A kosarad üres
Operations with vectors
The sum of vectors
Let us take triangle ABC along with vector \latex{\overrightarrow a} and vector \latex{\overrightarrow b}. Let us translate the triangle by \latex{\overrightarrow a}, then let us translate the resulting triangle \latex{ A’B’C’ } by \latex{\overrightarrow{b}}. (Figure 75)
Figure 75
\latex{ C }
\latex{ A }
\latex{ B }
\latex{ A' }
\latex{ B' }
\latex{ C' }
\latex{ C'' }
\latex{ B'' }
\latex{ A'' }
\latex{\overrightarrow b}
\latex{\overrightarrow a }
\latex{\overrightarrow a}
\latex{\overrightarrow a}
\latex{\overrightarrow a}
\latex{\overrightarrow b}
\latex{\overrightarrow b}
\latex{\overrightarrow b}
It can be realised that doing the two parallel translations in succession can be replaced by a single parallel translation which is defined by the vector
\latex{ \overrightarrow {AA\text{\textquotedblright}} = \overrightarrow {BB\text{\textquotedblright}} = \overrightarrow {CC\text{\textquotedblright}} }.
This remark makes it possible to define the sum of two vectors.
DEFINITION: The sum of vectors \latex{\overrightarrow a} and \latex{\overrightarrow b} (notation: \latex{\overrightarrow a+ \overrightarrow b}) is the vector of the parallel translation by which doing the parallel translations defined by \latex{\overrightarrow a} and in \latex{\overrightarrow b} succession can be replaced.
In Figure 75 it can also be seen that the sum of two vectors is independent of which directed line segments they are represented by.
Graphically we can add two vectors according to the definition as shown in Figure 76. In this case we take the two vectors (more precisely the directed line segments representing them) one after the other (the starting point of the second one is the end-point of the first one); the sum vector points from the starting point of the first vector to the end-point of the second vector. (Triangle rule of adding vectors.)
triangle rule
Figure 76
\latex{\overrightarrow a}
\latex{\overrightarrow b}
\latex{\overrightarrow a+\overrightarrow b}
Two vectors can also be added so that we take the directed line segments representing them from a common starting point. In this case the sum vector is the diagonal vector of the parallelogram formed by the two vectors with common starting point which starts from the common starting point. This is the parallelogram rule of adding vectors. (Figure 77)
Figure 77
\latex{\overrightarrow a}
\latex{\overrightarrow b}
\latex{\overrightarrow a+\overrightarrow b}
\latex{\overrightarrow b}
\latex{\overrightarrow a}
We have two notes in connection with the latter method:
It can only be applied if the two vectors are not parallel. In the case of parallel vectors the triangle rule should be followed.
It represents it very well that the sum of two vectors is independent of the order of addends, i.e. the addition of vectors is a commutative operation:
\latex{\overrightarrow a+\overrightarrow b = \overrightarrow b+\overrightarrow a}.
The addition of more vectors is done based on the triangle rule so that we take the vectors one after the other (the starting point of the next one is the end-point of the previous one); the sum vector points from the starting point of the first vector to the end-point of the last vector.
In Figure 78 it can also be seen that the addition of vectors is an associative operation, i.e. we can group the added vectors arbitrarily.
Figure 78
\latex{\overrightarrow a}
\latex{\overrightarrow b}
\latex{\overrightarrow c}
\latex{\overrightarrow d}
\latex{\overrightarrow a+\overrightarrow b+\overrightarrow c+\overrightarrow d}
\latex{(\overrightarrow a+\overrightarrow b)+\overrightarrow c}
\latex{\overrightarrow c+\overrightarrow d}
\latex{\overrightarrow a+\overrightarrow b}
\latex{\overrightarrow b+\overrightarrow c}
Example 1
Let us determine the following vectors in regular hexagon \latex{ ABCDEF } in Figure 79.
- \latex{\overrightarrow {AB}+\overrightarrow{BC}}
- \latex{\overrightarrow {EF}+\overrightarrow{ED}}
- \latex{\overrightarrow {FA}+\overrightarrow{AD}}
- \latex{\overrightarrow {EF}+\overrightarrow{DE}}
Solution
- \latex{\overrightarrow {AB}+\overrightarrow{BC}=\overrightarrow{AC}}
- \latex{\overrightarrow {EF}+\overrightarrow{ED}=\overrightarrow{EO}}
- \latex{\overrightarrow {FA}+\overrightarrow{AD}=\overrightarrow{FD}}
- \latex{\overrightarrow {EF}+\overrightarrow{DE}=\overrightarrow{DF}=-\overrightarrow{FD}}
Figure 79
\latex{ E }
\latex{ D }
\latex{ C }
\latex{ O }
\latex{ B }
\latex{ A }
\latex{ F }
The difference of two vectors
It seems to be practical to define the difference of two vectors so that on one hand it is in tune with the definition of the sum of vectors and on the other hand it is in tune with the concept of a negative vector. It means that a definition is needed based on which
\latex{(\overrightarrow a+\overrightarrow b)-\overrightarrow a=\overrightarrow b} and \latex{(\overrightarrow a+\overrightarrow b)-\overrightarrow b=\overrightarrow a},
and\latex{\overrightarrow a+(-\overrightarrow b)=\overrightarrow a- \overrightarrow b}.
Figure 80
\latex{\overrightarrow a}
\latex{\overrightarrow b}
\latex{-\overrightarrow b}
\latex{\overrightarrow a-\overrightarrow b}
\latex{\overrightarrow b}
\latex{\overrightarrow a+(-\overrightarrow b)=\overrightarrow a-\overrightarrow b}
DEFINITION: The difference of vectors \latex{\overrightarrow a} and \latex{\overrightarrow b} (notation: \latex{\overrightarrow a-\overrightarrow b}) is vector \latex{\overrightarrow a+(-\overrightarrow b)}. (Figure 80)
The difference vector (more precisely the directed line segment representing the difference vector) of two vectors is constructed graphically so that we take the two vectors from a common starting point, and the difference vector points from the end-point of the deduction vector to the end-point of the minuend vector. (Figure 81)
\latex{\overrightarrow a-\overrightarrow b}
\latex{\overrightarrow b}
\latex{\overrightarrow a}
Example 2
In parallelogram \latex{ ABCD } in Figure 82 let us determine vectors
- \latex{\overrightarrow {AB}-\overrightarrow{AD}};
- \latex{\overrightarrow {DC}-\overrightarrow{CB}};
- \latex{\overrightarrow {CD}-\overrightarrow{CB}};
- \latex{\overrightarrow {DC}-\overrightarrow{BC}}.
Solution
- \latex{\overrightarrow {AB}-\overrightarrow{AD}=\overrightarrow{DB}}
- \latex{\overrightarrow {DC}+\overrightarrow{CB}=\overrightarrow{DC}-\overrightarrow{DA}=\overrightarrow{AC}}
- \latex{\overrightarrow {CD}-\overrightarrow{CB}=\overrightarrow{BD}}
- \latex{\overrightarrow {DC}+\overrightarrow{BC}=\overrightarrow{AB}-\overrightarrow{AD}=\overrightarrow{DB}}
Figure 82
\latex{ D }
\latex{ C }
\latex{ B }
\latex{ A }
What will be the result if we add two opposite vectors or in other words if we subtract a vector from itself? According to the definitions we have given so far the magnitude (or absolute value) of the resulting vector will be \latex{ 0 }.
DEFINITION: The vector the magnitude (absolute value) of which is 0 is called zero vector. (Notation: \latex{\overrightarrow{0}})
According to the conventions the direction of \latex{\overrightarrow{0}} is arbitrary, i.e. it is unidirectional with any vector.
We mentioned it when having listed the properties of the parallel translation that if the axes of the two line reflections producing the translation coincide, then the translation is the identical transformation. By introducing the concept of zero vector it can also be said that the parallel translation by \latex{\overrightarrow{0}} is the identical transformation.
Multiplying a vector by a number
If \latex{\overrightarrow{a}} is an arbitrary vector, then \latex{\overrightarrow a+\overrightarrow a}, when determined by applying the triangle rule, is a vector which is unidirectional with \latex{\overrightarrow{a}}, and its magnitude is double the magnitude of \latex{\overrightarrow{a}}. For this reason it is practical to consider this vector as \latex{2\times \overrightarrow{a}}. (Figure 83)
We generalise the above observation when we define the product of a vector and of an arbitrary real number (scalar).
Let us consider a vector \latex{\overrightarrow{a}} and a real number \latex{\alpha}. (Now we use the letters of the Greek alphabet to denote real numbers so that we can distinguish them from the vectors.)
Let us consider a vector \latex{\overrightarrow{a}} and a real number \latex{\alpha}. (Now we use the letters of the Greek alphabet to denote real numbers so that we can distinguish them from the vectors.)
Figure 83
\latex{\overrightarrow a}
\latex{\overrightarrow a}
\latex{\overrightarrow a}
\latex{\overrightarrow a+\overrightarrow a=2\overrightarrow a}
DEFINITION:
- If \latex{\overrightarrow a \neq \overrightarrow 0}, then \latex{\alpha\times\overrightarrow a} is a vector the magnitude of which is \latex{\left|\alpha\right|\times\left|\overrightarrow a\right|}, and when \latex{\alpha\gt0} then it is unidirectional with \latex{\overrightarrow a}, and when \latex{\alpha\lt0} then it is in the opposite direction with \latex{\overrightarrow a}.
- If \latex{\overrightarrow a=\overrightarrow 0}, then \latex{\alpha\times\overrightarrow a=\overrightarrow 0} for any real number \latex{\alpha}.
Example 3
Let us take a vektor, and let us give the following vectors graphically.
- \latex{2\overrightarrow a}, \latex{\;3\overrightarrow a}, \latex{\;(2+3)\times\overrightarrow a=5\overrightarrow a};
- \latex{(-2)\times\overrightarrow a}, \latex{\;3\times(-2\overrightarrow a)}, \latex{\;(3\times(-2))\times\overrightarrow a=-6\overrightarrow a}
Figure 84
\latex{\overrightarrow a}
\latex{2\overrightarrow a}
\latex{3\overrightarrow a}
\latex{5\overrightarrow a}
Solution
The solutions of the example are shown in Figure 84-85.
Figure 85
\latex{\overrightarrow a}
\latex{-\overrightarrow a}
\latex{-2\overrightarrow a}
\latex{-2\overrightarrow a}
\latex{-2\overrightarrow a}
\latex{-2\overrightarrow a}
\latex{-\overrightarrow a}
\latex{-\overrightarrow a}
\latex{-\overrightarrow a}
\latex{-\overrightarrow a}
\latex{-\overrightarrow a}
\latex{-\overrightarrow a}
\latex{-6\overrightarrow a}
Example 4
Let us take \latex{\overrightarrow a} and \latex{\overrightarrow b} not parallel with it, and let us give the following vectors graphically:
\latex{\overrightarrow a+\overrightarrow b}, \latex{\;2\overrightarrow a+ 2\overrightarrow b}; \latex{\;2\times(\overrightarrow a+\overrightarrow b)}.
Figure 86
\latex{\overrightarrow a}
\latex{\overrightarrow b}
\latex{\overrightarrow a}
\latex{\overrightarrow a}
\latex{2\overrightarrow a}
\latex{\overrightarrow b}
\latex{\overrightarrow b}
\latex{2\overrightarrow b}
\latex{\overrightarrow a+\overrightarrow b}
\latex{\overrightarrow a+\overrightarrow b}
\latex{2\overrightarrow a+2\overrightarrow b}
Solution
The solutions of the example can be seen in Figure 86.
⯁ ⯁ ⯁
It can be shown that what we experienced while solving examples 3 and 4 are valid in general too, i.e. the below identities are satisfied when multiplying vectors by numbers (scalars):
- \latex{\alpha\times\overrightarrow a+ \beta\times\overrightarrow a=(\alpha+\beta)\times \overrightarrow a};
- \latex{\alpha\times(\beta\times \overrightarrow a)=(\alpha\times\beta)\times \overrightarrow a};
- \latex{\alpha\times(\overrightarrow a+\overrightarrow b)=\alpha\times\overrightarrow a+\alpha\times\overrightarrow b}.
Vectors in the coordinate system
In the Cartesian coordinate system we have the opportunity not only to give the vectors by drawing the directed line segments representing them but to characterise them by numbers, coordinates.
Example 5
In the Cartesian coordinate system the starting point of the directed line segment representing \latex{\overrightarrow a} is \latex{A(2; 3)}, its end-point is \latex{B(6; 5)}. Let us give the end-point of the directed line segment which also defines \latex{\overrightarrow a} and its starting point is point \latex{(0; 0)}.
Figure 87
\latex{B(6; 5)}
\latex{A(2; 3)}
\latex{(4; 2)}
\latex{ y }
\latex{ x }
-
\latex{ 1 }
\latex{ 1 }
-
Solution
In Figure 87 it can be seen that the coordinates of the end-point of the representative starting from the origin are: \latex{(4; 2)}.
⯁ ⯁ ⯁
Based on example 5 we can generalise:
If the starting point of a directed line segment producing a given vector is the origin, and its end-point is point \latex{(x_0; y_0)}, then the end-point of the directed line segment with starting point \latex{(a; b)} producing the same vector is point \latex{(x_0 + a; y_0 + b)}.
In the Cartesian coordinate system it is practical to represent every single vector by its representative the starting point of which is the origin, since this way the coordinates of the end-point of the directed line segment with the origin as its starting point unambiguously define the vector.
DEFINITION: In the Cartesian coordinate system the coordinates of a vector are the same as the coordinates of the end-point of its representative the starting point of which is the origin.
Notation: \latex{\overrightarrow a (a_1; a_2)} (Figure 88)
Notation: \latex{\overrightarrow a (a_1; a_2)} (Figure 88)
Figure 88
\latex{\overrightarrow a}
\latex{(a_1; a_2)}
\latex{ y }
\latex{ x }
-
\latex{ 1 }
\latex{ 1 }
-
Notes: Giving the coordinates of a vector is similar to selecting the fraction out of the different fraction forms of a rational number in which the numerator and the denominator are relative primes.
In the above definition we characterised a vector unambiguously by a point. It can be done the other way round too, to every point we can assign the vector which points from the origin to the point in question.
DEFINITION: In the Cartesian coordinate system the position vector of a point is the vector pointing from the origin to the point.
The coordinates of a point in the plane and the coordinates of the position vector of the point are the same.
Exercises
{{exercise_number}}. With the help of the vectors defined by the vertices of the cube in the figure do the operations below, i.e. give the sum vectors and the difference vectors with the help of the vertices of the cube.
\latex{ E }
\latex{ A }
\latex{ B }
\latex{ C }
\latex{ G }
\latex{ H }
\latex{ F }
\latex{ D }
- \latex{\overrightarrow{AB}+\overrightarrow{BC}}
- \latex{\overrightarrow{AD}+\overrightarrow{FG}}
- \latex{\overrightarrow{FB}-\overrightarrow{FG}}
- \latex{\overrightarrow{AB}-\overrightarrow{EH}}
- \latex{(\overrightarrow{AB}+\overrightarrow{CG})-\overrightarrow{EH}}
{{exercise_number}}. Take a triangle and vector different from the zero vector. Translate the triangle
- by \latex{2\overrightarrow{a}};
- by \latex{-\overrightarrow{a}};
- by \latex{\frac{1}{2}\overrightarrow{a}};
- by \latex{-\frac{3}{4}\overrightarrow{a}}.
{{exercise_number}}. In the Cartesian coordinate system \latex{\overrightarrow p} is the position vector of point \latex{P(4; 3)}. Give the coordinates of the end-point of the vector which is equal to \latex{\overrightarrow p} and the starting point of which is the point
- \latex{(1; 0)};
- \latex{(1; -1)};
- \latex{(3; 4)};
- \latex{(7; -2)};
- \latex{(-2; -3)};
- \latex{(a; b)}.
{{exercise_number}}. In the Cartesian coordinate system the vertices of triangle \latex{ ABC } are: \latex{A(0; 2)}, \latex{B(–4; –3)}, \latex{C(9; 2)}. Give the coordinates of the vector by which we translated triangle \latex{ ABC }into triangle \latex{ A’B’C’ }, if
- \latex{A'(5; 2)};
- \latex{C'(0; 0)};
- \latex{B'(-2; -1)};
{{exercise_number}}. Express \latex{\overrightarrow{AC}} and \latex{\overrightarrow{DB}} with the help of \latex{\overrightarrow{AB}} and \latex{\overrightarrow{AD}} in parallelogram \latex{ ABCD } shown in the figure.
\latex{ D }
\latex{ A }
\latex{ B }
\latex{ C }
