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clarify notation in 2 ways to address confusion in comments: (1) remove unnecessary grouping brackets, (2) change n^2 to ‖n‖^2 to denote vector magnitude squared
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edit approved Apr 24 at 1:57
Don Hatch
edit approved Apr 24 at 1:57
Don Hatch
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The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle

There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.

Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$, $\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as

  • $\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$$\gamma=(\vec{u}\times\vec{w})\cdot\vec{n}/\left\lVert\vec{n}\right\rVert^2$
  • $\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$$\beta=(\vec{w}\times\vec{v})\cdot\vec{n}/\left\lVert\vec{n}\right\rVert^2$
  • $\alpha=1-\gamma-\beta$

The coordinates of the projected point is

  • $P'=\alpha P_1+\beta P_2 +\gamma P_3$

The point $P'$ lies inside $T$ if

  • $0\leq\alpha\leq 1$,
  • $0\leq\beta\leq 1$, and
  • $0\leq\gamma\leq 1$.

The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle

There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.

Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$, $\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as

  • $\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$
  • $\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$
  • $\alpha=1-\gamma-\beta$

The coordinates of the projected point is

  • $P'=\alpha P_1+\beta P_2 +\gamma P_3$

The point $P'$ lies inside $T$ if

  • $0\leq\alpha\leq 1$,
  • $0\leq\beta\leq 1$, and
  • $0\leq\gamma\leq 1$.

The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle

There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.

Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$, $\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as

  • $\gamma=(\vec{u}\times\vec{w})\cdot\vec{n}/\left\lVert\vec{n}\right\rVert^2$
  • $\beta=(\vec{w}\times\vec{v})\cdot\vec{n}/\left\lVert\vec{n}\right\rVert^2$
  • $\alpha=1-\gamma-\beta$

The coordinates of the projected point is

  • $P'=\alpha P_1+\beta P_2 +\gamma P_3$

The point $P'$ lies inside $T$ if

  • $0\leq\alpha\leq 1$,
  • $0\leq\beta\leq 1$, and
  • $0\leq\gamma\leq 1$.
deleted 4 characters in body
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amWhy
amWhy
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The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle

There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.

Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$, $\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as

  • $\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$
  • $\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$
  • $\alpha=1-\gamma-\beta$

and theThe coordinates of the projected point is

  • $P'=\alpha P_1+\beta P_2 +\gamma P_3$

The point $P'$ lies inside $T$ if

  • $0\leq\alpha\leq 1$,
  • $0\leq\beta\leq 1$, and
  • $0\leq\gamma\leq 1$.

The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle

There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.

Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$, $\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as

  • $\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$
  • $\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$
  • $\alpha=1-\gamma-\beta$

and the coordinates of the projected point is

  • $P'=\alpha P_1+\beta P_2 +\gamma P_3$

The point $P'$ lies inside $T$ if

  • $0\leq\alpha\leq 1$,
  • $0\leq\beta\leq 1$, and
  • $0\leq\gamma\leq 1$.

The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle

There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.

Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$, $\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as

  • $\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$
  • $\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$
  • $\alpha=1-\gamma-\beta$

The coordinates of the projected point is

  • $P'=\alpha P_1+\beta P_2 +\gamma P_3$

The point $P'$ lies inside $T$ if

  • $0\leq\alpha\leq 1$,
  • $0\leq\beta\leq 1$, and
  • $0\leq\gamma\leq 1$.
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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Community Bot
Community Bot
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The question is a slight extension of the question given here: Check whether a point is within a 3D TriangleCheck whether a point is within a 3D Triangle

There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.

Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$, $\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as

  • $\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$
  • $\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$
  • $\alpha=1-\gamma-\beta$

and the coordinates of the projected point is

  • $P'=\alpha P_1+\beta P_2 +\gamma P_3$

The point $P'$ lies inside $T$ if

  • $0\leq\alpha\leq 1$,
  • $0\leq\beta\leq 1$, and
  • $0\leq\gamma\leq 1$.

The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle

There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.

Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$, $\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as

  • $\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$
  • $\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$
  • $\alpha=1-\gamma-\beta$

and the coordinates of the projected point is

  • $P'=\alpha P_1+\beta P_2 +\gamma P_3$

The point $P'$ lies inside $T$ if

  • $0\leq\alpha\leq 1$,
  • $0\leq\beta\leq 1$, and
  • $0\leq\gamma\leq 1$.

The question is a slight extension of the question given here: Check whether a point is within a 3D Triangle

There is an elegant solution to this given by W. Heidrich, Journal of Graphics, GPU, and Game Tools,Volume 10, Issue 3, 2005.

Let $\vec{u}=P_2-P_1$, $\vec{v}=P_3-P_1$, $\vec{n}=\vec{u}\times\vec{v}$, $\vec{w}=P-P_1$. We then have directly the barycentric coordinates of the projection $P'$ of $P$ onto $T$ as

  • $\gamma=\left[(\vec{u}\times\vec{w})\cdot\vec{n}\right]/\vec{n}^2$
  • $\beta=\left[(\vec{w}\times\vec{v})\cdot\vec{n}\right]/\vec{n}^2$
  • $\alpha=1-\gamma-\beta$

and the coordinates of the projected point is

  • $P'=\alpha P_1+\beta P_2 +\gamma P_3$

The point $P'$ lies inside $T$ if

  • $0\leq\alpha\leq 1$,
  • $0\leq\beta\leq 1$, and
  • $0\leq\gamma\leq 1$.
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Håkon Hægland
Håkon Hægland
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