1. Vectors – The Cross Product Lecture 13 Wed, Sep 24, 2003

2. The Cross Product The cross product of two 3D vectors u = (u 1, u 2, u 3 ) v = (v 1, v 2, v 3 ) is defined to be u  v = (u 2 v 3 – u 3 v 2, u 3 v 1 – u 1 v 3, u 1 v 2 – u 2 v 1 ).

3. The Cross Product u1u1 u2u2 u3u3 v1v1 v2v2 v3v3 u1u1 u2u2 u3u3 v1v1 v2v2 v3v3 u1u1 u2u2 u3u3 v1v1 v2v2 v3v3 = u 2 v 3 – u 3 v 2 = u 3 v 1 – u 1 v 2 = u 1 v 2 – u 2 v 1 +    + +

4. Properties of the Cross Product The cross product is a vector. The cross product is defined only for 3D vectors. u  v = –(v  u). u  v is perpendicular to both u and v. (tu)  v = u  (tv) = t(u  v). |u  v| = |u||v|sin .

5. Example: The Cross Product VectorDemo.cpp

6. Normals to Triangles Given a triangle PQR in 3D, find a unit normal to the surface. P Q R

7. Normals to Triangles Form the cross product of Q – P and R – P. P Q R R – P Q – P

8. Example: Triangle Normal Let P = (1, 1, 2), Q = (3, 1, 5), and R = (1, 0, 4). Q – P = (2, 0, 3). R – P = (0, –1, 2). (Q – P)  (R – P) = (3, –4, –2). What about (R – P)  (Q – P)?

9. Right-hand Coordinate System The right-hand 3D coordinate system. i  j = k j  k = i k  i = j x y z i j k

10. Interpolating Vectors Suppose P is a vertex at the junction of three faces with normal vectors n 1 = (.7071,.7071, 0). n 2 = (.5774,.5774,.5774). n 3 = (0,.7071,.7071). If the faces are approximations to a smooth surface, what should the normal n at P be?

11. Interpolating Vectors We should let n be the normalized value of n 1 + n 2 + n 3. n 1 + n 2 + n 3 = (1.2845, 1.9916, 1.2845). n = (.4765,.7388,.4765).