Question:
$A,B,C$ are three non-collinear points lying on a plane whose normal vector is $(\hat{i}+\hat{j}+\hat{k})$. If all the three coordinates of every point is an integer, then prove that the area of the triangle is necessarily an irrational number.
My attempt:
I don't understand how to do this problem. I know how such types of problems are dealt with in the 2D plane, but not in 3D space. Can anyone offer some help on the starting steps? Thanks!
Best Answer
I assume that the plane on which the points lay is normal to the vector $(1,1,1)$. After a translation, we can assume that one of the points is $(0,0,0)$ and the other two have integer coordinates: let them be $(a,b,c)$ and $(d,e,f)$. Notice that the area of the triangle can be seen as the determinant of the matrix having as columns $(a,b,c)$, $(d,e,f)$ and $(1,1,1)$, divided by twice the length of $(1,1,1)$, since it is normal to the other two points (the idea is that you are calculating the volume of a solid having as base a quadrilateral having twice the area you are interested in). The determinant of a matrix with integer coefficients is integer, and the length of $(1,1,1)$ is $\sqrt{3}$. So the area is irrational.