If four distinct points with position vectors a, b, c, d are coplanar, then [ a b c] equal…

Question

If four distinct points with position vectors $\vec a,\vec b,\vec c,\vec d$ are coplanar, then $[\vec a\ \vec b\ \vec c]$ equals:

Accepted Answer

Since $\vec a,\vec b,\vec c,\vec d$ coplanar, $(\vec b-\vec a),(\vec c-\vec b),(\vec d-\vec c)$ are coplanar (as displacement vectors within the plane), so $[(\vec b-\vec a)\ (\vec c-\vec b)\ (\vec d-\vec c)]=0$. Expanding this triple product and collecting terms yields
$$[\vec a\ \vec b\ \vec c] = [\vec d\ \vec c\ \vec a]+[\vec b\ \vec d\ \vec a]+[\vec c\ \vec d\ \vec b]$$

Solutions · 1