/**
* @brief
* Construct a plane with a point and a normal.
* @param p
* the point
* @param n
* the normal
* @throw std::domain_error
* if the normal vector is the zero vector
* @remark
* The plane normal is normalized if necessary.
* @remark
* Let \f$X\f$ be the point and \f$\vec{n}\f$ the unnormalized plane normal, then the plane equation is given by
* \f{align*}{
* \hat{n} \cdot P + d = 0, \hat{n}=\frac{\vec{n}}{|\vec{n}|}, d = -\left(\hat{n} \cdot X\right)
* \f}
* \f$X\f$ is on the plane as
* \f{align*}{
* &\hat{n} \cdot X + d\\
* =&\hat{n} \cdot X + -(\hat{n} \cdot X)\\
* =&\hat{n} \cdot X - \hat{n} \cdot X\\
* =&0
* \f}
*/
Plane3(const VectorType& p, const VectorType& n)
: _n(n), _d(0.0f) {
if (_n.normalize() == 0.0f) {
throw std::domain_error("normal vector is zero vector");
}
_d = -_n.dot(p);
}
/**
* @brief
* Construct a plane with three non-collinear points.
* @param a, b, c
* the point
* @throw std::domain_error
* if the points are collinear
* @remark
* Assume \f$a\f$, \f$b\f$ and \f$c\f$ are not collinear. Let \f$u = b - a\f$,
* \f$v = c - a\f$, \f$n = u \times v\f$, \f$\hat{n} = \frac{n}{|n|}\f$ and
* \f$d = -\left(\hat{n} \cdot a\right)\f$.
* @remark
* We show that \f$a\f$, \f$b\f$ and \f$c\f$ are on the plane given by the
* equation \f$\hat{n} \cdot p + d = 0\f$. To show this for \f$a\f$, rewrite
* the plane equation to
* \f{eqnarray*}{
* \hat{n} \cdot p + -(\hat{n} \cdot a) = 0\\
* \f}
* shows for \f$p=a\f$ immediatly that \f$a\f$ is on the plane.
* @remark
* To show that \f$b\f$ and \f$c\f$ are on the plane, the plane equation is
* rewritten yet another time using the bilinearity property of the dot product
* and the definitions of \f$d\f$ and \f$\hat{n}\f$ its
* \f{align*}{
* &\hat{n} \cdot p + -(\hat{n} \cdot a) \;\;\text{Def. of } d\\
* =&\hat{n} \cdot p + \hat{n} \cdot (-a) \;\;\text{Bilinearity of the dot product}\\
* =&\hat{n} \cdot (p - a) \;\;\text{Bilinearity of the dot product}\\
* =&\frac{n}{|n|} \cdot (p - a) \;\;\text{Def. of } \hat{n}\\
* =&(\frac{1}{|n|}n) \cdot (p - a) \;\;\\
* =&\frac{1}{|n|}(n \cdot (p - a)) \;\;\text{Compatibility of the dot product w. scalar multiplication}\\
* =&n \cdot (p - a) \;\;\\
* =&(u \times v) \cdot (p - a) \;\;\text{Def. of } n
* \f}
* Let \f$p = b\f$ then
* \f{align*}{
* &(u \times v) \cdot (b - a) \;\text{Def. of } u\\
* =&(u \times v) \cdot u\\
* =&0
* \f}
* or let \f$p = c\f$ then
* \f{align*}{
* &(u \times v) \cdot (c - a) \;\text{Def. of } u\\
* =&(u \times v) \cdot v\\
* =&0
* \f}
* as \f$u \times v\f$ is orthogonal to both \f$u\f$ and \f$v\f$.
* This shows that \f$b\f$ and \f$c\f$ are on the plane.
*/
Plane3(const VectorType& a, const VectorType& b, const VectorType& c) {
auto u = b - a;
if (u == VectorType::zero()) {
throw std::domain_error("b = a");
}
auto v = c - a;
if (u == VectorType::zero()) {
throw std::domain_error("c = a");
}
_n = u.cross(v);
if (0.0f == _n.normalize()) {
/* u x v = 0 is only possible for u,v != 0 if u = v and thus b = c. */
throw std::domain_error("b = c");
}
_d = -_n.dot(a);
}
/**
* @brief
* Construct a plane from a translation axis and a distance from the origin.
* @param t
* the translation axis
* @param d
* the distance from the origin
* @post
* Given an axis \f$\vec{t}\f$ and a distance from the origin \f$d\f$
* the plane equation is given by
* \f{align*}{
* \hat{n} \cdot p + d = 0, \hat{n}=\frac{\vec{t}}{|\vec{t}|}
* \f}
* @throw std::domain_error
* if the translation axis is the zero vector
*/
Plane3(const VectorType& t, const ScalarType& d)
: _n(t), _d(d) {
if (_n.normalize() == 0.0f) {
throw std::domain_error("axis vector is zero vector");
}
}
