If we have a triangle floating in $\mathbb{R}^3$, can we calculate the surface area based on projections to the $(x,y)$ and $(y,z)$ planes?

Question

If we have a triangle floating in $\mathbb{R}^3$, can we calculate the surface area based on projections to the $(x,y)$ and $(y,z)$ planes?

I ask this because I'm studying exterior forms at the moment and am trying to understand why any $k$ form can be written as a linear combination of $x_{i_1} \wedge .... \wedge x_{i_k}$ for increasing indicies

Answer 1

Using the cross product, you can verify that the Pythagorean Theorem generalizes to tell you that $$A^2 = A_{xy}^2 + A_{xz}^2 + A_{yz}^2,$$ where $A$ is the area of the (tilted) parallelogram and the other three are the areas of the respective projections.

This is actually an equation relating norms of $2$-vectors (the wedge products of $2$ vectors), and there's an analogous equation relating the norms of $k$-vectors (the wedge products of $k$ vectors).

Answer 2

Not if you only have two projections.

Consider a triangle parallel to the x-y plane. Its projections onto the x-z and y-z planes will just be a line segment.

So the third projection is necessary.