these are the cosines of the angles that the vector forms with the positive semi-axes of coordinates. Direction cosines uniquely specify the direction of the vector. If a vector has length 1, then its direction cosines are equal to its coordinates. In general, for a vector with coordinates ( a; b; c) direction cosines are equal:
where a, b, g are the angles made by the vector with the axes x, y, z respectively.
21) Decomposition of a vector in unit vectors. The unit vector of the coordinate axis is denoted by , the axes by , and the axes by (Fig. 1).
For any vector that lies in the plane, the following expansion takes place:
If the vector is located in space, then the expansion in unit vectors of the coordinate axes has the form:
22)Dot product two non-zero vectors and the number equal to the product of the lengths of these vectors and the cosine of the angle between them is called:
23)Angle between two vectors
If the angle between two vectors is acute, then their scalar product is positive; if the angle between the vectors is obtuse, then the scalar product of these vectors is negative. The scalar product of two nonzero vectors is equal to zero if and only if these vectors are orthogonal.
24) The condition of parallelism and perpendicularity of two vectors.
Condition for vectors to be perpendicular
Vectors are perpendicular if and only if their scalar product is zero. Given two vectors a(xa;ya) and b(xb;yb). These vectors will be perpendicular if the expression xaxb + yayb = 0.
25) Vector product of two vectors.
The vector product of two non-collinear vectors is a vector c=a×b that satisfies the following conditions: 1) |c|=|a| |b| sin(a^b) 2) c⊥a, c⊥b 3) Vectors a, b, c form a right-hand triplet of vectors.
26) Collinear and coplanar vectors..
Vectors are collinear if the abscissa of the first vector is related to the abscissa of the second in the same way as the ordinate of the first is to the ordinate of the second. Given two vectors a (xa;ya) And b (xb;yb). These vectors are collinear if xa = x b And y a = y b, Where R.
Vectors −→ a,−→b and −→ c are called coplanar, if there is a plane to which they are parallel.
27) Mixed product of three vectors. Mixed product of vectors- scalar product of vector a and the vector product of vectors b and c. Find the mixed product of vectors a = (1; 2; 3), b = (1; 1; 1), c = (1; 2; 1).
Solution:
1·1·1 + 1·1·2 + 1·2·3 - 1·1·3 - 1·1·2 - 1·1·2 = 1 + 2 + 6 - 3 - 2 - 2 = 2
28) The distance between two points on a plane. The distance between two given points is equal to the square root of the sum of the squared differences of the same coordinates of these points.
29) Division of a segment in this relation. If point M(x; y) lies on a line passing through two given points ( , ) and ( , ), and a relation is given in which point M divides the segment , then the coordinates of point M are determined by the formulas
If point M is the midpoint of the segment, then its coordinates are determined by the formulas
30-31. Slope of a straight line is called the tangent of the angle of inclination of this line. The slope of a straight line is usually denoted by the letter k. Then by definition
Equation of a straight line with slope has the form where k- straight line slope, b– some real number. Using the equation of a straight line with an angle coefficient, you can specify any straight line that is not parallel to the axis Oy(for a straight line parallel to the ordinate axis, the angular coefficient is not defined).
33. General equation of a straight line on a plane. Equation of the form 
There is general equation of a line Oxy. Depending on the values of constants A, B and C, the following special cases are possible:
C = 0, A ≠0, B ≠ 0 – the straight line passes through the origin
A = 0, B ≠0, C ≠0 (By + C = 0) - straight line parallel to the Ox axis
B = 0, A ≠0, C ≠ 0 (Ax + C = 0) – straight line parallel to the Oy axis
B = C = 0, A ≠0 – the straight line coincides with the Oy axis
A = C = 0, B ≠0 – the straight line coincides with the Ox axis
34.Equation of a line in segments on a plane in a rectangular coordinate system Oxy has the form where a And b- some non-zero real numbers. This name is not accidental, since the absolute values of numbers A And b equal to the lengths of the segments that the straight line cuts off on the coordinate axes Ox And Oy respectively (segments are counted from the origin). Thus, the equation of a line in segments makes it easy to construct this line in a drawing. To do this, you should mark the points with coordinates and in a rectangular coordinate system on the plane, and use a ruler to connect them with a straight line.
35. The normal equation of a line has the form
where is the distance from the straight line to the origin; – the angle between the normal to the line and the axis.
The normal equation can be obtained from the general equation (1) by multiplying it by the normalizing factor 
, the sign is opposite to the sign so that .
The cosines of the angles between the straight line and the coordinate axes are called direction cosines, – the angle between the straight line and the axis, – between the straight line and the axis:
Thus, the normal equation can be written in the form
Distance from point to a straight line determined by the formula 
36. The distance between a point and a line is calculated using the following formula: 
where x 0 and y 0 are the coordinates of the point, and A, B and C are coefficients from the general equation of the line
37. Reducing the general equation of a line to normal. The equation and the plane in this context do not differ from each other in anything other than the number of terms in the equations and the dimension of space. Therefore, first I will say everything about the plane, and at the end I will make a reservation about the straight line.
Let the general equation of the plane be given: Ax + By + Cz + D = 0.
;. we get the system: g;Mc=cosb, MB=cosa Let's bring it to normal form. To do this, we multiply both sides of the equation by the normalizing factor M. We get: Max+Mvu+MCz+MD=0. In this case MA=cos;.g;Mc=cosb, MB=cosa we obtain the system:
M2 B2=cos2b
M2 C2=cos2g
Adding up all the equations of the system, we get M*(A2 +B2+C2)=1 Now all that remains is to express M from here in order to know by which normalizing factor the original general equation must be multiplied to bring it to normal form:
M=-+1/ROOT KV A2 +B2 +C2
MD must always be less than zero, therefore the sign of the number M is taken opposite to the sign of the number D.
With the equation of a straight line, everything is the same, only from the formula for M you should simply remove the term C2.
| Ax + By + Cz + D = 0, |
38.General equation of the plane in space is called an equation of the form
Where A 2 + B 2 + C 2 ≠ 0 .
In three-dimensional space in the Cartesian coordinate system, any plane is described by an equation of the 1st degree (linear equation). And conversely, any linear equation defines a plane.
40.Equation of a plane in segments. In a rectangular coordinate system Oxyz in three-dimensional space an equation of the form 
, Where a, b And c– non-zero real numbers are called equation of the plane in segments. Absolute values of numbers a, b And c equal to the lengths of the segments that the plane cuts off on the coordinate axes Ox, Oy And Oz respectively, counting from the origin. Sign of numbers a, b And c shows in which direction (positive or negative) the segments are plotted on the coordinate axes
41) Normal plane equation.
The normal equation of a plane is its equation written in the form
where , , are the direction cosines of the plane normal, e
p is the distance from the origin to the plane. When calculating the direction cosines of the normal, it should be assumed that it is directed from the origin to the plane (if the plane passes through the origin, then the choice of the positive direction of the normal is indifferent).
42) Distance from a point to a plane.Let the plane be given by an equation and a point be given. Then the distance from the point to the plane is determined by the formula
 |
Proof. The distance from a point to a plane is, by definition, the length of the perpendicular drawn from the point to the plane
Angle between planes
Let the planes and be specified by the equations and , respectively. You need to find the angle between these planes.
The planes, intersecting, form four dihedral angles: two obtuse and two acute or four right angles, and both obtuse angles are equal to each other, and both acute angles are also equal to each other. We will always look for an acute angle. To determine its value, we take a point on the line of intersection of the planes and at this point in each of
planes, we draw perpendiculars to the intersection line.
Denote by alpha, beta and gamma the angles formed by vector a with the positive direction of the coordinate axes (see Fig. 1). The cosines of these angles are called the direction cosines of the vector a.
Instructions
Since the coordinates a in the Cartesian rectangular coordinate system are equal to the projections of the vector onto the coordinate axes, then
a1 = |a|cos(alpha), a2 = |a|cos(beta), a3 = |a|cos(gamma). From here:
cos (alpha)=a1||a|, cos(beta) =a2||a|, cos(gamma)= a3/|a|.
In this case |a|=sqrt(a1^2+ a2^2+ a3^2). Means
cos (alpha)=a1|sqrt(a1^2+ a2^2+ a3^2), cos(beta) =a2|sqrt(a1^2+ a2^2+ a3^2),
cos(gamma)= a3/sqrt(a1^2+ a2^2+ a3^2).
It should be noted the main property of direction cosines. The sum of the squares of the direction cosines of a vector is equal to one.
Indeed, cos^2(alpha)+cos^2(beta)+cos^2(gamma)=
= a1^2|(a1^2+ a2^2+ a3^2)+ a2^2|(a1^2+ a2^2+ a3^2)+ a3^2/(a1^2+ a2^2+ a3^2) =
=(a1^2+ a2^2+ a3^2)|(a1^2+ a2^2+ a3^2) = 1.
First way
Example: given: vector a=(1, 3, 5). Find its direction cosines.
Solution. In accordance with what we found, we write out:
|a|= sqrt(ax^2+ ay^2+ az^2)=sqrt(1+9 +25)=sqrt(35)=5.91.
Thus, the answer can be written in the following form:
(cos(alpha), cos(beta), cos(gamma))=(1/sqrt(35), 3/sqrt(35), 5/(35))=(0.16;0.5;0, 84).
Second way
When finding the direction cosines of vector a, you can use the technique of determining the cosines of angles using the scalar product. In this case, we mean the angles between a and the direction unit vectors of rectangular Cartesian coordinates i, j and k. Their coordinates are (1, 0, 0), (0, 1, 0), (0, 0, 1), respectively.
It should be recalled that the scalar product of vectors is defined as follows.
If the angle between the vectors is φ, then the scalar product of two winds (by definition) is a number equal to the product of the moduli of the vectors and cosφ. (a, b) = |a||b|cos f. Then, if b=i, then (a, i) = |a||i|cos(alpha),
or a1 = |a|cos(alpha). Further, all actions are performed similarly to method 1, taking into account coordinates j and k.
Question 6.
Cross product: definition and properties. Area of a parallelogram and a triangle. Expression of the scalar product in terms of coordinates. Examples.
In a rectangular coordinate system of three-dimensional space vector product of two vectors 
And 
is a vector , where are the coordinate vectors.
Let the vector be given. Unit vector in the same direction as 
(unit vector 
) is found by the formula:
.
Let the axis 
forms angles with the coordinate axes 
.Direction cosines of the axis 
The cosines of these angles are called:. If the direction 
given by a unit vector 
, then the direction cosines serve as its coordinates, i.e.:
.
The direction cosines are related to each other by the relation:
If the direction 
given by an arbitrary vector 
, then find the unit vector of this vector and, comparing it with the expression for the unit vector 
, get:
Dot product
Dot product
two vectors 
And 
is a number equal to the product of their lengths and the cosine of the angle between them: 
.
The scalar product has the following properties:
Hence, 
.
Geometric meaning of the dot product: scalar product of a vector and a unit vector 
equal to the projection of the vector 
to the direction determined 
, i.e. 
.
The following table of multiplication of unit vectors follows from the definition of the scalar product: 
:
.
If vectors are given by their coordinates 
And 
, i.e. 
,
, then, multiplying these vectors scalarly and using the multiplication table of unit vectors, we obtain the expression for the scalar product 
through vector coordinates:
.
Vector artwork
Cross product of a vector
to vector 
called a vector 
, the length and direction of which are determined by the conditions:
The vector product has the following properties:
From the first three properties it follows that the vector multiplication of a sum of vectors by a sum of vectors obeys the usual rules for multiplying polynomials. You just need to make sure that the order of the factors does not change.
The basic vectors are multiplied as follows:
If 
And 
, then taking into account the properties of the vector product of vectors, we can derive a rule for calculating the coordinates of the vector product from the coordinates of the factor vectors:
If we take into account the above rules for multiplying unit vectors, then:
A more compact form of writing an expression for calculating the coordinates of the vector product of two vectors can be constructed by introducing the concept of a determinant of a matrix.
Let us consider the special case when the vectors 
And 
belong to the plane 
, i.e. they can be represented as 
And 
.
If the coordinates of the vectors are written in table form as follows: 
, then we can say that a square matrix of the second order is formed from them, i.e. size 
, consisting of two rows and two columns. Each square matrix is associated with a number, which is calculated from the elements of the matrix according to certain rules and is called a determinant. The determinant of a second-order matrix is equal to the difference between the products of the elements of the main diagonal and the secondary diagonal:
.
In this case:
The absolute value of the determinant is thus equal to the area of the parallelogram constructed on the vectors 
And 
, both on the sides.
If we compare this expression with the vector product formula (4.7), then:
|
|
This expression is a formula for calculating the determinant of a third-order matrix from the first row.
Thus:
Determinant of a third-order matrix is calculated as follows:
and is the algebraic sum of six terms.
The formula for calculating the determinant of a third-order matrix is easy to remember if you use ruleSarrus, which is formulated as follows:
Each term is the product of three elements located in different columns and different rows of the matrix;
The products of elements forming triangles with a side parallel to the main diagonal have a plus sign;
The products of elements belonging to the secondary diagonal and two products of elements forming triangles with a side parallel to the secondary diagonal have a minus sign.
