope: $$ Vol\left( {{\bf V}_n^{\left( m \right)} } \right) = \left\| {{\bf V}_n^{\left( m \right)} } \right\| = \sqrt {\sum\limits_k {V_k ^2 } } $$ which is invariant under any unitary transformation.
There are many alternative representations for the volume, mainly the Gramian. But to focus onto your question, let's rewrite $\bf V$ as $$ \begin{array}{l} {\bf V}_n^{\left( m \right)} = {\bf x}_1 \wedge {\bf x}_2 \wedge \cdots \wedge {\bf x}_n = \left( {{\bf x}_1 \wedge {\bf x}_2 \wedge \cdots \wedge {\bf x}_{n - 1} } \right) \wedge {\bf x}_n = \\ = \left( {{\bf x}_1 \wedge {\bf x}_2 \wedge \cdots \wedge {\bf x}_{n - 1} } \right) \wedge {\bf x}_{n\, \bot } = {\bf V}_{n - 1}^{\left( m \right)} \wedge \left\| {{\bf x}_{n\, \bot } } \right\|{\bf u}_{n\; \bot } = {\bf V}_{n - 1}^{\left( m \right)} \wedge \left\| {{\bf x}_n \cdot {\bf u}_{n\; \bot } } \right\|{\bf u}_{n\; \bot } \\ \end{array} $$ Here ${\bf x}_{n\, \bot }$ indicates the projection of ${\bf x}_n$ onto the sub-space orthogonal to the sub-space spanned by ${\bf x}_1 , \ldots , {\bf x}_{n-1}$ (the rejection of ${\bf x}_n$), and ${\bf u}_{n\, \bot }$ the corresponding unit vector.
Therefore $$ Vol\left( {{\bf V}_n^{\left( m \right)} } \right) = \left\| {{\bf V}_n^{\left( m \right)} } \right\| = \left\| {{\bf V}_{n - 1}^{\left( m \right)} } \right\|\;\left\| {{\bf x}_n \cdot {\bf u}_{n\; \bot } } \right\| = {\rm "base"}\, \times \,{\rm "height"} $$