## 03. Iloczyn skalarny - ang. (The scalar product)

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The scalar product

We have already got to know a little bit about vectors. We know what they are and what they can be used for. We also know that on the vectors we can perform actions such as adding and subtracting, and multiplying the vector by the number which is a scalar. Now we can see that the vector can be multiplied by another vector, and this can be performed in two ways. Here we will learn the multiplication, which results in giving a number (scalar) and it is not a vector. Such a product is called scalar product.

Scalar product of vectors $\vec{a}$ and $\vec{b}$ is called a number (scalar) defined by the formula

${\vec{a} \circ \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot \cos\varphi,}$

where $\varphi$is the angle between the vectors $\vec{a}$i $\vec{b}$ with a common beginning.

Figure 1. Illustration of the scalar product

The scalar product of two vectors is therefore equal to the product of their lengths multiplied by the cosine of the angle between them. At the same time

$\vec{a} \circ \vec{b} = |a| \cdot |b| \cdot \cos\varphi = |b| \cdot |a| \cdot \cos\varphi = \vec{b} \circ \vec{a}.$

THE PROPERTY of scalar product:

The alternation right of the scalar product (W1)

${\vec{a} \circ \vec{b} = \vec{b} \circ \vec{a}.}$

The communication right with scalar multiplication by a scalar (W2)

$k\cdot(\vec{a}\circ\vec{b})=(k\cdot\vec{a})\circ\vec{b}.$

The separation right (W3)

$\vec{a}\circ(\vec{b}+\vec{c})=\vec{a}\circ\vec{b}+\vec{a}\circ\vec{c}.$

Example

Calculate the scalar product of vectors with given lengths of $|\vec{a}|=\sqrt{10}$i $|\vec{b}|=\sqrt{40}$ attached at the same point, knowing that the angle between them is $60^{\circ}$.

Solution:

$\vec{a} \circ \vec{b} = \sqrt{10}\cdot\sqrt{40}\cdot\cos 60^{\circ} =\sqrt{400}\cdot\frac{1}{2}=10.$

Example

Calculate the scalar product $\vec{a} \circ \vec{a}$.

Solution:

$\vec{a}^2 = \vec{a} \circ \vec{a} = |\vec{a}|\cdot |\vec{a}|\cdot\cos 0^{\circ} = |\vec{a}|^2.$

• Scalar product $\vec{a}\circ\vec{a}$we mark $\vec{a}^2$and call a square vector.

• Square vector is equal to the square of its length, which is$\vec{a}^2=|\vec{a}|^2$.

To calculate the value of the scalar products we need to know the length of the two vectors and the angle between them. There is a second way to calculate the scalar product using the  coordinates of vectors.

Scalar product of vectors in the plane $\vec{a}$$\vec{b}$respectively with  the coordinates $[x_a,y_a]$ and $[x_b,y_b]$we call the number (scalar ) defined by the formula

$\vec{a} \circ \vec{b} = x_a \cdot x_b + y_a \cdot y_b.$

a) on the plane                                                               b) in the space
Figure 2. The versors of the coordinates sets

In a similar way, we can also calculate the dot product of vectors in three-dimensional space.

Scalar product of vectors in three-dimensional space $\vec{a}=[x_a,y_a,z_a]$i $\vec{b}=[x_b,y_b,z_b]$ we call a number (scalar) defined by the formula

$\vec{a} \circ \vec{b} = x_a \cdot x_b + y_a \cdot y_b + z_a \cdot z_b.$

Since we know that the square of the length of the vector $\vec{a}=[x_a,y_a]$ is equal to the product of the scalar, ie

$|\vec{a}|^2 = \vec{a} \circ \vec{a} = x_a \cdot x_a + y_a \cdot y_a = x_a^2+y_a^2$

then we have already familiar formula for the length of the vector $\vec{a}$

$|\vec{a}|=\sqrt{x_a^2+y_a^2}.$

Similar relations appear for vectors in three-dimensional space

$|\vec{a}|^2 = \vec{a} \circ \vec{a} = x_a \cdot x_a + y_a \cdot y_a + z_a \cdot z_a = x_a^2+y_a^2+z_a^2$

so the length of the vector $\vec{a}$ we can calculate depending on$|\vec{a}|=\sqrt{x_a^2+y_a^2+z_a^2}.$

Example

Calculate the scalar product of vectors $\vec{a}=[1,-3]$i $\vec{b}=[-6,2]$ acting at the same point.

Solution:

This time we will use the second method of calculating the scalar product.

$\vec{a} \circ \vec{b} = 1 \cdot (-6) + (-3) \cdot 2 = (-6) + (-6) = -12.$

Example:

Calculate the scalar product of vectors $\vec{a}=[1,-3,3]$i $\vec{b}=[-6,2,4]$ acting at the same point.

Solution:

This time we will also use the second method of calculating the scalar product.

$\vec{a} \circ \vec{b} = 1 \cdot (-6) + (-3) \cdot 2 + 3 \cdot 4 = (-6) + (-6) + 12 = 0.$

It should be noted that if the vectors are perpendicular, and the angle between them is $90^{\circ}$ then $\cos 90^{\circ}=0$, and at the same time the scalar product is 0. A similar conclusion can also be drawn in the opposite direction, that is, if the scalar product of vectors is zero it means that these vectors are perpendicular.

• The condition of orthogonality of vectors, ie $\vec{a} \circ \vec{b} = 0,$jeżeli $\vec{a} \perp \vec{b}$.

Let’s move in our discussion one step further. If the angle between the vectors is sharp, ie $\varphi\in (0^{\circ}, 90^{\circ})$ then $\cos\varphi > 0$(cosine of the angles in the first quarter has the positive value) and therefore the whole scalar product $\vec{a} \circ \vec{b} = |a| \cdot |b| \cdot \cos\varphi > 0$, as the length of the vectors are always positive.

If we think in a similar way we can find that if the angle between the vectors is open, ie $\varphi\in (90^{\circ}, 180^{\circ})$ then $\cos\varphi < 0$(cosine angles in the second quarter are of negative values) and therefore the whole scalar product $\vec{a} \circ \vec{b} = |a| \cdot |b| \cdot \cos\varphi < 0$.

The summary of two different ways of calculating the scalar product allows us to calculate the angle, and actually cosine of the angle between vectors with known coordinates.

The angle $\varphi$ between vectors $\vec{a}=[x_a,y_a]$i $\vec{b}=[x_b,y_b]$can be calculated as follows

$\cos\varphi=\frac{\vec{a}\circ\vec{b}}{|\vec{a}|\cdot|\vec{b}|}=\frac{x_a \cdot x_b + y_a \cdot y_b}{|\vec{a}|\cdot|\vec{b}|}.$

Later in order to simplify the recording and increase the readability instead of $\vec{a}\circ\vec{b}$we will write in a shorter way $\vec{a}\vec{b}$.

Example

Calculate $(2\vec{a}-3\vec{b})5\vec{c}$.

Solution:

calar multiplication can be performed as usual multiplication of polynomials linear, so using the qualities of W2 i W3 we have:

$(2\vec{a}-3\vec{b})5\vec{c}=10\vec{a}\vec{c}-15\vec{b}\vec{c}.$

Example

Calculate $(2\vec{a}-3\vec{b})(5\vec{c}+6\vec{d})$.

Solution:

Doing the activities in the same way as in the previous example, we obtain:

$(2\vec{a}-3\vec{b})(5\vec{c}+6\vec{d})=10\vec{a}\vec{c}-15\vec{b}\vec{c}+12\vec{a}\vec{d}-18\vec{b}\vec{d}.$

Example

Calculate $(\vec{a}-\vec{b})^2$.

Solution:

Figure 3. The difference of vectors

As before we can calculate:

$(\vec{a}-\vec{b})^2=(\vec{a}-\vec{b})(\vec{a}-\vec{b})=\vec{a}^2-2\vec{a}\vec{b}+\vec{b}^2.$

Bearing in mind that the square vector is equal to the square of its length, and how we calculate the scalar product, we can write:

$|\vec{a}-\vec{b}|^2=|\vec{a}|^2-2|\vec{a}||\vec{b}|\cos\varphi+|\vec{b}|^2,$

where $\varphi$ is an angle between vectors $\vec{a}$i $\vec{b}$.

Let’s mark the difference of vectors $\vec{a}-\vec{b}$ from the previous example as $\vec{c}$then the last result can be written as:

$|\vec{c}|^2=|\vec{a}|^2-2|\vec{a}||\vec{b}|\cos\varphi+|\vec{b}|^2.$

Now we can look at the layout of our vectors as the triangle whose sides are equal in length, respectively to the lengths of individual vectors, ie $|\vec{a}|=a$, $|\vec{b}|=b$oraz $|\vec{c}|=c$then we obtain

$c^2=a^2+b^2-2ab\cos\varphi,$

where $\varphi$is the angle between the sides of a triangle with lengths $a$ and $b$.

Figure 4. The illustration of the cosine rule

his compound is commonly known as the cosine theory or it can be called Carnot's theory. This theory allows to calculate the length of the third side knowing the length of two sides of a triangle and the angle between them and to calculate the angle between any two sides knowing the length of the three sides of the triangle. In addition, it is worth noting that if the angle between the sides (vectors) is $90^{\circ}$then this rule takes the form of a well-known Pythagorean theory. In other words, the Pythagorean theory is a special case of a more general cosine theory.

Let's look at another concept discussed earlier. Vector lying on the axis, whose length is equal to the unit of length and the direction is consistent with the direction of the axis is called the axis or unit vector of the axis. The unit vectors of another axis $OX$, $OY$, $OZ$ of the coordinate system is denoted by $\vec{i}$, $\vec{j}$oraz $\vec{k}$. So the vector$\vec{a}$ long $|\vec{a}|=a$for example, lying on the axis $OX$can be written as $a\vec{i}$, that is $\vec{a}=a\vec{i}$.

Note also that

$\vec{i}^2=\vec{i}\circ\vec{i}=1\cdot 1\cdot\cos 0^{\circ}=1, \;\;\;\;\;\vec{j}^2=\vec{j}\circ\vec{j}=1\cdot 1\cdot\cos 0^{\circ}=1$

and

$\vec{i}\circ\vec{j}=\vec{j}\circ\vec{i}=1\cdot1\cdot\cos 90^{\circ}=0.$

However, in the case of three-dimensional space

$\vec{i}\circ\vec{i}=1 \;\;\;\;\; \vec{j}\circ\vec{j}=1 \;\;\;\;\; \vec{k}\circ\vec{k}=1$

and

$\vec{i}\circ\vec{j}=\vec{j}\circ\vec{i}=0 \;\;\;\;\; \vec{i}\circ\vec{k}=\vec{k}\circ\vec{i}=0 \;\;\;\;\; \vec{j}\circ\vec{k}=\vec{k}\circ\vec{j}=0.$

Figure 5. Vector’s coordinates

Marking by $\vec{i}$, $\vec{j}$, $\vec{k}$ versors of axis we obtain

$\vec{a_x}=a_x\vec{i} \;\;\;\;\; \vec{a_y}=a_y\vec{j} \;\;\;\;\; \vec{a_z}=a_z\vec{k}$

and since $\vec{a}=\vec{a_x}+\vec{a_y}+\vec{a_z}$, so

$\vec{a}=a_x\vec{i}+a_y\vec{j}+a_z\vec{k}$

what as we know  we can write briefly

$\vec{a}=[a_x,a_y,a_z].$

Now we calculate the scalar product of vectors $\vec{a}=a_x\vec{i}+a_y\vec{j}+a_z\vec{k}$i $\vec{b}=b_x\vec{i}+b_y\vec{j}+b_z\vec{k}$.

$\vec{a}\circ\vec{b}=(a_x\vec{i}+a_y\vec{j}+a_z\vec{k})\circ(b_x\vec{i}+b_y\vec{j}+b_z\vec{k})=a_xb_x+a_yb_y+a_zb_z.$

Interesting information and notes

Scalar product is widely used in economics. Take two vectors. The first vector $\bf x$ should represent a basket of goods $x_1$, $x_2$, $x_3$, and the second vector $\bf c$should represent prices of $c_1$, $c_2$, $c_3$ goods from this basket. Then, to calculate the value of the basket you can use the scalar product.

${\bf c}\circ{\bf x}=c_1x_1+c_2x_2+c_3x_3.$

Scalar product commonly occurs in physics. We can use it to describe e.g. work $$W$$ defined as the product of force $\vec{F}$, changes $\vec{L}$ and the cosine of the angle $\varphi$between the vector of force and the vector of displacement.

{/tex}W=|\vec{F}| \cdot |\vec{L}| \cdot \cos\varphi {tex}

or in a short way using vector notation

{/tex}W=\vec{F} \circ \vec{L}. {tex}

Another example can be the description of the flow of electrostatic of the force field (which is the Gauss's law, one of the basic laws of electromagnetism). The flux of the intensity of electrostatic field $\Phi$ is the scalar product of the vector of the intensity field $\vec{E}$ and the vector of the surface $\vec{S}$

{/tex}\Phi=\vec{E} \circ \vec{S}, {tex}

or a little longer

{/tex}\Phi=|\vec{E}| \cdot |\vec{S}| \cdot \cos\varphi {tex}

where $\varphi$ is the angle between the vector of the intensity field and the space vector. Of course, the examples of physics can be multiplied indefinitely, but let’s stop at these few. It's hard to imagine actually the classical mechanics or quantum mechanics without the usage of this kind of writing.

Literature:

Bronsztejn IN, Siemiendiajew, Mathematics, Encyclopedic handbook, McGraw-Hill, London, 1999

Empacher AB, Vulture Z., Żakowska A., W. Zakowski, A small mathematical dictionary, Wiedza Powszechna, Warsaw 1975

Antoniewicz R., Misztal A., Mathematics for economics student , PWN, Warsaw, 2003

Jurlewicz T., Skoczylas Z., Algebra and analytic geometry, GIS Publishing House, Wroclaw, 2011

CHECK what you KNOW

1. Scalar product is marked by the symbol:

• $\cdot$
• $\times$

• $\circ$.

2. The scalar product is:

• vector
• the number
• angle.

.

3. The angle between the vectors$\vec{a}=[1,-3,4]$ and $\vec{b}=[-1,0,-2]$ is

• right

• acute

• open..

Student’s note:

1. Till nowadays, perhaps we defined the product very differently, for example by using the symbols "$\cdot$", "$\circ$" or "$\times$". Now we have to do it with more precision because, in fact, each of these three symbols mean other activities such as "$\cdot$" - the product of the numbers, "$\circ$" - the scalar product of vectors, and "$\times$" - the vector product of vectors and we will talk about it in the next section.
2. $\vec{a}\circ\vec{b}=0 \Leftrightarrow \varphi=90^{\circ}$(right angle); $\vec{a}\circ\vec{b}>0 \Leftrightarrow \varphi<90^{\circ}$(kąt ostry), $\vec{a}\circ\vec{b}<0 \Leftrightarrow \varphi>90^{\circ}$(obtuse angle)
3. In order to, knowing the coordinates of the vertices of the triangle, establish which triangle we have to determine we have to calculate the coordinates of the individual vectors and then calculate the scalar products between each pair of vectors (note that the tested vectors should be engaged in the same top of the triangle. If the first tested angle is right or open, this means that the triangle is right-angled or open-angled respectively. Otherwise, it is necessary to examine the next angle. If another angle is right or open it decides about the case. If this angle was also acute then we need to examine a third angle so we can ultimately determine with which triangle we are dealing with.
4. Sometimes vectors are denoted by bold for example, $\bf a$i $\bf b$, then the scalar product would be written as${\bf a}\circ{\bf b}$.
6. Try to identify as many other ways of the scalar product to describe o known rights, claims, correlations.

Teacher's notes:

1. Scalar product, for example in the economy can be determined in other ways such as $({\bf c,x})$ or $({\bf c|x})$.
2. The basket does not have to consist of three good, it is worth generalizing to$n$ goods. Then, with vector $\bf x$dóbr $x_1$, $x_2$, $\dots$, $x_n$ and vector$\bf c$the prices of goods of this basket $c_1$, $c_2$, $\dots$, $c_n$ the value of the basket is calculated

$({\bf c|x})=c_1x_1+c_2x_2+ \cdots +c_nx_n.$

3. Explain the concepts of such terms as  goods and a basket of goods..

4. In economics,$\bf x$ is called a vector of goods while $\bf c$a covector of the prices of the basket.

5. It is worth introducing the symbol of a sum, then

$({\bf c|x})=\sum^n_{i=1}c_ix_i.$

6. Discuss how to identify a well-known type of triangle vertices.
7. Give other examples of using of scalar products (especially in physics or economics).
8. Encourage students to watch the applet