Proving that a face diagonal of a cube is orthogonal to a body diagonal

Question

If we consider the following cube, I have found in an exercise:

Prove that $(FH)$ and $(AG)$ are orthogonal.

But I don't think that's even true.

Answer 1

Let $\vec{i}:=\overrightarrow{BC},\,\vec{j}:=\overrightarrow{BA},\,\vec{k}:=\overrightarrow{BF}$ so $\overrightarrow{FH}=\vec{i}+\vec{j},\,\overrightarrow{AG}=\vec{i}-\vec{j}+\vec{k}$ are orthogonal. But we don't need a specific coordinate system to note $\overrightarrow{FH}$ is orthogonal to $\overrightarrow{AG}$, because $\overrightarrow{FH}$ is orthogonal to $\overrightarrow{AC}$, which is $\overrightarrow{AG}$'s projection onto a plane containing $\overrightarrow{FH}$ normal to $\overrightarrow{CG}$.

Answer 2

Let $I$ be the midpoint of $HD$ and $J$ be the one of $BF$ and $a$ the length of the cube's edge. Then $IJ$ is parallel to $FH$ and meets $AG$ in $M$. Consider the triangle $GMI$. Now the square of $GI$ is $\frac54a^2$, the square of $MI$ is $a^2/2$ and the square of $MG$ is $\frac34a^2$, all by Pythagoras. Observe that the the squares of $MG$ and $MG$ sum up to the square of $GI$. By Pythagoras again, $\angle GMI$ is $\pi/2$.