Given a linear transformation $\,T:I^3\!\subset \mathbb{R}^3 \to \mathbb{R}^3$ that maps the unit cube that is invertible, i.e., $\det\left(T\right) \neq 0$, I know that it defines a parallelepiped that preserves or reverses it's orientation depending on the sign of $\det\left(T\right)$ and stretches its volume by the factor $\left\lvert\, \det\left(T\right)\, \right\rvert$. I want to calculate the areas of the sides of such parallelepiped.

To do this, I'm thinking of seeing where the transformation maps the vertices of $\,2$ adjacent faces of the cube, calculate the distance between the new vertices and then the height. Is all of this correct? Is there an easier way to calculate this?

Thanks in advance


You can calculate the area of a parallelogram using the cross product. So the surface is $S=2(|T \vec{1}_x \times T \vec{1}_y| + |T \vec{1}_x \times T \vec{1}_z| + |T \vec{1}_y \times T \vec{1}_z|)$ where $\vec{1}_x$ is the unit vector in the $x$ direction etc.

  • $\begingroup$ If I'm not mistaken, the parallelepiped is made of 2 parallelograms (not necessarily equal) as adjacent sides so, which one of those is given by $S$? $\endgroup$ – Luis Vera Mar 2 '16 at 13:04
  • $\begingroup$ One face has surface area $|T\vec{1}_x\times T\vec{1}_y|$, another face has $|T\vec{1}_x\times T\vec{1}_z|$ and the third face has $|T\vec{1}_y\times T\vec{1}_z|$, which together are half of the parallelepiped's total surface area. $\endgroup$ – Wouter Mar 2 '16 at 13:09
  • $\begingroup$ that's right, it gives 3 different parallelograms, thanks! $\endgroup$ – Luis Vera Mar 2 '16 at 13:11

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