My question is why vector$OM={1\over 3} \(OB+OC+OD)$and $OA '$ can be expressed as the form $2OM-OA$
A simple question about vector and geometry
1
$\begingroup$
linear-algebra
geometry
1 Answers
0
$OM = \frac{OA + OB + OC}3$ because the midpoint of a polyhedron is the "average" of those points, thus you sum them and divide by their quantity (which is here $3$).
The reason why you can put OA' = 2 OM - OA is because you can write $ OA = OM - (OM - OA). $ Therefore "reflecting through the point" may be see as OA' = OM + (OM - OA) = 2 OM - OA. Since the tetrahedron is reqular, the vector $OA - OM$ is orthogonal to the triangle $BCD$, so that the reflection of the point $OA$ to the otherside just corresponds to changing the $-$ for a $+$ in the writing of $OA = OM - (OM - OA)$.
Hope that helps,