Let $A, B, C, D$ be four points in space. Prove
$$|AB|^2 + |CD|^2 -|BC|^2 - |AD|^2 = 2\cdot \vec{AC}\cdot \vec{DB}$$
Clearly,
$$AB = B-A$$
$$CD = D-C$$
$$AD = D-A$$
If I directly substitute the values, ignoring the distance operator, I get:
$$|AB|^2 + |CD|^2 -|BC|^2 - |AD|^2=(B-A)^2 + (D-C)^2 - (C-B)^2 - (D-A)^2$$
$$=B^2 + A^2 -2BA + D^2 + C^2 -2DC -C^2 - B^2 + 2CB - D^2 - A^2 + 2DA$$
$$= 2(CB + DA - DC - BA)$$
$$= 2(B-C+A-D-C+D-A+B) = 4(B - C) = 4\vec{AB}$$
But I'm not sure I can ignore the distance operator like that. Specifically, can I state that:
$$|AB|^2 = (B-A)^2$$
can you give reasons?
You can do your initial substitution, interpreting $\overline{AB}=\overline{B}-\overline{A}$, where $\overline{A}$ denotes a vector from the origin to the point A. You can also do your second step, because $$\|\overline{AB}\|^2=\|\overline{B}-\overline{A}\|^2=(\overline{B}-\overline{A})\cdot(\overline{B}-\overline{A})=\overline{B}\cdot \overline{B}+\overline{A}\cdot\overline{A}-\overline{B}\cdot\overline{A}-\overline{A}\cdot \overline{B}$$
However in the end you are trying to convert $BA$ to $B-A$ a second time, which is no good because $\overline{B}\cdot\overline{A}\neq \overline{AB}$; the former is a scalar and the latter is a vector.
In short, you need to keep quite clear which letters stand for points and which stand for vectors. What you should do is leave the LHS as a big mess, then work from the RHS. Simplify $$2(\overline{AC}\cdot\overline{DB})=2(\overline{C}-\overline{A})\cdot(\overline{D}-\overline{B})$$