Your cart is empty
The dot product
In the field of sciences new processes prove to be capable of surviving only if they provide better results and their use is simpler than that of their ancestors. Numerous results of mathematics are used in the different fields of life. With the help of these we can get answers to the questions arising day by day.
However it also often happened that mathematics was inspired by other fields of science. Such an example is the concept of work, learnt in physics.
\latex{W=F\times s\times\cos{\alpha}}
\latex{\alpha}
\latex{ F }
\latex{ s }
Figure 9
We know that work is done if there is a displacement when force is applied. The quantity can be determined with the following formula: \latex{W = F\times s\times \cos{\alpha}}. (Figure 9)
The force \latex{ F } and the displacement \latex{ s } appearing in this formula are directed quantities, because both have a direction and a length. The size of the angle included between the two vectors is denoted by \latex{\alpha}. However the result obtained, the work does not have a direction, i.e. it is a scalar.
It seemed to be logical to introduce an operation which had not been applied for vectors before, among other things, because of this relation.
Let \latex{\overrightarrow{a}} and \latex{\overrightarrow{b}} be two vectors. Their angle of inclination can be obtained if we take them from a common starting point and we determine the convex angle included between them. (Figure 10)
\latex{\overrightarrow a}
\latex{\overrightarrow b}
\latex{\overrightarrow b}
\latex{\overrightarrow a}
\latex{\alpha}
\latex{\alpha}
Figure 10
So the following holds for the angle of inclination of the two vectors:
\latex{0^{\circ}\leq \alpha\leq 180^{\circ}}.
We assign a number to the two vectors according to the following rule.
DEFINITION: The dot product (or scalar product) of the two vectors is the real number obtained by multiplying the magnitudes of the two vectors and the cosine of the angle included between them:
\latex{\overrightarrow a \times \overrightarrow b=\left| \overrightarrow a \right|\times\left| \overrightarrow b \right|\times \cos{\alpha}}.
\latex{\overrightarrow a \times \overrightarrow b=\left| \overrightarrow a \right|\times\left| \overrightarrow b \right|\times \cos{\alpha}}.
The name “product” implies that the dot product bears some of the characteristics of the multiplication we became familiar with on the set of real numbers.
The following are implied by the definition directly:
- \latex{\overrightarrow a\times\overrightarrow b=\overrightarrow b\times\overrightarrow a }, i.e. the dot product is commutative.
- \latex{(\lambda\times\overrightarrow{a})\times\overrightarrow b=\lambda\times(\overrightarrow a\times \overrightarrow b)=\lambda\times\overrightarrow a\times\overrightarrow b=\overrightarrow a\times(\lambda\times\overrightarrow b),} where \latex{\lambda} denotes an arbitrary real number.
- \latex{\overrightarrow a\times\overrightarrow a=\left|\overrightarrow a\right|\times\left|\overrightarrow a\right|\times\cos{0^{\circ}}=\left|\overrightarrow a\right|^2}. Instead of the notation \latex{\overrightarrow a \times \overrightarrow a} the notation \latex{\overrightarrow a^2} is often used. It implies that \latex{\left|\overrightarrow a\right|=\sqrt{\overrightarrow a^2}} i.e. the length of a vector equals the square root of its square.
On the contrary it can easily be shown that the associative property, we got familiar with for the real numbers, does not hold, i.e. in general:
\latex{(\overrightarrow a\times\overrightarrow b)\times\overrightarrow c\neq \overrightarrow a\times(\overrightarrow b\times \overrightarrow c)}.
The explanation for it is that the dot products of the two vectors in parentheses are real numbers respectively, and thus the expression on the left-hand-side gives a vector pointing to the direction of \latex{\overrightarrow c}, and the expression on the right-hand-side gives a vector pointing to the direction of \latex{\overrightarrow a} as a result.
Example 1
Let us give vectors \latex{\overrightarrow a}, \latex{\overrightarrow b} and \latex{\overrightarrow c} so that the following equality holds:
\latex{(\overrightarrow a\times\overrightarrow b)\times\overrightarrow c= \overrightarrow a\times(\overrightarrow b\times \overrightarrow c)}.
Solution
- The equality holds for every vector triplets that are parallel with each other. This condition is sufficient but not necessary. (Figure 11)
- It is enough to require for equality that \latex{\overrightarrow a} and \latex{\overrightarrow c} are parallel. (Figure 12)
Obviously we can look for further special cases.
\latex{\overrightarrow a}
\latex{\overrightarrow b}
\latex{\overrightarrow c}
Figure 11
\latex{\overrightarrow a}
\latex{\overrightarrow c}
\latex{\overrightarrow b}
Figure 12
Example 2
Let us examine the conditions when is it satisfied that the dot product of two vectors is 0, i.e. when \latex{\overrightarrow a\times\overrightarrow b = 0}.
Solution
Since \latex{\overrightarrow{a}\times\overrightarrow{b}=\left|\overrightarrow{a}\right|\times\left|\overrightarrow{b}\right|\times\cos{\alpha}}, this product will be equal to \latex{ 0 } if and only if \latex{\cos \alpha = 0}, or if there is a vector among the vectors the magnitude of which is \latex{ 0 }, i.e. it is the zero vector.
We know that \latex{\cos{\alpha} = 0} is satisfied only in the case of \latex{\alpha = 90^{\circ}\;\; (0^{\circ} \leq \alpha \leq 180^{\circ})}, and that the zero vector is perpendicular to every vector, therefore we can formulate the following theorem.
THEOREM: The dot product of two vectors is \latex{ 0 } if and only if they are perpendicular to each other.
Since \latex{\cos \alpha} is positive in the case of acute angles, and it is negative for obtuse angles, the dot product will be positive if \latex{\alpha} is an acute angle, and it will be negative if \latex{\alpha} is an obtuse angle. (Figure 13)
\latex{\overrightarrow a\times \overrightarrow b\geq 0\Leftrightarrow0^{\circ}\leq\alpha\leq90^{\circ}}
\latex{\overrightarrow a\times \overrightarrow b\lt0\Leftrightarrow90^{\circ}\lt\alpha\leq180^{\circ}}
\latex{\overrightarrow a}
\latex{\overrightarrow a}
\latex{\overrightarrow b}
\latex{\overrightarrow b}
\latex{\alpha}
\latex{\alpha}
Figure 13
Example 3
Let \latex{\left|\overrightarrow a\right|=3}, \latex{\left|\overrightarrow b\right|=4}, we also know that \latex{\overrightarrow a\times \overrightarrow b=-6\times\sqrt3.} What is the measure of the angle included between the two?
Solution
Since \latex{\overrightarrow{a}\times\overrightarrow{b}=\left|\overrightarrow{a}\right|\times\left|\overrightarrow{b}\right|\times\cos{\alpha}},
\latex{-6\times\sqrt3=3\times4\times\cos \alpha.}
This implies that:
\latex{\cos{\alpha}=-\frac{\sqrt3}{2}},
which means that the angle included between the two vectors will be \latex{150^{\circ}}.Example 4
Let us consider two vectors one of which is the unit vector, i.e. \latex{\left|\overrightarrow e\right|=1}.
Let us examine what we get in the case of acute angles and what in the case of obtuse angles if we multiply these vectors.
Solution
Based on the figures we can establish that the dot product of a vector and the unit vector is equal to the signed projection of the vector on the straight line of the unit vector. (Figure 14)
\latex{\overrightarrow a}
\latex{\overrightarrow e}
\latex{\alpha}
\latex{\left|\overrightarrow a\right|\times \cos{\alpha}\gt0}
\latex{\left|\overrightarrow a\right|\times \cos{\alpha}\lt0}
\latex{\alpha}
\latex{\overrightarrow e}
\latex{\overrightarrow a}
Figure 14
Thus we can also produce projection vectors, which coincide the straight line of the unit vector, and their length is exactly equal to the orthogonal projection of the \latex{\overrightarrow a} given. (Figure 15)
In this case the formula generating the projection vector \latex{\overrightarrow{OA'}} is: \latex{\overrightarrow{OA'}=(\overrightarrow a\times \overrightarrow e)\times \overrightarrow e}, where \latex{\overrightarrow a\times\overrightarrow e} in the parentheses denotes a real factor, therefore the parentheses have an important role here and they cannot be omitted.
With the help of these one of the important properties, the distributivity of the dot product can be proven, i.e.:
\latex{(\overrightarrow a+\overrightarrow b)\times \overrightarrow c=\overrightarrow a\times\overrightarrow c+\overrightarrow b\times\overrightarrow c}.
\latex{OA'=(\overrightarrow a\times \overrightarrow e)\times\overrightarrow e}
\latex{\overrightarrow a}
\latex{\overrightarrow e}
\latex{ A }
\latex{ A' }
\latex{ O }
Figure 15
distributivity
distributivity
Example 5
Let us prove that the diagonals of a parallelogram are perpendicular to each other if and only if the parallelogram is a rhombus.
Solution
Let the two vectors starting from one of the vertices of the parallelogram be \latex{\overrightarrow a} and \latex{\overrightarrow b}. With the help of these the two diagonals can be expressed in the form of \latex{\overrightarrow a+\overrightarrow b} and \latex{\overrightarrow a-\overrightarrow b} , according to the figure. (Figure 16)
Let us consider the dot product of the vectors defined by the two diagonals:
Let us consider the dot product of the vectors defined by the two diagonals:
\latex{(\overrightarrow a+\overrightarrow b)\times(\overrightarrow a-\overrightarrow b)=\overrightarrow a^2-\overrightarrow b^2=\left|\overrightarrow a\right|^2-\left|\overrightarrow b\right|^2}.
This product will be 0, i.e. the diagonals will be perpendicular to each other if and only if
\latex{\left|\overrightarrow a\right|=\left|\overrightarrow b\right|}.
And it means that the lengths of the sides of the parallelogram are equal, i.e. it is a rhombus.
\latex{\overrightarrow a+\overrightarrow b}
\latex{\overrightarrow a-\overrightarrow b}
\latex{\overrightarrow a}
\latex{\overrightarrow b}
Figure 16
Exercises
- Give the dot product of the following vectors:
- \latex{\left|\overrightarrow a\right|=2,\;\;\left|\overrightarrow b\right|=3}, the angle of inclination of the two vectors is: \latex{\alpha=30^{\circ}};
- \latex{\left|\overrightarrow a\right|=1,\;\;\left|\overrightarrow b\right|=1}, the angle of inclination of the two vectors is: \latex{\alpha=45^{\circ}};
- \latex{\left|\overrightarrow a\right|=\frac 1 2,\;\;\left|\overrightarrow b\right|=2}, the angle of inclination of the two vectors is: \latex{\alpha=90^{\circ}};
- \latex{\left|\overrightarrow a\right|=\sqrt2,\;\;\left|\overrightarrow b\right|=\sqrt{50}}, the angle of inclination of the two vectors is: \latex{\alpha=120^{\circ}}.
- Give the angle of inclination of the vectors if the following is given:
- \latex{\left|\overrightarrow a\right|=2,\;\;\left|\overrightarrow b\right|=3}, and \latex{\overrightarrow a\times\overrightarrow{b}=3};
- \latex{\left|\overrightarrow a\right|=2,\;\;\left|\overrightarrow b\right|=2}, and \latex{\overrightarrow a\times\overrightarrow{b}=2\times\sqrt 2};
- \latex{\left|\overrightarrow a\right|=2,\;\;\left|\overrightarrow b\right|=3}, and \latex{\overrightarrow a\times\overrightarrow{b}=0};
- \latex{\left|\overrightarrow a\right|=2,\;\;\left|\overrightarrow b\right|=\frac 1 2}, and \latex{\overrightarrow a\times\overrightarrow{b}=-\frac{\sqrt3}{2}}.
- Is it possible that force is applied and there is displacement, however no work is done?
- In which cases can the equality \latex{(\overrightarrow a\times\overrightarrow b)\times\overrightarrow c=(\overrightarrow a\times\overrightarrow c)\times\overrightarrow b} be satisfied?
- Take the position vectors \latex{\overrightarrow a}, \latex{\overrightarrow b} and \latex{\overrightarrow b} from the centroid of a regular triangle pointing to the vertices; we know that these are unit vectors. Determine the following quantities:
- \latex{\overrightarrow a\times \overrightarrow b};
- \latex{(\overrightarrow a\times \overrightarrow b)\times\overrightarrow c};
- \latex{(\overrightarrow a\times \overrightarrow b)\times(\overrightarrow c\times\overrightarrow a)};
- \latex{(\overrightarrow a\times \overrightarrow b)\times\overrightarrow c+(\overrightarrow b\times\overrightarrow c)\times \overrightarrow a}.
- Give the angle of inclination of the unit vectors \latex{\overrightarrow a} and \latex{\overrightarrow b} if
- the vectors \latex{\overrightarrow a+ \overrightarrow b} and \latex{\overrightarrow a- \overrightarrow b} are perpendicular to each other;
- the vectors \latex{\overrightarrow a+ \overrightarrow b} and \latex{\overrightarrow a- 2\times\overrightarrow b} are perpendicular to each other;
- the vectors \latex{\overrightarrow a+ 2\times\overrightarrow b} and \latex{2\times\overrightarrow a- \overrightarrow b} are perpendicular to each other;
- the vectors \latex{-\overrightarrow a+ \overrightarrow b} and \latex{\overrightarrow a- 3\times\overrightarrow b} are perpendicular to each other.
- The unit vectors \latex{\overrightarrow a} and \latex{\overrightarrow b} are given and the angle included between them is \latex{60^{\circ}}. Give the projection vector of the vector \latex{\overrightarrow a} pointing to the direction of \latex{\overrightarrow b}.
- Verify that the sum of squares of the lengths of the diagonals of the parallelogram is equal to the sum of squares of the lengths of the sides.
- Verify that the sum of squares of the lengths of the medians of a triangle is equal to \latex{\frac3 4} of the sum of squares of the lengths of its sides.
- Verify that the dot product is distributive.