# The area of a triangle in $\mathbb{R}^3$ is $\sqrt{S_1^2+S_2^2+S_3^2}$, where each $S_i$ is the area of a projection onto a coordinate plane

Prove that if the three vertices of triangle $ABC$ are $A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3)$,
then the area of triangle $ABC=S=\sqrt{S_1^2+S_2^2+S_3^2}$, where $$S_1=\frac{1}{2}\begin{vmatrix}y_1&z_1&1\\y_2&z_2&1\\y_3&z_3&1\end{vmatrix} \qquad S_2=\frac{1}{2}\begin{vmatrix}x_1&z_1&1\\x_2&z_2&1\\x_3&z_3&1\end{vmatrix} \qquad S_3=\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}$$

Here, $S_1$, $S_2$, $S_3$ are the areas of the projections of $\triangle ABC$ onto the $yz$, $zx$, $xy$ planes, respectively.

This formula is given in my book, but I do not know how this formula is derived. I tried, but failed. I googled and found this, but could not understand it: https://www.johndcook.com/blog/2016/08/23/area-of-a-triangle-and-its-projections/

• You can always use Heron's formula to calculate the area of $\triangle ABC$ from the lengths of its edges; and you can demonstrate the equality by comparing that result with what you get from adding the squared-determinants together. (You can also formulate the problem in terms of vector cross products, but if you don't (yet) know why or how the cross product gives you area, the direct approach with Heron's formula may be better.) – Blue May 17 '18 at 13:00
• If $\alpha,\beta,\gamma$ are angles between the plane of the triangle and coordinate planes, then $S_1=S\cos\alpha$, $S_2=S\cos\beta$ and $S_3=S\cos\gamma$. Now you have to prove $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$. For this, consider the normal on the plane of the triangle. – SMM May 17 '18 at 13:03