Consider the cube $ABCDEFGH$.
Determine a straight line that is orthogonal to both $(AF)$ and $(AH)$. Justify (geometric method).
Is it possible to justify like that:
If we consider the plane $(AFH)$, then the lines $(AF)$ and $(AH)$ are both lines of the plane that intersect. A orthogonal line of the plane is $(EC)$, is this line orthogonal to $(AF)$ and $(AH)$.
Introducing the coordinate system $(A, \vec{AD}, \vec{AB}, \vec{AE})$, then
$E=(0,0,1)$ and $C=(1,1,0)$, thus $\vec{EC} = (1,1,-1)$.
$A=(0,0,0)$ and $H=(1,0,1)$ thus $\vec{AH} = (1,0,1)$.
If we consider the scalar product of $\vec{AH}$ and $\vec{EC}$, then $\vec{AH} \cdot \vec{EC} = 0$, thus $(AH) \perp (EC)$
$F = (0,1,1)$, $\vec{AF} = (0,1,1)$ Scalar product of $\vec{EC}$ and $\vec{AF}$: $\vec{EC} \cdot \vec{AF} = 0$, thus $(AF) \perp (EC)$
It is the normal of a plane passing points A, F and H"
$A=(0. 0. 0)$
$F=(0, 1,1)$
$H=(1, 0, 1)$
The equation of the plane is:
${\begin{vmatrix}x-x_A&y-y_A&z-z_A\\x_F-x_A&y_F-y_A&z_F-z_A\\x_H-x_A&y_H-y_A&z_H-z_A\end{vmatrix}}$=${\begin{vmatrix}x&y&z\\0&1&1\\1&0&1 \end{vmatrix}}=x+y-z=0$
So the orthogonal line to AF and AH is $N=(1, 1, -1)$
You can also use the vector product of vectors $AF=(0, 1, 1)$ and $AH=(1, 0, 1)$
$N=\big(\begin{vmatrix}y_F&z_F\\y_H&z_H\end{vmatrix}, \begin{vmatrix}z_F&x_F\\z_H&x_H\end{vmatrix}, \begin{vmatrix}x_F&y_F\\x_H&y_H\end{vmatrix}\big)=(1, 1, -1)$