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The Cross Product of Two Vectors In Space Written by Dr. Julia Arnold Associate Professor of Mathematics Tidewater Community College, Norfolk Campus, Norfolk, VA With Assistance from a VCCS LearningWare Grant In this lesson you will learn •how to find the cross product of two vectors •how to find an orthogonal vector to a plane defined by two vectors •how to find the area of a parallelogram given two vectors •how to find the volume of a parallelepiped given three vectors Objective 1 Finding the cross product of two vectors The Cross Product The cross product of two vectors, denoted as the dot product, represents a vector. ab , unlike The cross product is defined to be for a a1 ,a2 , a3 and b b1 ,b2 , b3 a b a2b3 a3b2 , a3b1 a1b3 , a1b2 a2b1 You are probably wondering if there is an easy way to remember this. The easy way is to use determinants of size 3 x 3. a a1 ,a2 , a3 and b b1 ,b2 , b3 a b a2b3 a3b2 , a3b1 a1b3 , a1b2 a2b1 Let’s set up a 3 x 3 determinant as follows: 1. First use the unit vectors i , j, and k as the first row of the determinant. 2. Use row 2 for the components of a and row 3 for the components of b. i j k a1 a2 a3 i a2b3 j a3b1 k a1b2 i a3b2 j a1b3 k a2b1 b1 b2 b3 a2b3 a3b2 i a3b1 a1b3 j a1b2 a2b1 k Find the cross product for the vectors below. Do the problem before clicking again. a 2,4,5 i j k 2 4 5 1 2 1 and b 1,2,1 i 4 1 j 51 k 2 2 i 5 2 j 2 1 k 41 4 10i 51 2 j 4 4 k 6i 7 j 8k Since the cross product is determined by using determinants, we can understand the algebraic properties of the Cross Product which are: • u v v u Which would come from the fact that if you interchange two rows of a determinant you negate the determinant. • • u v w uv uw c u v cu v u cv • u 0 0 u 0 • uu 0 • u v w u v w Objective 2 find an orthogonal vector to a plane defined by two vectors Now that you can do a cross product the next step is to see why this is useful. Let’s look at the 3 vectors from the last problem a 2,4,5 , b 1,2,1 and a b 6,7,8 What is the dot product of a 2,4,5 with a b 6,7,8 And b 1,2,1 with a b 6,7,8 If you answered 0 in both cases, you?would be correct. Recall that whenever two non-zero vectors are perpendicular, their dot product is 0. Thus the cross product creates a vector perpendicular to the vectors a and b. Orthogonal is another name for perpendicular. Show that u v w u v w is true for u 1,2,3 , v 1,1,0 , and w 5,0,1 Solution: u 1,2,3 , v 1,1,0 , and w 5,0,1 The left hand side i j 1 1 k 0 i 0 0 (5k j 0) i j 5k 5 1 0 1,2,3 1,1,5 1 2 15 14 The right hand side i j k 1 2 3 0 3 j k (2k 0 3i ) 3i 3 j k 1 1 0 3,3,1 5,0,1 15 0 1 14 Geometric Properties of the Cross Product Let u and v be nonzero vectors and let theta) be the angle between u and v 1. u v is orthogonal to both u and v u v u (the Greek letter v Remember: orthogonal means perpendicular to Geometric Properties of the Cross Product Let u and v be nonzero vectors and let theta) be the angle between u and v 2. u v u v u v u (the Greek letter v sin Proof: Let u u1 ,u 2 , u 3 and v v1 , v 2 , v 3 u v u 2 v 3 u 3 v 2 , u 3 v1 u1v 3 , u1v 2 u 2 v1 uv u 2 v 3 u 3 v 2 2 u 3 v1 u1v 3 2 u1v 2 u 2 v1 2 This is the left hand side Let u u1 ,u 2 , u 3 and v v1 ,v 2 , v 3 u v sin u12 u 2 2 u 3 2 v12 v 2 2 v 3 2 1 cos2 Re call cos uv u v 2 so cos2 2 u v u u v 2 u v sin u v 1 2 u 2 v 2 v u v 2 u u v 2 v uv 2 2 u 2 v uv Now all we need to do is show that the stuff under the radical is the same as the square of the magnitude of the cross product. In other words, equate the radicands. 2 Let u u1 , u 2 , u3 and v v1 , v 2 , v3 u2v3 u3v 2 u3v1 u1v 3 u1v2 u2v 1 uv u 2 2 2 2 2 2 u v (u12 u22 u32 )(v12 v22 v32 ) 2 v uv 2 u12v12 u12v22 u12v32 u22v12 u22v22 u22v32 u32v12 u32v22 u32v32 (u v) 2 u1v1 u2v2 u3v3 u1v1 u2v2 u3v3 u12 v12 2u1v1u2 v2 2u1v1u3v3 u22v22 2u2v2u3v3 u32 v32 u 2 2 v (u v) 2 u12 v12 u12v22 u12v32 u22v12 u22v22 u22v32 u32v12 u32v22 u32v32 u12 v12 2u1v1u2v2 2u1v1u3v3 u22v22 2u2v2u3v3 u32v32 u12v22 u12v32 u22v12 u22v32 u32v12 u32v22 2u1v1u2v2 2u1v1u3v3 2u2v2u3v3 u12v22 2u1v1u2 v2 u22v12 u12 v32 2u1v1u3v3 u32v12 u22v32 2u2 v2u3v3 u32v22 u1 v2 u2 v1 u1 v3 u3 v1 u2v3 u3 v2 2 2 2 Which concludes the proof (Sometimes proofs are not hard but some do require patience.) 3. u v 0 if and only if u and v are multiples of each other 4. u v area of the paralleogram having u and v as adjacent sides. Proof: The area of a parallelogram is base times height. A = bh ||u|| y ||v|| sin = y/||u|| ||u||sin = y = height ||v|| y = ||v|| ||u||sin = u v Example problem for property 1 Find 2 unit vectors perpendicular to a= 2 i - j + 3k and b = -4i + 2j - k. Solution The two given vectors define a plane. The Cross Product of the vectors is perpendicular to the plane and is proportional to one of the desired unit vectors. To make its length equal to one, we simply divide by its magnitude: . Find 2 unit vectors perpendicular to a= 2 i - j + 3k and b = -4i + 2j - k. Solution i j k 2 1 3 (i 12 j 4k ) (6i 2 j 4k ) (5i 10 j ) 4 2 1 5i 10 j 25 100 125 5 5 Divide the vector by its m agnitude 5 10 1 2 5 2 5 5i 10 j i j i j i j 5 5 5 5 5 5 5 5 5 5 1 For a second unit vector simply multiply the answer by -1 5 2 5 i j 5 5 Objective 3 find the area of a parallelogram given two vectors Example for property 4 4. u v area of the paralleogram having u and v as adjacent sides. Area of a Parallelogram via the Cross Product z Show that the following 4 points define a parallelogram and then find the A area. A(4,4,6), B(4,14,6), C(1,11,2), D(1,1,2) x B C y D Solution: First we find the vectors of two adjacent sides: AB 4 4, 4 14,6 6 0, 10,0 AD 4 1, 4 1,6 2 3, 3,4 Now find the vectors associated with the opposite sides: DC 1 1,1 11,2 2 0, 10,0 AB DB 1 4,11 14,2 6 3, 3,4 AD This shows that opposite sides are associated with the same vector, hence parallel. Thus the figure is of a parallelogram. Continued The area is equal to the magnitude of the cross product of vectors representing two adjacent sides: Area = |AB XAD| i j k 0 10 0 40 i 0 j 0k 0 i 0 j 30k 40 i 0 j 30k 3 3 4 40 i 30k (40) 2 302 1600 900 2500 50 The area of the parallelogram is 50 square units. Area of a Triangle via the Cross Product Since the area of a triangle is based on the area of a parallelogram, it follows that the area would be ½ of the cross product of vectors of two adjacent sides. Find the area of the triangle whose vertices are P(4,4,6), Q(5,16,-2) R(1,1,2) Area parallelogram) = |PQ x PR|. The area of the triangle is half of this. z P R Q x Area of a Triangle via the Cross Product Continued P(4,4,6), Q(5,16,-2) R(1,1,2) Area parallelogram) = |PQ x PR| But what if we choose RP and RQ? Would the result be the same? Let’s do both and see! z P R PQ 5 4,16 4,2 6 1,12,8 PR 1 4,1 4,2 6 3,3,4 Q x RP 4 1,4 1,6 2 3,3,4 RQ 5 1,16 1,2 2 4,15,4 PQ 5 4,16 4,2 6 1,12,8 RP 4 1,4 1,6 2 3,3,4 PR 1 4,1 4,2 6 3,3,4 RQ 5 1,16 1,2 2 4,15,4 i j k 1 12 8 48 i 24 j 3k 24 i 4 j 36k 72 i 28 j 33k 3 3 4 1 1 1 1 72 i 28 j 33k (72) 2 282 332 5184 784 1089 7075 2 2 2 2 i j k 3 3 4 12 i 16 j 45k 60 i 12 j 12k 72 i 28 j 33k 4 15 4 Since this is the same vector, the magnitude would be the same also. z P R Q x The area of the triangle is ½ 7057 42 square units. The triple scalar product is defined as: u vw For u u1 i u 2 j u3 k , v v1 i v2 j v3 k , and w w1 i w2 j w3 k u1 u 2 u v w v1 v2 u3 v3 w1 w2 w3 Objective 4 how to find the volume of a parallelepiped given three vectors The triple scalar product is defined as: AB ·(AC x AD). A parallelepiped is a 6 sided figure whose sides are parallelograms. The volume of the parallelepiped can be found using the absolute value of the triple scalar product and 3 adjacent vectors. Why would the triple scalar product be the volume of the figure? Why would the triple scalar product be the volume of the figure? B Since the base is a parallelogram we could represent its area by the cross product ||AC x AD|| Recall that the area of a parallelepiped is the base times the height. The height would be equivalent to: C A D proj AC AD AB The triple scalar product is defined as: AB ·(AC x AD). base times the height B ||AC x AD|| proj AC AD AB AC AD projAC AD AB AC AD C A D AB AC AD AB AC AD AC AD The absolute value insures a positive answer for the volume. Volume of a Parallelepiped via the Scalar Triple Product: Find the volume of the parallelepiped with adjacent edges AB, AC, and AD, where the points are A(4, -3, -2), B(2, 0, 5), C(-3, 2, 1), and D(1, 3, 2). The volume is given by the scalar triple product: AB · (AC X AD). First we need the three vectors: AB = [2 - 4]i + [0 - (-3)]j + [5 - (-2)]k = -2i + 3j + 7k. AC = [-3 - 4]j + [2 - (-3)]j + [1 - (-2)]k = -7i + 5j + 3k AD = [1 - 4]i + [3 - (-3)]j + [2 - (-2)]k = -3i + 6j + 4k First, find the cross product: = i[20 - 18] - j[-28 - (-9)] + k[-42 - (-15)] = 2i + 19j - 27k Now form the dot product to get the volume: Volume = |AB · (2i + 19j - 27k)| = | (-2i + 3j + 7k) · (2i + 19j - 27k)| = | 4 + 57 - 189 | = 136 cubic units From:http://www.jtaylor1142001.net/calcjat/Solutions/VCro ssProduct/VCPVolParallelepiped.htm Finally, for all of you potential physicists, a real world application of the cross product. Torque is defined by Webster as “a twisting or wrenching effect or moment exerted by a force acting at a distance on a body, equal to the force multiplied by the perpendicular distance between the line of action of the force and the center of rotation at which it is exerted”. M If a vector force F is applied at point B of the vector AB where both vectors are in the same plane, then M is the moment of the force F about the point B. ||M|| will measure the tendency of the vector AB to rotate counterclockwise according to the right hand rule about an axis directed along the vector M. A great example of torque is tightening a bolt with a wrench. A few observations can be made: The farther away from the bolt that the force is applied the greater the magnitude of the torque. Thus, a longer wrench produces greater torque for a given amount of force. Second, we can see that the angle which produces the largest torque would be when = 90o. Recall: force lengthvector force length sin where is the angle between them. Example: Suppose you have a 12 inch wrench and you apply a 20 lb force at an angle of 30 degrees. What is the torque in foot-pounds at the bolt? What is the maximum torque that can be applied to this bolt? Solution: force lengthvector force length sin where is the angle between them. 1 201sin 20 10 foot pounds 6 2 The maximum torque is applied when 201sin 201 20 foot pounds 2 2 For comments on this presentation you may email the author Dr. Julia Arnold at [email protected]