Given points P0(4,0,3), P1(3,3,4), P2(3,0,3) in 3-space, and a viewer at point V

Question

Given points P0(4,0,3), P1(3,3,4), P2(3,0,3) in 3-space, and a viewer at point V(2,2,2), give a geometric test to determine whether, from V ’s location, the vertices of triangle P0P1P2 appear in clockwise or counterclockwise order.

(Note that if P0P1P2 appears in counterclockwise order to the viewer, the normal of triangle formed by P0P1P2 is facing to the viewer using right hand rule and the viewer can see the triangle. It is equivalent to say the normal and the vector formed by (viewer – P0) both point to the same side. Note that the vector can also be formed by (viewer – P1) or (viewer – P2).)

Explanation / Answer

Viewer = (2,2,2)

P0 = (4,0,3)

P1 = (3,3,4)

P2 = (3,0,3)

Viewer - P0 = (2,2,2)-(4,0,3)

Viewer - P0 = (-2,2,-1)

Now, we will calculate dot product for P1 and P2 with viewer-P0. If both have same sign then they are on same side, otherwise they are on opposite side.

(Viewer - P0)•(P1) = (-2,2,-1)•(3,3,4)

(Viewer - P0)•(P1) = -6+6-4 = -4

(Viewer - P0)•(P2) = (-2,2,-1)•(3,0,3)

(Viewer - P0)•(P2) = -6+0-3 = -9

So they both have negative sign.

So it is counterclockwise order to the viewer.

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