Designing a teaching situation: The cross product of two vectors concept (Geometry Textbook for 12nd Grade, Chapter 3, Lesson 2)

JOURNAL OF SCIENCE OF HNUE Vol. 57, No. 1, pp. 3-7 DESIGNING A TEACHING SITUATION: THE CROSS PRODUCT OF TWO VECTORS CONCEPT (Geometry Textbook for 12nd Grade, Chapter 3, Lesson 2) Bui Van Nghi Hanoi National University of Education Nguyen Tien Trung(∗) University of Education Publishing House, HNUE (∗)Email: trungnt@hnue.edu.vn Abstract. The cross product of two vectors is important concept which is applied in solving many problems when using the spatial coordinate method. By applying the Theory of Didactical Situations in Mathematics, the authors designed teaching situation that they call “the cross product of two vectors concept” in which students learn through situations that involve action, communication and validation. Keywords: Theory of Didactical Situations in Mathematics, cross product. 1. Introduction Making use of Theory Didactical Situations in Mathematics, we can design teaching situations that will encourage active learning in students. The cross prod- uct of two vectors is a new and somewhat special concept that is used by high school mathematics teachers. This concept can be used to solve many problems that involve the spatial coordinate method. In this paper, the authors describe a teaching situa- tion that they designed which makes use of the cross product concept to introduce a positive learning activity to students. 2. Content On basis of the Theory of Didactical Situations in Mathematics and other theories that involve active teaching methods [1, 2, 3, 8], the authors designed a scenario of a teaching situation to develop the concept of cross product of two vectors. The scenario involves the following actions: 3 Bui Van Nghi and Nguyen Tien Trung 2.1. Action 1 (Situation of action) [1] The teacher asks students to solve problem: “In space, there exists plane (ABC) with A (2;−1; 3) , B (4; 0; 1) , C (1; 1; 0). Let determine normal vector of (ABC)”. Solution: We have two vectors −→ AB = (2; 1;−2) , −→AC = (−1; 2;−3) with their direc- tion parallel to a plane (ABC). The two vectors−→ AB, −→ AC are not in the same direction, meaning that A,B,C are not along one line. So, the three points A,B,C define the plane (ABC). Let ~n(ABC) = (x; y; z) ( ~n(ABC) 6= ~0 ) be the normal vector of plane (ABC). Therefore, we have:{ ~n(ABC)⊥−→AB ~n(ABC)⊥−→AC ⇔ { ~n(ABC). −→ AB = 0 ~n(ABC). −→ AC = 0 ⇔ { 2x+ y − 2z = 0 (1) −x+ 2y − 3z = 0 (2) ⇔ { 6x+ 3y − 6z = 0 (3) −2x+ 4y − 6z = 0 (4) Subtracting (4) from (3), we have −8x+ y = 0⇔ y = 8x. Choosing x = 1; y = 8→ z = 5 and so ~n(ABC) = (1; 8; 5) being a vector which is right-angle with two vectors −→ AB, −→ AC or ~n(ABC) = (1; 8; 5) is a normal vector of plane (ABC). Therefore, we know to determine a vector which is right-angled with two given vectors. 2.2. Action 2 (Situation of action) [1] Can we make a general example? Problem: In space Oxyz, define a vector which is right-angled with two vectors ~u (a1; a2; a3) ;~v (b1; b2; b3). If a student knew the way to solve the same problem, he would do like this: Let vector ~n = (x; y; z) be right-angled with two vectors ~u;~v, therefore, we have the conditions: { ~n⊥~u ~n⊥~v ⇔ { ~n . ~u = 0 ~n .~v = 0 ⇔ { a1x+ a2y + a3z = 0 (1) b1x+ b2y + b3z = 0 (2) ⇔ { a1b3x+ a2b3y + a3b3z = 0 (3) a3b1x+ a3b2y + a3b3z = 0 (4) 4 Designing a teaching situation: the cross product of two vectors concept... Subtracting (4) from (3) and we have: (a3b1 − a1b3) x+ (a3b2 − a2b3) y = 0 ⇔ (a3b1 − a1b3) x = (a2b3 − a3b2) y. So we choose y = a3b1 − a1b3; x = a2b3 − a3b2 and apply it in (1), we have z = a1b2 − a2b1 and ~n = (a2b3 − a3b2; a3b1 − a1b3; a1b2 − a2b1) = (∣∣∣∣ a2 a3b2 b3 ∣∣∣∣ ; ∣∣∣∣ a3 a1b3 b1 ∣∣∣∣ ; ∣∣∣∣ a1 a2b1 b2 ∣∣∣∣ ) (∗) As a result, the vector sought is ~n = (a2b3 − a3b2; a3b1 − a1b3; a1b2 − a2b1) = (∣∣∣∣ a2 a3b2 b3 ∣∣∣∣ ; ∣∣∣∣ a3 a1b3 b1 ∣∣∣∣ ; ∣∣∣∣ a1 a2b1 b2 ∣∣∣∣ ) . It should be noted that if students subtract (3) from (4), we will have a normal vector: ~n′ = (a3b2 − a2b3; a1b3 − a3b1; a2b1 − a1b2) = −~n(∗∗). With this process, student will have discovered that both vectors ~n, ~n′ are normal vectors of plane (α). They are parallel to each other and can be chosen as a normal vector of plane (α). 2.3. Action 3 (Situation of action, comunication and validation) Students apply this formula to solve some problems. Exercise 1. Define a vector which is right-angled with two vectors ~u = (3; 0;−1) and ~v = (3; 3;−2). Solution: Applying formula (*) we have ~n = (3; 3; 9). We can choose vector 1 3 ~n = (1; 1; 3), a vector which is right-angled with the two given vectors. Exercise 2. Define a vector which is right-angled with the two vectors ~u = (1; 2;−1) and ~v = (−2; 4; 2). Solution: Applying formula (*) we have ~n = (0; 0; 0). So in this situation, when applying formula (*) we can not find a vector which is different from a null vector which is right-angled with two given vectors. Teacher’s guides : Certainly there are many vectors which are right-angled with the two given vectors, such as ~κ = (2;−1; 0) ; ~̟ (0; 1; 2) ; etc. We conversely see that the two vectors ~u, ~v lie along the same direction. It is not difficult to see that the two vectors ~u, ~v are parallel if and only if ~n = (0; 0; 0). Exercise 3. In a space with an axis of coordinates Oxyz, let there be four points: A (−1; 0; 3) , B (2; 1; 1) , C (2;−3; 5) , D (−1;−1; 4). Let consider whether these four points exist in one plane or not. 5 Bui Van Nghi and Nguyen Tien Trung Solution: It is easy to see that the two vectors −→ AB = (3; 1;−2) ,−→AC = (3;−3; 2) are not in the same direction. So, the three points A,B,C define a plane (ABC). We need to check whether point D belongs to plane (ABC) or not. Teacher leads students to discuss a way to find a solution to the problem: + If D ∈ (ABC) then vector −−→AD = (0;−1; 1) is right-angled with a normal vector on plane (ABC) (because vector −−→ AD has a direction which is parallel to or belongs to plane (ABC)). + If D /∈ (ABC), vector −−→AD is not right-angled with a normal vector of plane (ABC). According to formula (*), the normal vector of plane (ABC) is ~n(ABC) =− 1 4 (∣∣∣∣ 1 − 2−3 2 ∣∣∣∣ ; ∣∣∣∣ −2 32 3 ∣∣∣∣ ; ∣∣∣∣ 3 13 − 3 ∣∣∣∣ ) =− 1 4 (−4;−12;−12) = (1; 3; 3) . Then, we have ~n(ABC). −−→ AD = 0→ D ∈ (ABC). After doing the above actions, the teacher leads his students to find and pro- pose a formula and natural cross products of the two vectors by themselves: - Applying formula (*) or (**), we can define a vector which is right-angled to two given vectors. - Formulas (*) or (**) allow us to define the vector from two given vectors. That vector is called the cross product of two vectors (or vector product), sign [~u,~v]. The cross product of two vectors is a vector which is defined by the formula [~u,~v] = (a2b3 − a3b2; a3b1 − a1b3; a1b2 − a2b1) = (∣∣∣∣ a2 a3b2 b3 ∣∣∣∣ ; ∣∣∣∣ a3 a1b3 b1 ∣∣∣∣ ; ∣∣∣∣ a1 a2b1 b2 ∣∣∣∣ ) . - The two vectors ~u,~v are in the same direction if and only if the cross product of them is vector ~0. Here, with the presentation as above, we still have the concept of a cross product of two vectors which is in the same direction: the cross product of two vectors is vector ~0. - Four points A,B,C,D are in a space that belongs to one plane if[−→ AB, −→ AC ] . −−→ AD = 0. The four points A,B,C,D exist in a space that does not belong to one plane if [−→ AB, −→ AC ] . −−→ AD 6= 0. 6 Designing a teaching situation: the cross product of two vectors concept... 3. Conclusion In terms of the Theory of Didactical Situations in Mathematics, by doing designed actions, students play a role in the learning process by discovering the essential formula. Also, students will discover that the formula which they have created can be applied when seeking solutions to other problems. In the process of studying, discovering and adapting the formula, students are studied more positively by action and communication. This helps students learn how to solve problems to determine planes and lines in space and helps students can easily develop concepts related to normal vectors of planes and the direction of vectors along a line. REFERENCES [1] Guy Brousseau, 2002. Theory of Didactical situations in Mathematics, Vol. 19, Kluwer Academic Publishers. [2] Annie Bessot et al., 2009. 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