3 points on a plane acquired empirically:
Point $P: (65.143, 407.19, 1.000)$
Point $Q: (624.608, 359.191,1.000)$
Point $R: (69.766, 82.885, 1.000)$
In the following equation that solves for an intersect point between a ray(line) and a plane, how do I calculate $\vec v$, the direction vector, from the three planar points from above?
$$t = \frac{-(n\cdot p + d)}{v\cdot n}$$
Vector/Line the above equation uses:
$$p = po + t\vec v$$
Vector plane equation used above: $$\vec n \cdot \vec p + d = 0$$ $$\vec n \cdot (\vec po + t\vec v)+ d = 0$$ $$(\vec po\cdot \vec n)+(t\vec v\cdot\vec n)= -d $$ $$ t\vec v \cdot \vec n=-(\vec po \cdot \vec n +d)$$
$$t = \frac{=-(\vec po \cdot \vec n +d)}{\vec n\cdot v}$$
Solving for t denominator:
float denom = n.dot(v);
v(l,m,n) = ?