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.

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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. Basic Factors of the Didactic Mathematics
(Vietnamese-French book), Ho Chi Minh National University Publishing House.
[3] Nguyen Ba Kim, 2011.Method of teaching Mathematics. University of Education
Publishing House, HNUE.
[4] Tran Van Hao et al., 2009. Geometry textbook for 12nd grade. Education Pub-
lishing House, Hanoi.
[5] Bui Van Nghi, 2008. Method of teaching Mathematics in detail. University of
Education Publishing House, HNUE.
[6] Bui Van Nghi, 2009.Applying theory of teaching to teach Mathematics. University
of Education Publishing House, HNUE.
[7] Doan Quynh et al., 2009. Geometry textbook for 12nd grade (advanced). Educa-
tion Publishing House, Hanoi.
[8] Virginia M. Warfield, 2006. Invitation to Didactique. University of Washington,
Seattle, Washington.
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