Section 9.4 - The Cross Product The cross product of two vectors u and v with being the angle between them (0 ≤ ≤ )is u v |u||v|sinn̂ where n̂ is a unit vector in a direction mutually orthogonal (perpendicular) to both u and v as determined by the right-hand rule. Note that the cross product yields a vector. Hence, it is also sometimes called the vector product. Further note that any two vectors that are not parallel determine a plane. The cross product, therefore, finds a vector that is perpendicular to this plane. Because n̂ is a unit vector the magnitude of u v is |u v| |u||v|sin Example: If |a| 30, b 10, and the angle between the vectors is 30∘ we have a b |a| b sin30∘ 3010 1 150 2 Furthermore, the magnitude of the cross product is the same as the area of a parallelogram determined by the corresponding vectors (see top of page 670). Two vectors u and v are parallel if and only if u v 0. Note that since any vector u is parallel to itself, u u 0 for any vector u. Note that v u −u v. Properties of the cross product (see page 669). Find the following: 1. î î ______ 2. ̂ ̂ ______ 3. k̂ k̂ ______ 4. î ̂ ______ 5. ̂ k̂ ______ 6. k̂ î ______ 7. ̂ î ______ 8. k̂ ̂ ______ We will use determinants to determine the cross product of vectors that are given in component form. See discussion on pages 670-672. If u u1î u2̂ u3k̂ and v v1î v2̂ v3k̂ then using the distributive laws (Properties 3, 3, & 4 on page 669) we have u v u1î u2̂ u3k̂ v1î v2̂ v3k̂ u1î v1î v2̂ v3k̂ u2̂ v1î v2̂ v3k̂ u3k̂ v1î v2̂ v3k̂ u1î v1î u1î v2̂ u1î v3k̂ u2̂ v1î u2̂ v2̂ u2̂ v3k̂ u3k̂ v1î u3k̂ v2̂ u3k̂ v3k̂ u1v1î î u1v2î ̂ u1v3 î k̂ u2v1̂ î u2v2̂ ̂ u2v3 ̂ k̂ u3v1 k̂ î u3v2 k̂ ̂ u3v3 k̂ k̂ u1v2k̂ − u1v3̂ − u2v1k̂ u2v3î u3v1̂ − u3v2î u2v3 − u3v2 î u3v1 − u1v3 ̂ u1v2 − u2v1 k̂ u2v3 − u3v2 î − u1v3 − u3v1 ̂ u1v2 − u2v1 k̂ (This is the form we get using determinants) 1 9. Calculate 〈2,1,−1 〈−3,2,1 10. Find a vector perpendicular to the plane that includes the points P1,−1,0, Q2,1,−1,and R−1,1,2. What is the area of the triangle formed by these three points? Find a unit vector that is perpendicular to the plane that includes these points (this vector is called the unit normal vector to the plane). Application: Torque. When we turn a bolt by applying a force F to a wrench, the torque (or moment) we produce acts along the axis of the bolt to drive the bolt forward (using the right-hand rule). The magnitude of the torque depends on how far on the wrench the force is applied and on how much of the force is perpendicular to the wrench at the point of application. The number we use to measure the torque’s magnitude is the product of the length of the lever (or moment) arm r and the scalar component of F perpendicular to r. Thus, the magnitude of the torque is || r F |r| F sin where is the angle between r and F. The product u v w is called the triple scalar product of u, v,andw (in that order). Note from the dot product we have u v w |u v||w |cos where is the angle between uv and w. The absolute value of the triple scalar product is the volume of the parallelepiped (parallelogram-sided box) determined by u, v,andw. |u v w ||u v||w ||cos| The number |u v| is the area of the base parallelogram. The number |w ||cos| is the parallelepiped’s height. Because of this geometry, the triple scalar product is sometimes called the box product. By treating the planes of v and w and of w and u as the base planes of the parallelepiped determined by u, v,andw, we see that u v w v w u w u v Since the dot product is commutative, we also have u v w u v w so that the dot and cross may be "interchanged" without altering the value of the triple scalar product. Homework for Section 9.4: Page 674 #’s 1, 3, 4, 5-21 odd 2.