In Computer Graphics, in order to tell if a [[Triangle]] is facing the [[Canonical Camera|Camera]], you use its [[Triangle Orientation|Orientation Vector]] $\vec n$. Typically, a [[../Math/Counter-Clockwise|Counter-Clockwise]] [[Triangle Orientation|Ordering of Vertices]] is used. >[!seealso] Related Topic: [[Back-face Culling]] From some observer [[../Math/Point|Point]] $E$, to determinate if it is possible to see the front of the [[Triangle]] $\triangle PQR$: Let $P'$ be any [[../Math/Point|Point]] in $\triangle PQR$. If the [[Triangle]] is facing the observer, then the [[../Math/Vector|Vector]] $E-P'$ must be less than $90\degree$ apart from the [[Triangle Orientation|Orientation Vector]] If the two vectors are less than $90\degree$ apart, then the [[Cosine]] of the angle between them must be $0$, which means we can leverage the [[../Math/Dot Product|Dot Product]] as it is outwardly proportional to the [[Cosine]] of the angle between the two [[../Math/Vector|Vectors]]. $ \huge \begin{align} (E-P')\cdot \vec m >0 &\iff E \text{ can see the front of the Triangle}\\ (E-P')\cdot \vec m = 0 &\iff E \text{ can see the edge of the Triangle}\\ (E-P')\cdot \vec m < 0 &\iff E \text{ can see the back of the Triangle} \end{align} $