Calculus 3 Section 12.4 Reading Assignment: The Cross Product

Answer Only Exercise 3 by using a screenshot provided Calculus Pearson textbook. Make sure you read these three questions very carefully and see on what it is asking for and what is really about. Please be very serious careful with this assignment of exercise #3.

References: Thomas' Calculus: Early Transcendentals | Calculus | Calculus | Mathematics | Store | Pearson+

Exercise 3. Read Example 3 (p. 739). Explain how the geometric meaning of the cross product helps solve this example. We won't be covering the last two subsections, but they might be worth taking a look at. The subsection "Torque" (p. 739-740) covers a physical application of the cross product. While the last subsection "Triple or Box Product" (p. 740 -741) goes over a geometric way that the dot product and cross product interact with each other. `To elaborate, it's easier to compute the dot product and magnitudes of the two vectors rather than computing the cross product and then taking its magnitude. Another issue is that arcsin does not give obtuse angles, which means that this harder method will also give you the wrong answer if the angle is greater than 90 degrees.Chapter 12 Vectors and Geometry of Space 12.4 The Cross Product 739 Solution The vector PQ X PR is perpendicular to the plane because it is perpendicular to both vectors. In terms of components, PQ = (2 - Di + (1 + 1)j + (-1 - 0jk = i + 2j - k PR = (-1 - 1)i + (1 + 1)j + (2 - 0)k = -2i + 2j + 2k i j PQ X PR = 1 2 -2 2 ! - 13 21- 1-2 - 21 + 1-2 3 /k = 61 + 6k. EXAMPLE 3 Find the area of the triangle with vertices P(1, -1, 0), ((2, 1, -1), and R(-1, 1, 2) (Figure 12.32). Solution The area of the parallelogram determined by P. Q. and R is |PQ x PR) = 161 + 6k| Values from Example 2 = V(63+ (6) = V2.36 = 6V/2. The triangle's area is half of this, or 3V 2. EXAMPLE 4 Find a unit vector perpendicular to the plane of P(1. -1, 0). Q(2, 1, -1). and R(-1. 1. 2). Solution Since PQ X PR is perpendicular to the plane, its direction n is a unit vector perpendicular to the plane. Taking values from Examples 2 and 3, we have PQ X PR _ 61 + 6k |PQ X PRI 6V2 VI For ease in calculating the cross product using determinants, we usually write vectors in the form v = uji + wj + v;k rather than as ordered triples v = (v1, 12, 1).Torque When we turn a bolt by applying a force F to a wrench (Figure 12.33), we produce a torque that causes the bolt to rotate. The torque vector points in the direction of the axis Torque of the bolt according to the right-hand rule (so the rotation is counterclockwise when viewed from the rip of the vector). The magnitude of the torque depends on how far out 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 arm r and the scalar component of F perpendicular to r. In the notation of Figure 12.33, Magnitude of torque vector = r| |F| sine, Component of F or r X F . If we let n be a unit vector along the axis of the bolt in the direction of the Perpendicular to Is ... torque, then a complete description of the torque vector is r X F, or Its length is | F| sin A. F Torque vector = r X F = (r| |F| sing) n. Recall that we defined u X v to be 0 when u and v are parallel. This is consistent with the FIGURE 12.33 The torque vector torque interpretation as well. If the force F in Figure 12.33 is parallel to the wrench, mean- describes the tendency of the force F to ing that we are trying to turn the bolt by pushing or pulling along the line of the wrench's drive the bolt forward. handle, the torque produced is zero.Chapter 12 Vectors and Geometry of Space 740 Chapter 12 Vectors and the Geometry of Space 3 ft bar EXAMPLE 5 The magnitude of the torque generated by force F at the pivot point P in Figure 12.34 is 20 1b |PQ X F| = [PQ| |F| sin 70" = (3)(20)(0.94) = 56.4 ft-lb. magnitude force In this example the torque vector is pointing out of the page toward you. FIGURE 12.34 The magnitude of the torque exerted by F at P is about 56.4 ft-lb (Example 5). The bar rotates counter- Triple Scalar or Box Product clockwise around P. The product (u x v) . w is called the triple scalar product of u, v, and w (in that order). As you can see from the formula [( x v) w = [ux v/w/cose). the absolute value of this product is the volume of the parallelepiped (parallelogram-sided box) determined by u, v, and w (Figure 12.35). The number u X v is the area of the base parallelogram. The number |w |cos #| is the parallelepiped's height. Because of this geometry, (u x v) . w is also called the box product of u, v. and w. Height = /w/ cas @ Area of base Volume = area of base . height -(ux v) . w/ FIGURE 12.35 The number (u X v). w| is the volume of a parallelepiped.By treating the planes of v and w and of w and u as the base planes of the parallelepi- ped determined by u, v, and w, we see that (uxv) . w = (v x w) . u = (w xu) * v. The dot and cross may be interchanged Since the dot product is commutative, we also have in a triple scalar product without altering its value. (uxv) . w = u (v X w). The triple scalar product can be evaluated as a determinant: (u X V) . W = k . W = WI + Wa 14 WI WaChapter 12 Vectors and Geometry of Space 12.4 The Cross Product 741 Calculating the Triple Scalar Product as a Determinant WI (u X V) . W 11 Any two rows of a matrix can be inter- changed without changing the absolute EXAMPLE 6 Find the volume of the box (parallelepiped) determined by value of the determinant. So we can take u = i + 2j - k. v = -21 + 3k, and w = 7j - 4k. the vectors u, v. w in any order when calculating the absolute value of the Solution Using the rule for calculating a 3 X 3 determinant, we find triple product. (u X V) 'W -2 -4 + (-1) -23. 0 7 The volume is |(u X v) . w| = 23 units cubed