Το καλάθι σου είναι άδειο.
Dot product in the coordinate system
We know that we can precisely mark out a star in the sky or we can unambiguously determine a place on Earth if we give the coordinates that can be assigned to the given object in the corresponding coordinate system.
These positions can be given also with the help of the corresponding position vectors. However the position vectors can be also expressed by the linear combination of the base vectors of the coordinate system, and these base vectors can unambiguously be given with the help of the coordinates corresponding to the point. (Figure 17)
According to this an arbitrary position vector can be given in terms of the base vectors in the following form:
\latex{\vec{v}=x\bold{i} + y\bold{j}}.
\latex{y}
\latex{y\times \bold{j}}
\latex{\bold{j}}
\latex{0}
\latex{\bold{i}}
\latex{x \times\bold{i}}
\latex{x}
\latex{\vec{v}}
\latex{P\left(x;y \right)}
Figure 17
The one-to-one correspondence implies that both the position vector and the point marked out by it can be described with the coordinate pair \latex{\left(x; y\right)}. It can be expressed with notations in the following form:
\latex{P\left(x;y\right) \quad} or \latex{\quad \vec{v}\left(x;y\right)}
The length of a position vector \latex{\vec{v}} can be obtained easily if we apply the Pythagorean theorem for the right-angled triangle shown in the figure. (Figure 18)
It implies that
\latex{y}
\latex{x}
\latex{x}
\latex{\vec{v}}
\latex{y}
Figure 18
\latex{|\vec{v}|=\sqrt{x^2+y^2}}.
Example 1
Let us calculate the length of the vector \latex{\vec{a}\left(8;-6\right)}.
Solution I
If we represent the vector with the origin as the starting point in the coordinate system, then the coordinates of its end-point are \latex{\left(8; –6\right)}. (Figure 19)
In the figure it can be seen that the length of \latex{\vec{a}} is the length of the hypotenuse of a right-angled triangle with \latex{6}- and \latex{8}-unit-long legs. Let us apply the Pythagorean theorem.
\latex{|\vec{a}|=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10}.
Solution II
As we could see it earlier the dot product of a vector with itself – based on the definition – is equal to the square of the magnitude of the vector. Thus
\latex{|\vec{a}|^2=\vec{a} \times \vec{a} = 8^2+\left(-6\right)^2=100}, which implies \latex{|\vec{a}|=10.}
\latex{y}
\latex{0}
\latex{x}
\latex{-6}
\latex{-1}
\latex{8}
\latex{6}
\latex{\left(8;-6\right)}
\latex{\vec{a}\left(8;-6\right)}
\latex{8}
\latex{1}
Figure 19
In general:
The length (magnitude) of the vector \latex{\vec{a} \left(a_1; a_2\right)} is:
\latex{|\vec{a}|=\sqrt{{a_1}^2+{a_2}^2}}.
We know that base vectors are unit vectors and perpendicular to each other. It implies that their dot product is \latex{0}.
It is known that the operations of vectors can also be defined with the help of coordinates.
Let us examine the relationship between the coordinates and the dot product of the vectors.
Let us consider two vectors in the coordinate system according to the figure (Figure 20).
By applying the characteristics of the dot product operation:
\latex{\vec{a} \times \vec{b}=\left(a_1\bold{i}+a_2\bold{j}\right)\times\left(b_1\bold{i}+b_2\bold{j}\right)=}
\latex{=a_1 b_1\bold{i}\times\bold{i}+a_1 b_2\bold{i}\times\bold{j}+a_2 b_1\bold{j}\times\bold{i}+a_2 b_2\bold{j}\times\bold{j}}.
\latex{=a_1 b_1\bold{i}\times\bold{i}+a_1 b_2\bold{i}\times\bold{j}+a_2 b_1\bold{j}\times\bold{i}+a_2 b_2\bold{j}\times\bold{j}}.
Let us use now that \latex{\bold{i}^2 = \bold{j}^2 = 1} and \latex{\bold{i} \times \bold{j} = 0}, therefore the values of the two middle terms in the right-hand-side expression are \latex{0}, i.e. the dot product has the following simple form in terms of the coordinates:
\latex{\vec{a}\left(a_1;a_2\right)}
\latex{\vec{b}\left(b_1;b_2\right)}
\latex{\alpha}
\latex{y}
\latex{x}
Figure 20
\latex{\vec{a}\times\vec{b} = a_1 b_1 + a_2 b_2}.
Let us apply the definition of the dot product:
\latex{\vec{a}\times\vec{b} = |\vec{a}| \times |\vec{b}| \times \cos \alpha = \sqrt{{a_1}^2+{a_2}^2}\times\sqrt{{b_1}^2 \times {b_2}^2}\times \cos \alpha}.
The result makes it possible to determine the angle of inclination of two vectors in the coordinate system since:
\latex{\sqrt{{a_1}^2+{a_2}^2}\times\sqrt{{b_1}^2 \times {b_2}^2}\times \cos \alpha=a_1 b_1 + a_2 b_2}.
It implies that it is true for the angle of inclination of two vectors that:
\latex{\cos \alpha = \frac{a_1 b_1 + a_2 b_2}{\sqrt{{a_1}^2+{a_2}^2}\times\sqrt{{b_1}^2 \times {b_2}^2}}}.
Example 2
Let us determine the value of \latex{x}, if we know that the vectors \latex{\vec{a}\left(3; 2\right)} and \latex{\vec{b}\left(x; -2\right)} are perpendicular to each other.
Solution
Since the two vectors are perpendicular, their dot product is \latex{0}.
By expressing it with the help of the coordinates:
\latex{\vec{a} \times \vec{b}=3x+2 \times \left(-2\right)=3x -4 = 0}.
Expressing \latex{x} from the equation implies that \latex{x=\frac{4}{3}}, i.e.: \latex{\vec{b}\left(\frac{4}{3};-2\right)}.
Example 3
Let us give the angle included between the vectors \latex{\vec{a}\left(3;0\right)} and \latex{\vec{b}\left(1;\sqrt{3}\right)}.
Solution
Let us write down the dot product of the vectors with the help of the coordinates and then with the help of the definition:
\latex{\vec{a} \times \vec{b}=3 \times 1 + 0 \times \sqrt{3}=3};
\latex{\vec{a} \times \vec{b}= \sqrt{9} \times \sqrt{1 + 3} \times \cos \alpha = 6 \times \cos \alpha}.
Since the two values need to be equal to each other:
\latex{6 \times \cos \alpha =3}, i.e. \latex{\cos \alpha = \frac{1}{2}\left(0°\le \alpha \le 180°\right)}.
It implies that the measure of the angle included between the two vectors is: \latex{\alpha = 60º}. (Figure 21)
\latex{y}
\latex{0}
\latex{x}
\latex{\vec{e}}
\latex{60°}
\latex{\frac{1}{2}}
Figure 21
Example 4
Let us verify that the following inequality holds for arbitrary real numbers:
\latex{ab+cd \le \sqrt{a^2+c^2}\times \sqrt{b^2+d^2}}.
Solution
Let us realise that the expressions of the inequality can also be found in the dot product of the vectors.
Let us consider the following two vectors: \latex{\vec{u}\left(a;c\right)} and \latex{\vec{v}\left(b;d\right)}. Let us express the dot product of these both ways:
\latex{\vec{u}\times \vec{v} = ab+cd = \sqrt{a^2+c^2}\times \sqrt{b^2+d^2}\times \cos \alpha \le \sqrt{a^2+c^2}\times \sqrt{b^2+d^2}}
When writing down the inequality we considered the range of the cosine function, i.e. that \latex{\cos \alpha\le1}. It verifies our statement.
It is also worth examining under which conditions the equality holds. It is also true if \latex{\cos \alpha = 1}, i.e. \latex{\alpha = 0º}. For the vectors written down this condition means that they are unidirectional vectors, i.e. one of them can be derived from the other one by multiplying it with a suitable positive real number.
Let this real number be \latex{\lambda\ne0}.
Then \latex{\vec{u}\left(a;c\right)=\lambda \times \vec{v}\left(b;d\right)}. Thus it is true for the corresponding coordinates that
\latex{a=\lambda \times b} and \latex{c = \lambda\cdot d}, which implies \latex{\frac{a}{b}=\frac{c}{d}}
as the condition of the equality, if none of the denominators is \latex{0}.
It can easily be seen that in the case of \latex{b = 0 \space a = 0}, and in the case of \latex{d = 0 \space c = 0} equality is obtained.
Notes:
- Any vector of the space can unambiguously be expressed as the linear combination of the base vectors of a spatial coordinate system. (Figure 22)
For the base vectors \latex{\bold{i}}, \latex{\bold{j}} and \latex{\bold{k}} it is true that
\latex{\bold{i}^2 =\bold{j}^2=\bold{k}^2=1};
\latex{\bold{i}\times \bold{j}=\bold{i}\times \bold{k}=\bold{j}\times \bold{k}=0}.
\latex{\bold{i}\times \bold{j}=\bold{i}\times \bold{k}=\bold{j}\times \bold{k}=0}.
- The length (magnitude) of a vector \latex{\vec{v}} is:
\latex{|\vec{v}|=\sqrt{x^2+y^2+z^2}}.
- The dot product of two vectors is as follows if \latex{\vec{v}=x_1 \times \bold{i}+y_1 \times \bold{j}+z_1\times \bold{k}}, \latex{\vec{u}=x_2 \times \bold{i}+y_2 \times \bold{j}+z_2\times \bold{k}}:
\latex{z}
\latex{z\cdot \bold{k}}
\latex{\bold{j}}
\latex{\bold{i}}
\latex{x \cdot\bold{i}}
\latex{x}
\latex{\vec{v}}
\latex{P\left(x;y;z \right)}
\latex{\bold{k}}
\latex{y \cdot\bold{j}}
\latex{y}
Figure 22
\latex{\vec{u}\times \vec{v} = |\vec{u}|\times |\vec{v}|\times \cos \phi},
where \latex{\phi} is the angle included between the two vectors, or
\latex{\vec{u}\times \vec{v}=x_1 \times x_2+y_1 \times y_2 + z_1 \times z_2}.
- With the method applied in example \latex{4} it can be verified that it is true for arbitrary real numbers \latex{x_1, x_2, y_1, y_2, z_1, z_2} that:
\latex{\sqrt{{x_1}^2+{y_1}^2+{z_1}^2} \times \sqrt{{x_2}^2+{y_2}^2+{z_2}^2} \ge x_1 \times x_2 + y_1 \times y_2 + z_1 \times z_2}.
Exercises
{{exercise_number}}. The vectors \latex{\vec{a}\left(3; 1\right)} and \latex{\vec{b}\left(2; -1\right)} are given. Determine the r esults of the following operations:
- \latex{\vec{a}\times \vec{b}};
- \latex{\left(\vec{a}+\vec{b}\right) \times \vec{a}};
- \latex{\left(\vec{a}-2 \times \vec{b}\right)^2};
- \latex{\left(\vec{a}+\vec{b}\right) \times \left(\vec{a}-\vec{b}\right)}.
{{exercise_number}}. Determine the angle of inclination of the following two vectors:
- \latex{\vec{a}\left(3;1\right)} and \latex{\vec{b}\left(-1;3\right)};
- \latex{\vec{a}\left(3;1\right)} and \latex{\vec{b}\left(-3;-1\right)};
- \latex{\vec{a}\left(3;1\right)} and \latex{\vec{b}\left(2;-1\right)};
- \latex{\vec{a}\left(3;1\right)} and \latex{\vec{b}\left(-2;-1\right)}.
{{exercise_number}}. Give the value of the \latex{x}-coordinate in the following vectors so that the given vectors will be perpendicular to each other.
- \latex{\vec{a}\left(3;1\right)} and \latex{\vec{b}\left(1;x\right)}
- \latex{\vec{a}\left(3;1\right)} and \latex{\vec{b}\left(2x-2;-1\right)}
- \latex{\vec{a}\left(3;3x-5\right)} and \latex{\vec{b}\left(2;-1\right)}
- \latex{\vec{a}\left(x^2-x+1;1\right)} and \latex{\vec{b}\left(-2;-1\right)}
{{exercise_number}}. Give the coordinates of the unit vector unidirectional with the vector \latex{\vec{a}\left(3; 4\right)}.
{{exercise_number}}. Two points are given in the coordinate system: \latex{A\left(1; 5\right)} and \latex{B\left(6; 2\right)}. Give the angles of the triangle \latex{OAB}, where \latex{O} denotes the centre of the coordinate system.
{{exercise_number}}. Give the measure of the work done by the force \latex{\vec{F}\left(3; 7\right)}, if there is a displacement of \latex{\vec{s}\left(15; 6\right)}.
{{exercise_number}}. Verify the following inequalities. When is the equality satisfied?
- \latex{4a+3b\le5 \times \sqrt{a^2 + b^2}}
- \latex{4a + 3b \le \sqrt{a^2+9}\times \sqrt{b^2+16}}