V dot N is |V||N|cos(theta), where theta is the angle between N and V. If V dot N is greater than 0, then theta must be less than 90, so P must be on the same half-plane as N. If V dot N is 0, then theta is 90, so P must be on the line. If V dot N is negative, then theta is between 90 and 180 (exclusive), so P must be on the opposite half-plane as N.
shadaj
As a student in 189, I understood this by relating it to perceptrons, where a linear boundary is created to separate two classes. In this case, the two classes represent whether a point can lie inside the triangle or not (from the perspective of the line we are querying).
Staffviviehn
Great connections from both of the comments above!
V dot N is |V||N|cos(theta), where theta is the angle between N and V. If V dot N is greater than 0, then theta must be less than 90, so P must be on the same half-plane as N. If V dot N is 0, then theta is 90, so P must be on the line. If V dot N is negative, then theta is between 90 and 180 (exclusive), so P must be on the opposite half-plane as N.
As a student in 189, I understood this by relating it to perceptrons, where a linear boundary is created to separate two classes. In this case, the two classes represent whether a point can lie inside the triangle or not (from the perspective of the line we are querying).
Great connections from both of the comments above!