All formulas on the topic scalar product of vectors. How to find the scalar product of vectors. Definition of the scalar product of vectors in terms of coordinates
Scalar product vectors (hereinafter referred to as SP). Dear friends! The mathematics exam includes a group of problems for solving vectors. We have already considered some problems. You can see them in the "Vectors" category. In general, the theory of vectors is simple, the main thing is to study it consistently. Calculations and actions with vectors in the school mathematics course are simple, the formulas are not complicated. Look into . In this article, we will analyze tasks on the joint venture of vectors (included in the exam). Now "immersion" in the theory:
H To find the coordinates of a vector, you need to subtract from the coordinates of its endcorresponding coordinates of its beginning
And further:
*Vector length (modulus) is defined as follows:
These formulas must be memorized!!!
Let's show the angle between the vectors:
It is clear that it can vary from 0 to 180 0(or in radians from 0 to Pi).
We can draw some conclusions about the sign of the scalar product. The lengths of vectors are positive, obviously. So the sign of the scalar product depends on the value of the cosine of the angle between the vectors.
Possible cases:
1. If the angle between the vectors is sharp (from 0 0 to 90 0), then the cosine of the angle will have a positive value.
2. If the angle between the vectors is obtuse (from 90 0 to 180 0), then the cosine of the angle will have a negative value.
*At zero degrees, that is, when the vectors have the same direction, the cosine is equal to one and, accordingly, the result will be positive.
At 180 o, that is, when the vectors have opposite directions, the cosine is equal to minus one,and the result will be negative.
Now the IMPORTANT POINT!
At 90 o, that is, when the vectors are perpendicular to each other, the cosine is zero, and hence the joint venture is zero. This fact (consequence, conclusion) is used in solving many problems where we are talking about relative position vectors, including in the tasks included in the open bank of tasks in mathematics.
We formulate the statement: the scalar product is equal to zero if and only if the given vectors lie on perpendicular lines.
So, the formulas for the SP vectors are:
If the coordinates of the vectors or the coordinates of the points of their beginnings and ends are known, then we can always find the angle between the vectors:
Consider the tasks:
27724 Find the inner product of vectors a and b .
We can find the scalar product of vectors using one of two formulas:
The angle between the vectors is unknown, but we can easily find the coordinates of the vectors and then use the first formula. Since the beginnings of both vectors coincide with the origin, the coordinates of these vectors are equal to the coordinates of their ends, that is
How to find the coordinates of a vector is described in.
We calculate:
Answer: 40
Find the coordinates of the vectors and use the formula:
To find the coordinates of a vector, it is necessary to subtract the corresponding coordinates of its beginning from the coordinates of the end of the vector, which means
We calculate the scalar product:
Answer: 40
Find the angle between vectors a and b . Give your answer in degrees.
Let the coordinates of the vectors have the form:
To find the angle between vectors, we use the formula for the scalar product of vectors:
Cosine of the angle between vectors:
Hence:
The coordinates of these vectors are:
Let's plug them into the formula:
The angle between the vectors is 45 degrees.
Answer: 45
In the case of a plane problem, the scalar product of vectors a = (a x ; a y ) and b = (b x ; b y ) can be found using the following formula:
a b = a x b x + a y b y
The formula for the scalar product of vectors for spatial problems
In the case of a spatial problem, the scalar product of vectors a = (a x ; a y ; a z ) and b = (b x ; b y ; b z ) can be found using the following formula:
a b = a x b x + a y b y + a z b z
Dot product formula of n-dimensional vectors
In the case of an n-dimensional space, the scalar product of vectors a = (a 1 ; a 2 ; ... ; a n ) and b = (b 1 ; b 2 ; ... ; b n ) can be found using the following formula:
a b = a 1 b 1 + a 2 b 2 + ... + a n b n
Properties of the Dot Product of Vectors
1. The scalar product of a vector with itself is always greater than or equal to zero:
2. The scalar product of a vector with itself is equal to zero if and only if the vector is equal to the zero vector:
a a = 0<=>a = 0
3. The scalar product of a vector by itself is equal to the square of its modulus:
4. The operation of scalar multiplication is communicative:
5. If the scalar product of two non-zero vectors is equal to zero, then these vectors are orthogonal:
a ≠ 0, b ≠ 0, a b = 0<=>a ┴ b
6. (αa) b = α(a b)
7. The operation of scalar multiplication is distributive:
(a + b) c = a c + b c
Examples of tasks for calculating the scalar product of vectors
Examples of calculating the scalar product of vectors for plane problems
Find the scalar product of the vectors a = (1; 2) and b = (4; 8).
Solution: a b = 1 4 + 2 8 = 4 + 16 = 20.
Find the scalar product of vectors a and b if their lengths |a| = 3, |b| = 6, and the angle between the vectors is 60˚.
Solution: a · b = |a| |b| cos α = 3 6 cos 60˚ = 9.
Find the inner product of vectors p = a + 3b and q = 5a - 3 b if their lengths |a| = 3, |b| = 2, and the angle between the vectors a and b is 60˚.
Solution:
p q = (a + 3b) (5a - 3b) = 5 a a - 3 a b + 15 b a - 9 b b =
5 |a| 2 + 12 a · b - 9 |b| 2 \u003d 5 3 2 + 12 3 2 cos 60˚ - 9 2 2 \u003d 45 +36 -36 \u003d 45.
An example of calculating the scalar product of vectors for spatial problems
Find the scalar product of the vectors a = (1; 2; -5) and b = (4; 8; 1).
Solution: a b = 1 4 + 2 8 + (-5) 1 = 4 + 16 - 5 = 15.
An example of calculating the dot product for n-dimensional vectors
Find the scalar product of the vectors a = (1; 2; -5; 2) and b = (4; 8; 1; -2).
Solution: a b = 1 4 + 2 8 + (-5) 1 + 2 (-2) = 4 + 16 - 5 -4 = 11.
13. The cross product of vectors and a vector is called third vector , defined as follows:
2) perpendicular, perpendicular. (1"")
3) the vectors are oriented in the same way as the basis of the entire space (positively or negatively).
Designate: .
physical meaning vector product
is the moment of force relative to the point O; is radius is the vector of force application point, then
moreover, if transferred to the point O, then the triple must be oriented as a vector of the basis.
1. Definition and simple properties. Let us take non-zero vectors a and b and put them aside from an arbitrary point O: OA = a and OB = b. The value of the angle AOB is called the angle between the vectors a and b and is denoted (a,b). If at least one of the two vectors is zero, then the angle between them, by definition, is considered right. Note that, by definition, the angle between vectors is at least 0 and at most . Moreover, the angle between two non-zero vectors is equal to 0 if and only if these vectors are codirectional and equal to if and only if they are in opposite directions.
Let us check that the angle between the vectors does not depend on the choice of point O. This is obvious if the vectors are collinear. Otherwise, we set aside from an arbitrary point O 1 vectors O 1 A 1 = a and o 1 IN 1 = b and note that triangles AOB and A 1 ABOUT 1 IN 1 are equal on three sides, because |OA| = |O 1 A 1 | = |a|, |OB| = |O 1 IN 1 | = |b|, |AB| = |A 1 IN 1 | = |b–а|. Therefore, the angles AOB and A 1 ABOUT 1 IN 1 are equal.
Now we can give the main thing in this paragraph
(5.1) Definition. The scalar product of two vectors a and b (denoted by ab) is the number 6 , equal to the product of the lengths of these vectors and the cosine of the angle between the vectors. Briefly speaking:
ab = |a||b|cos (a,b).
The operation of finding the scalar product is called the scalar multiplication of vectors. The scalar product aa of a vector with itself is called the scalar square of this vector and is denoted a 2 .
(5.2) The scalar square of a vector is equal to the square of its length.
If |a| 0, then (a,a) = 0, whence a 2 = |a||a|cos0 = |a| 2 . If a = 0, then a 2 = |a| 2 = 0.
(5.3) Cauchy's inequality. The modulus of the scalar product of two vectors does not exceed the product of moduli of factors: |ab| |a||b|. In this case, equality is achieved if and only if the vectors a and b are collinear.
By definition |ab| = ||a||b|cos (a,b)| = |a||b||cos (a,b)| |a||b. This proves the Cauchy inequality itself. Now let's notice. that for non-zero vectors a and b equality in it is achieved if and only if |cos (a,b)| = 1, i.e. at (a,b) = 0 or (a,b) = . The latter is equivalent to the fact that the vectors a and b are co-directed or oppositely directed, i.e. collinear. If at least one of the vectors a and b is zero, then they are collinear and |ab| = |a||b| = 0.
2. Basic properties of scalar multiplication. These include the following:
(CS1) ab = ba (commutativity);
(CS2) (xa)b = x(ab) (associativity);
(CS3) a(b+c) = ab + ac (distributivity).
The commutativity here is obvious, because ab = ba. Associativity for x = 0 is also obvious. If x > 0 then
(ha)b = |ha||b|cos (xa,b) = |x||a||b|cos (xa,b) = x|a||b|cos (a,b) = x(ab),
for (xa, b) = (a,b) (from the codirection of the vectors xa and a - Fig. 21). If x< 0, then
(xa)b = |x||a||b|cos (хa,b) = –х|а||b|(–cos (a,b)) = x|a||b|cos (a,b) = x(ab),
for (xa, b) = – (a,b) (from the opposite direction of the vectors xa and a - Fig.22). Thus, associativity is also proven.
Proving distributivity is more difficult. For this we need such
(5.4) Lemma. Let a be a non-zero vector parallel to the line l and b an arbitrary vector. Then the orthogonal projectionb" of the vector b to the line l is equal to 
.
1)
<
/2. Then the vectors a and 
co-directed (Fig. 23) and
b"
=
=
=
.
2) > /2 . Then the vectors a andb"oppositely directed (Fig. 24) and
b"
=
–
=
–
=
.
3)
=
/2. Thenb"
=
0 and ab
=
0, whenceb"
=
=
0.
We now prove the distributivity of (CS3). It is obvious if the vector a is zero. Let a
0. Then draw a line l
||
a, and denote byb" Andc" orthogonal projections of the vectors b and c onto it, and throughd" be the orthogonal projection of the vector d = b + c onto it. By Theorem 3.5d"
=
b"+
c". Applying Lemma 5.4 to the last equality, we obtain the equality 
=
. Multiplying it scalarly by a, we find that 
2
=
, whence ad = ab+ac, which was to be proved.
The properties of scalar multiplication of vectors proved by us are similar to the corresponding properties of multiplication of numbers. But not all properties of multiplication of numbers carry over to scalar multiplication of vectors. Here are typical examples:
1
2) If ab = ac, then this does not mean that b = c, even if the vector a is non-zero. Example: b and c are two different vectors of the same length, forming equal angles with the vector a (Fig. 25).
3) It is not true that always a(bc) = (ab)c: if only because the validity of such an equality for bc, ab 0 implies that the vectors a and c are collinear.
3. Orthogonality of vectors. Two vectors are called orthogonal if the angle between them is right. Orthogonality of vectors is indicated by the icon .
When we defined the angle between vectors, we agreed to consider the angle between the zero vector and any other vector as a straight line. Therefore, the zero vector is orthogonal to any. This agreement allows us to prove such
(5.5) Sign of orthogonality of two vectors. Two vectors are orthogonal if and only if their dot product is 0.
Let a and b be arbitrary vectors. If at least one of them is zero, then they are orthogonal, and their scalar product is equal to 0. Thus, in this case the theorem is true. Let us now assume that both given vectors are non-zero. By definition, ab = |a||b|cos (a,b). Since by our assumption the numbers |a| and |b| are not equal to 0, then ab = 0 cos (a, b) = 0 (a, b) = /2, which was to be proved.
The equality ab = 0 is often taken as the definition of orthogonality of vectors.
(5.6) Corollary. If the vector a is orthogonal to each of the vectors a 1 , …, A P , then it is also orthogonal to any of their linear combinations.
It suffices to note that from the equality aa 1 = … = aa P = 0 implies the equality a(x 1 A 1 + … +x P A P ) = x 1 (ah 1 ) + … + x P (ah P ) = 0.
From Corollary 5.6 it is easy to derive the school criterion for the perpendicularity of a line and a plane. Indeed, let some line MN be perpendicular to two intersecting lines AB and AC. Then the vector MN is orthogonal to the vectors AB and AC. Let us take any straight line DE in the plane ABC. The vector DE is coplanar to the noncollinear vectors AB and AC, and therefore expands in them. But then it is also orthogonal to the vector MN, that is, the lines MN and DE are perpendicular. It turns out that the line MN is perpendicular to any line from the plane ABC, which was to be proved.
4. Orthonormal bases. (5.7) Definition. A basis of a vector space is said to be orthonormal if, firstly, all its vectors have unit length and, secondly, any two of its vectors are orthogonal.
Vectors of an orthonormal basis in three-dimensional space are usually denoted by the letters i, j and k, and on the vector plane by the letters i and j. Taking into account the sign of orthogonality of two vectors and the equality of the scalar square of a vector to the square of its length, the orthonormality conditions for the basis (i,j,k) of the space V 3 can be written like this:
(5.8) i 2 = j 2 = k 2 = 1 , ij = ik = jk = 0,
and the basis (i,j) of the vector plane as follows:
(5.9) i 2 = j 2 = 1 , ij = 0.
Let the vectors a and b have in the orthonormal basis (i,j,k) the spaces V 3 coordinates (a 1 , A 2 , A 3 ) and (b 1 b 2 ,b 3 ) respectively. Thenab = (A 1 i+A 2 j+A 3 k)(b 1 i+b 2 j+b 3 k) = a 1 b 1 i 2 +a 2 b 2 j 2 +a 3 b 3 k 2 +a 1 b 2 ij+a 1 b 3 ik+a 2 b 1 ji+a 2 b 3 jk+a 3 b 1 ki+a 3 b 2 kj = a 1 b 1 +a 2 b 2 +a 3 b 3 . This is how the formula for the scalar product of vectors a (a 1 ,A 2 ,A 3 ) and b(b 1 ,b 2 ,b 3 ) given by their coordinates in the orthonormal basis of the space V 3 :
(5.10) ab = a 1 b 1 + a 2 b 2 + a 3 b 3 .
For vectors a(a 1 ,A 2 ) and b(b 1 ,b 2 ) given by their coordinates in an orthonormal basis on the vector plane, it has the form
(5.11) ab = a 1 b 1 + a 2 b 2 .
Let us substitute b = a into formula (5.10). It turns out that in the orthonormal basis a 2 = a 1 2 + a 2 2 + a 3 2 . Because a 2 = |a| 2 , we get such a formula for finding the length of the vector a (a 1 ,A 2 ,A 3 ) defined by its coordinates in the orthonormal basis of the space V 3 :
(5.12) |a| = 
.
On the vector plane, by virtue of (5.11), it takes the form
(5.13) |a| = 
.
Substituting b = i, b = j, b = k into formula (5.10), we obtain three more useful equalities:
(5.14) ai = a 1 , aj = a 2 , ak = a 3 .
The simplicity of coordinate formulas for finding the scalar product of vectors and vector length is the main advantage of orthonormal bases. For non-orthonormal bases, these formulas are, generally speaking, incorrect, and their application in this case is a gross mistake.
5. Direction cosines. Take in an orthonormal basis (i,j,k) the spaces V 3 vector a(a 1 ,A 2 ,A 3 ). Thenai = |a||i|cos (a,i) = |a|cos (a, i).On the other hand, ai = a 1 according to the formula 5.14. It turns out that
(5.15) a 1 = |a|cos (a, i).
and, likewise,
A 2 = |a|cos (a,j), and 3 = |a|cos (a, k).
If the vector a is unit, these three equalities take on a particularly simple form:
(5.16) A 1 = cos (a, i),A 2 = cos (a, j),A 3 = cos (a, k).
The cosines of the angles formed by a vector with the vectors of an orthonormal basis are called direction cosines of this vector in the given basis. As formulas 5.16 show, the coordinates of a unit vector in an orthonormal basis are equal to its direction cosines.
From 5.15 it follows that a 1 2 + a 2 2 + a 3 2 = |a| 2 (cos 2 (a,i)+cos 2 (a,j)+cos 2 (a, k)). On the other hand, a 1 2 + a 2 2 + a 3 2 = |a| 2 . It turns out that
(5.17) the sum of the squared direction cosines of a nonzero vector is equal to 1.
This fact is useful for solving some problems.
(5.18) Problem. The diagonal of a rectangular parallelepiped forms with two of its edges coming out of the same vertex angles of 60 . What angle does it form with the third edge coming out of this vertex?
Consider an orthonormal basis of the space V 3
, whose vectors are represented by the edges of the parallelepiped coming out of the given vertex. Since the diagonal vector forms angles of 60 with two vectors of this basis
, the squares of two of its three direction cosines are equal to cos 2
60
= 1/4. Therefore, the square of the third cosine is 1/2, and this cosine itself is 1/ 
. So the desired angle is 45
.
Angle between vectors
Consider two given vectors $\overrightarrow(a)$ and $\overrightarrow(b)$. Let us set aside the vectors $\overrightarrow(a)=\overrightarrow(OA)$ and $\overrightarrow(b)=\overrightarrow(OB)$ from an arbitrarily chosen point $O$, then the angle $AOB$ is called the angle between the vectors $\overrightarrow( a)$ and $\overrightarrow(b)$ (Fig. 1).
Picture 1.
Note here that if the vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ are codirectional, or one of them is a zero vector, then the angle between the vectors is equal to $0^0$.
Notation: $\widehat(\overrightarrow(a),\overrightarrow(b))$
The concept of the scalar product of vectors
Mathematically, this definition can be written as follows:
The scalar product can be zero in two cases:
If one of the vectors will be a zero vector (Since then its length is zero).
If the vectors are mutually perpendicular (i.e. $cos(90)^0=0$).
Note also that the inner product is greater than zero if the angle between these vectors is acute (because $(cos \left(\widehat(\overrightarrow(a),\overrightarrow(b))\right)\ ) >0$), and less than zero if the angle between these vectors is obtuse (since $(cos \left(\widehat(\overrightarrow(a),\overrightarrow(b))\right)\ )
The concept of the scalar square is related to the concept of the scalar product.
Definition 2
The scalar square of the vector $\overrightarrow(a)$ is the scalar product of this vector with itself.
We get that the scalar square is
\[\overrightarrow(a)\overrightarrow(a)=\left|\overrightarrow(a)\right|\left|\overrightarrow(a)\right|(cos 0^0\ )=\left|\overrightarrow(a )\right|\left|\overrightarrow(a)\right|=(\left|\overrightarrow(a)\right|)^2\]
Calculation of the scalar product by the coordinates of vectors
In addition to the standard way of finding the value of the dot product, which follows from the definition, there is another way.
Let's consider it.
Let the vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ have coordinates $\left(a_1,b_1\right)$ and $\left(a_2,b_2\right)$, respectively.
Theorem 1
The scalar product of the vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ is equal to the sum of the products of the corresponding coordinates.
Mathematically, this can be written as follows
\[\overrightarrow(a)\overrightarrow(b)=a_1a_2+b_1b_2\]
Proof.
The theorem has been proven.
This theorem has several implications:
Corollary 1: Vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ are perpendicular if and only if $a_1a_2+b_1b_2=0$
Corollary 2: The cosine of the angle between the vectors is $cos\alpha =\frac(a_1a_2+b_1b_2)(\sqrt(a^2_1+b^2_1)\cdot \sqrt(a^2_2+b^2_2))$
Properties of the Dot Product of Vectors
For any three vectors and a real number $k$, the following is true:
$(\overrightarrow(a))^2\ge 0$
This property follows from the definition of a scalar square (Definition 2).
displacement law:$\overrightarrow(a)\overrightarrow(b)=\overrightarrow(b)\overrightarrow(a)$.
This property follows from the definition of the inner product (Definition 1).
Distributive law:
$\left(\overrightarrow(a)+\overrightarrow(b)\right)\overrightarrow(c)=\overrightarrow(a)\overrightarrow(c)+\overrightarrow(b)\overrightarrow(c)$. \end(enumerate)
By Theorem 1, we have:
\[\left(\overrightarrow(a)+\overrightarrow(b)\right)\overrightarrow(c)=\left(a_1+a_2\right)a_3+\left(b_1+b_2\right)b_3=a_1a_3+a_2a_3+ b_1b_3+b_2b_3==\overrightarrow(a)\overrightarrow(c)+\overrightarrow(b)\overrightarrow(c)\]
Combination law:$\left(k\overrightarrow(a)\right)\overrightarrow(b)=k(\overrightarrow(a)\overrightarrow(b))$. \end(enumerate)
By Theorem 1, we have:
\[\left(k\overrightarrow(a)\right)\overrightarrow(b)=ka_1a_2+kb_1b_2=k\left(a_1a_2+b_1b_2\right)=k(\overrightarrow(a)\overrightarrow(b))\]
An example of a problem for calculating the scalar product of vectors
Example 1
Find the inner product of vectors $\overrightarrow(a)$ and $\overrightarrow(b)$ if $\left|\overrightarrow(a)\right|=3$ and $\left|\overrightarrow(b)\right|= 2$, and the angle between them is $((30)^0,\ 45)^0,\ (90)^0,\ (135)^0$.
Solution.
Using Definition 1, we get
For $(30)^0:$
\[\overrightarrow(a)\overrightarrow(b)=6(cos \left((30)^0\right)\ )=6\cdot \frac(\sqrt(3))(2)=3\sqrt( 3)\]
For $(45)^0:$
\[\overrightarrow(a)\overrightarrow(b)=6(cos \left((45)^0\right)\ )=6\cdot \frac(\sqrt(2))(2)=3\sqrt( 2)\]
For $(90)^0:$
\[\overrightarrow(a)\overrightarrow(b)=6(cos \left((90)^0\right)\ )=6\cdot 0=0\]
For $(135)^0:$
\[\overrightarrow(a)\overrightarrow(b)=6(cos \left((135)^0\right)\ )=6\cdot \left(-\frac(\sqrt(2))(2)\ right)=-3\sqrt(2)\]
If in the problem both the lengths of the vectors and the angle between them are presented "on a silver platter", then the condition of the problem and its solution look like this:
Example 1 Vectors are given. Find the scalar product of vectors if their lengths and the angle between them are represented by the following values:
Another definition is also valid, which is completely equivalent to definition 1.
Definition 2. The scalar product of vectors is a number (scalar) equal to the product of the length of one of these vectors and the projection of another vector onto the axis determined by the first of these vectors. Formula according to definition 2:
We will solve the problem using this formula after the next important theoretical point.
Definition of the scalar product of vectors in terms of coordinates
The same number can be obtained if the multiplied vectors are given by their coordinates.
Definition 3. The dot product of vectors is the number equal to the sum of the pairwise products of their respective coordinates.
On surface
If two vectors and in the plane are defined by their two Cartesian coordinates
then the dot product of these vectors is equal to the sum of the pairwise products of their respective coordinates:
.
Example 2 Find the numerical value of the projection of the vector onto the axis parallel to the vector.
Solution. We find the scalar product of vectors by adding the pairwise products of their coordinates:
Now we need to equate the resulting scalar product to the product of the length of the vector and the projection of the vector onto an axis parallel to the vector (in accordance with the formula).
We find the length of the vector as Square root from the sum of squares of its coordinates:
.
Write an equation and solve it:
Answer. The desired numerical value is minus 8.
In space
If two vectors and in space are defined by their three Cartesian rectangular coordinates
,
then the scalar product of these vectors is also equal to the sum of the pairwise products of their respective coordinates, only there are already three coordinates:
.
The task of finding the scalar product in the considered way is after analyzing the properties of the scalar product. Because in the task it will be necessary to determine what angle the multiplied vectors form.
Properties of the Dot Product of Vectors
Algebraic properties
1. (commutative property: the value of their scalar product does not change from changing the places of multiplied vectors).
2. 
(associative property with respect to a numerical factor: the scalar product of a vector multiplied by some factor and another vector is equal to the scalar product of these vectors multiplied by the same factor).
3. 
(distributive property with respect to the sum of vectors: the scalar product of the sum of two vectors by the third vector is equal to the sum of the scalar products of the first vector by the third vector and the second vector by the third vector).
4. (scalar square of a vector greater than zero) if is a nonzero vector, and , if is a zero vector.
Geometric Properties
In the definitions of the operation under study, we have already touched on the concept of an angle between two vectors. It's time to clarify this concept.
In the figure above, two vectors are visible, which are brought to a common beginning. And the first thing you need to pay attention to: there are two angles between these vectors - φ 1 And φ 2 . Which of these angles appears in the definitions and properties of the scalar product of vectors? The sum of the considered angles is 2 π and therefore the cosines of these angles are equal. The definition of the dot product includes only the cosine of the angle, not the value of its expression. But only one corner is considered in the properties. And this is the one of the two angles that does not exceed π ie 180 degrees. This angle is shown in the figure as φ 1 .
1. Two vectors are called orthogonal And the angle between these vectors is a right (90 degrees or π /2 ) if the scalar product of these vectors is zero :
.
Orthogonality in vector algebra is the perpendicularity of two vectors.
2. Two non-zero vectors make up sharp corner (from 0 to 90 degrees, or, what is the same, less π dot product is positive .
3. Two non-zero vectors make up obtuse angle (from 90 to 180 degrees, or, what is the same - more π /2 ) if and only if dot product is negative .
Example 3 Vectors are given in coordinates:
.
Calculate the dot products of all pairs of given vectors. What angle (acute, right, obtuse) do these pairs of vectors form?
Solution. We will calculate by adding the products of the corresponding coordinates.
We got a negative number, so the vectors form an obtuse angle.
We got a positive number, so the vectors form an acute angle.
We got zero, so the vectors form a right angle.
We got a positive number, so the vectors form an acute angle.
.
We got a positive number, so the vectors form an acute angle.
For self-test, you can use online calculator Dot product of vectors and cosine of the angle between them .
Example 4 Given the lengths of two vectors and the angle between them:
.
Determine at what value of the number the vectors and are orthogonal (perpendicular).
Solution. We multiply the vectors according to the rule of multiplication of polynomials:
Now let's calculate each term:
.
Let's compose an equation (equality of the product to zero), give like terms and solve the equation:
Answer: we got the value λ = 1.8 , at which the vectors are orthogonal.
Example 5 Prove that the vector 
orthogonal (perpendicular) to vector
Solution. To check orthogonality, we multiply the vectors and as polynomials, substituting the expression given in the problem condition instead of it:
.
To do this, you need to multiply each term (term) of the first polynomial by each term of the second and add the resulting products:
.
As a result, the fraction due is reduced. The following result is obtained:
Conclusion: as a result of multiplication, we got zero, therefore, the orthogonality (perpendicularity) of the vectors is proven.
Solve the problem yourself and then see the solution
Example 6 Given the lengths of vectors and , and the angle between these vectors is π /4 . Determine at what value μ vectors and are mutually perpendicular.
For self-test, you can use online calculator Dot product of vectors and cosine of the angle between them .
Matrix representation of the scalar product of vectors and the product of n-dimensional vectors
Sometimes, for clarity, it is advantageous to represent two multiplied vectors in the form of matrices. Then the first vector is represented as a row matrix, and the second - as a column matrix:
Then the scalar product of vectors will be the product of these matrices :
The result is the same as that obtained by the method we have already considered. We got one single number, and the product of the matrix-row by the matrix-column is also one single number.
In matrix form, it is convenient to represent the product of abstract n-dimensional vectors. Thus, the product of two four-dimensional vectors will be the product of a row matrix with four elements by a column matrix also with four elements, the product of two five-dimensional vectors will be the product of a row matrix with five elements by a column matrix also with five elements, and so on.
Example 7 Find Dot Products of Pairs of Vectors
,
using matrix representation.
Solution. The first pair of vectors. We represent the first vector as a row matrix, and the second as a column matrix. We find the scalar product of these vectors as the product of the row matrix by the column matrix:
Similarly, we represent the second pair and find:
As you can see, the results are the same as for the same pairs from example 2.
Angle between two vectors
The derivation of the formula for the cosine of the angle between two vectors is very beautiful and concise.
To express the dot product of vectors
(1)
V coordinate form, first we find the scalar product of orts. The scalar product of a vector with itself is by definition:
What is written in the formula above means: the scalar product of a vector with itself is equal to the square of its length. The cosine of zero is equal to one, so the square of each orth will be equal to one:
Since the vectors
are pairwise perpendicular, then the pairwise products of the orts will be equal to zero:
Now let's perform the multiplication of vector polynomials:
We substitute in the right side of the equality the values of the corresponding scalar products of the orts:
We get the formula for the cosine of the angle between two vectors:
Example 8 Given three points A(1;1;1), B(2;2;1), C(2;1;2).
Find an angle.
Solution. We find the coordinates of the vectors:
,
.
Using the formula for the cosine of an angle, we get:
Hence, .
For self-test, you can use online calculator Dot product of vectors and cosine of the angle between them .
Example 9 Given two vectors
Find the sum, difference, length, dot product and the angle between them.
2.Difference
