Area of inverse image of triangle under orthogonal projection

Question

Let $T$ be the triangle in $\mathbb{R}^2$ with vertices $(0,0)$, $(1,0)$, $(\frac12, \frac12)$. Find the area of the inverse image of $T$ under the orthogonal projection $\phi(x,y,z)=(x,y)$.

So I find the following:

$\psi(0,0) = (0,0,1)$, $\psi(1,0) = (1,0,0)$, $\psi(\frac12,\frac12) = (\frac12,\frac12,\frac1{\sqrt2})$

Then the area is $\frac12\cdot$(length of one side)$\cdot$(length of another side)$\cdot\sin\theta$ and $\theta = \cos^{-1}\left(\frac{\text{dot product of two sides}}{\text{norms of two sides}}\right)$. However, I get some mess with lots of square roots.

What am I doing wrong?

Answer 1

Given $A(0,0,1);\;B(1,0,0);\;C (\frac12,\frac12,\frac1{\sqrt2})$

define two vectors $\vec u=B-A;\;\vec v=C-B$

we get $$\vec u=(1,0,-1);\;\vec v=(-\frac12,\frac12,\frac1{\sqrt2})$$ the area of the triangle is $$Area(\Delta)=\frac12||\vec u\times \vec v||$$ the cross product can be computed by the determinant of this formal matrix $$\left( \begin{array}{ccc} \vec i & \vec j & \vec k \\ 1 & 0 & -1 \\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{\sqrt{2}} \\ \end{array} \right)$$ giving $$\vec u\times \vec v= \frac{\vec i}{2}+\frac{\vec j}{2}-\frac{\vec j}{\sqrt{2}}+\frac{\vec k}{2}$$ where $\{\vec i,\vec j,\vec k\}$ is a base of $\mathbb{R}^3$.

In components $$\vec u\times \vec v= \left(\frac12,\frac12-\frac{1}{\sqrt 2},\frac12\right)$$ and $$Area(\Delta)=\frac12||\vec u\times \vec v||=\frac{1}{4} \sqrt{5-2 \sqrt{2}}\approx 0.3684$$

Answer 2

Erect vertical stalks on the three given points. Then any triangle $\triangle\subset{\mathbb R}^3$ with its vertices on the three stalks can be viewed as an inverse image of $T$ under $\phi$. Therefore you cannot talk about "the" inverse image of $T$.

Now any such $\triangle$ is lying in a plane $\Pi$ not parallel to the $z$-axis. If you want to know the area of $\triangle$ you have to know the angle $\theta$ between $\Pi$ and the $(x,y)$-plane, or what amounts to the same thing, the value $\cos\theta=\bigl|n_\Pi\cdot(0,0,1)\bigr|$, whereby $n_\Pi$ is the unit normal of $\Pi$. You then have $${\rm area}(\triangle)={1\over\cos\theta}\>{\rm area}(T)\ .$$