Corner φ general equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0, calculated by the formula:
Corner φ between two lines given canonical equations(x-x 1)/m 1 = (y-y 1)/n 1 and (x-x 2)/m 2 = (y-y 2)/n 2, calculated by the formula:
Distance from point to line
Each plane in space can be represented as a linear equation called general equation plane
Special cases.
o If in equation (8) , then the plane passes through the origin.
o When (,) the plane is parallel to the axis (axis, axis), respectively.
o When (,) the plane is parallel to the plane (plane, plane).
Solution: use (7)
Answer: general plane equation.
Example.
A plane in the rectangular coordinate system Oxyz is given by the general equation of the plane 
. Write down the coordinates of all normal vectors of this plane.
We know that the coefficients of the variables x, y and z in the general equation of a plane are the corresponding coordinates of the normal vector of this plane. Therefore, the normal vector of a given plane 
has coordinates. The set of all normal vectors can be defined as:
Write the equation of the plane if in the rectangular coordinate system Oxyz in space it passes through the point 
, A 
is the normal vector of this plane.
We present two solutions to this problem.
From the condition we have . We substitute this data into the general equation of the plane passing through the point:
Write the general equation of a plane parallel to the coordinate plane Oyz and passing through the point 
.
A plane that is parallel to the coordinate plane Oyz can be given by a general incomplete plane equation of the form . Since the point 
belongs to the plane by condition, then the coordinates of this point must satisfy the equation of the plane, that is, the equality must be true. From here we find. Thus, the required equation has the form.
Solution. The cross product, by definition 10.26, is orthogonal to the vectors p and q. Consequently, it is orthogonal to the desired plane and the vector can be taken as its normal vector. Let's find the coordinates of vector n:
that is 
. Using formula (11.1), we obtain
By opening the brackets in this equation, we arrive at the final answer.
Answer: 
.
Let's rewrite the normal vector in the form and find its length:
According to the above:
Answer: 
Parallel planes have the same normal vector. 1) From the equation we find the normal vector of the plane:.
2) Let’s compose the equation of the plane using the point and normal vector: 
Answer:
Vector equation of a plane in space
Parametric equation of a plane in space
Equation of a plane passing through a given point perpendicular to a given vector
Let a rectangular Cartesian coordinate system be given in three-dimensional space. Let us formulate the following problem:
Write an equation for a plane passing through this point M(x 0, y 0, z 0) perpendicular to the given vector n = ( A, B, C} .
Solution. Let P(x, y, z) is an arbitrary point in space. Dot P belongs to the plane if and only if the vector MP = {x − x 0, y − y 0, z − z 0) orthogonal to the vector n = {A, B, C) (Fig. 1).
Having written the condition for the orthogonality of these vectors (n, MP) = 0 in coordinate form, we get:
|
A(x − x 0) + B(y − y 0) + C(z − z 0) = 0 |
Equation of a plane using three points
In vector form
In coordinates
Mutual arrangement of planes in space
– general equations two planes. Then:
1) if 
, then the planes coincide;
2) if 
, then the planes are parallel;
3) if or , then the planes intersect and the system of equations
(6)
are the equations of the straight line of intersection of these planes.
|
Solution: We compose the canonical equations of the line using the formula: Answer: |
We take the resulting equations and mentally “pinch off”, for example, the left piece: . Now let's equate this piece to any number(remember that there was already a zero), for example, to one: . Since , then the other two “pieces” should also be equal to one. Essentially, you need to solve the system: |
Compose parametric equations of the following straight lines: 
Solution: Direct specified canonical equations and at the first stage you should find some point belonging to the line and its direction vector.
a) From the equations 
remove the point and the direction vector: . You can choose another point (how to do this is described above), but it is better to take the most obvious one. By the way, to avoid mistakes, always substitute its coordinates into the equations.
Let's create parametric equations for this line: 
The convenience of parametric equations is that they make it very easy to find other points on a line. For example, let's find a point whose coordinates, say, correspond to the value of the parameter: 
Thus: b) Consider the canonical equations 
. Selecting a point here is not difficult, but treacherous: (be careful not to confuse the coordinates!!!). How to remove the guide vector? You can speculate about what this line is parallel to, or you can use a simple formal technique: “Y” and “Z” are in the proportion, so let’s write down the direction vector , and put a zero in the remaining space: .
Let's compose the parametric equations of the straight line: 
c) Let’s rewrite the equations in the form , that is, “zet” can be anything. And if by any, then let, for example, . Thus, the point belongs to this line. To find the direction vector, we use the following formal technique: in the original equations there are “x” and “y”, and in the direction vector at these places we write zeros: . In the remaining space we put unit: . Instead of one, any number except zero will do.
Let's write down the parametric equations of the straight line: 
Let straight lines be given in space l And m. Through some point A of space we draw straight lines l 1 || l And m 1 || m(Fig. 138).
Note that point A can be chosen arbitrarily; in particular, it can lie on one of these lines. If straight l And m intersect, then A can be taken as the point of intersection of these lines ( l 1 = l And m 1 = m).
Angle between non-parallel lines l And m is the value of the smallest of adjacent angles formed by intersecting lines l 1 And m 1 (l 1 || l, m 1 || m). The angle between parallel lines is considered equal to zero.
Angle between straight lines l And m denoted by \(\widehat((l;m))\). From the definition it follows that if it is measured in degrees, then 0° < \(\widehat((l;m)) \) < 90°, and if in radians, then 0 < \(\widehat((l;m)) \) < π / 2 .
Task. Given a cube ABCDA 1 B 1 C 1 D 1 (Fig. 139).
Find the angle between straight lines AB and DC 1.
Straight lines AB and DC 1 crossing. Since straight line DC is parallel to straight line AB, the angle between straight lines AB and DC 1, according to definition, is equal to \(\widehat(C_(1)DC)\).
Therefore, \(\widehat((AB;DC_1))\) = 45°.
Direct l And m are called perpendicular, if \(\widehat((l;m)) \) = π / 2. For example, in a cube
Calculation of the angle between straight lines.
The problem of calculating the angle between two straight lines in space is solved in the same way as in a plane. Let us denote by φ the magnitude of the angle between the lines l 1 And l 2, and through ψ - the magnitude of the angle between the direction vectors A And b these straight lines.
Then if
ψ <90° (рис. 206, а), то φ = ψ; если же ψ >90° (Fig. 206.6), then φ = 180° - ψ. Obviously, in both cases the equality cos φ = |cos ψ| is true. According to the formula (the cosine of the angle between non-zero vectors a and b is equal to the scalar product of these vectors divided by the product of their lengths) we have
$$ cos\psi = cos\widehat((a; b)) = \frac(a\cdot b)(|a|\cdot |b|) $$
hence,
$$ cos\phi = \frac(|a\cdot b|)(|a|\cdot |b|) $$
Let the lines be given by their canonical equations
$$ \frac(x-x_1)(a_1)=\frac(y-y_1)(a_2)=\frac(z-z_1)(a_3) \;\; And \;\; \frac(x-x_2)(b_1)=\frac(y-y_2)(b_2)=\frac(z-z_2)(b_3) $$
Then the angle φ between the lines is determined using the formula
$$ cos\phi = \frac(|a_(1)b_1+a_(2)b_2+a_(3)b_3|)(\sqrt((a_1)^2+(a_2)^2+(a_3)^2 )\sqrt((b_1)^2+(b_2)^2+(b_3)^2)) (1)$$
If one of the lines (or both) is given by non-canonical equations, then to calculate the angle you need to find the coordinates of the direction vectors of these lines, and then use formula (1).
Task 1. Calculate the angle between lines
$$ \frac(x+3)(-\sqrt2)=\frac(y)(\sqrt2)=\frac(z-7)(-2) \;\;and\;\; \frac(x)(\sqrt3)=\frac(y+1)(\sqrt3)=\frac(z-1)(\sqrt6) $$
Direction vectors of straight lines have coordinates:
a = (-√2 ; √2 ; -2), b = (√3 ; √3 ; √6 ).
Using formula (1) we find
$$ cos\phi = \frac(|-\sqrt6+\sqrt6-2\sqrt6|)(\sqrt(2+2+4)\sqrt(3+3+6))=\frac(2\sqrt6)( 2\sqrt2\cdot 2\sqrt3)=\frac(1)(2) $$
Therefore, the angle between these lines is 60°.
Task 2. Calculate the angle between lines
$$ \begin(cases)3x-12z+7=0\\x+y-3z-1=0\end(cases) and \begin(cases)4x-y+z=0\\y+z+1 =0\end(cases) $$
Behind the guide vector A take the first straight line vector product normal vectors n 1 = (3; 0; -12) and n 2 = (1; 1; -3) planes defining this line. Using the formula \(=\begin(vmatrix) i & j & k \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end(vmatrix) \) we get
$$ a==\begin(vmatrix) i & j & k \\ 3 & 0 & -12 \\ 1 & 1 & -3 \end(vmatrix)=12i-3i+3k $$
Similarly, we find the direction vector of the second straight line:
$$ b=\begin(vmatrix) i & j & k \\ 4 & -1 & 1 \\ 0 & 1 & 1 \end(vmatrix)=-2i-4i+4k $$
But using formula (1) we calculate the cosine of the desired angle:
$$ cos\phi = \frac(|12\cdot (-2)-3(-4)+3\cdot 4|)(\sqrt(12^2+3^2+3^2)\sqrt(2 ^2+4^2+4^2))=0 $$
Therefore, the angle between these lines is 90°.
Task 3. In the triangular pyramid MABC, the edges MA, MB and MC are mutually perpendicular (Fig. 207);
their lengths are respectively 4, 3, 6. Point D is the middle [MA]. Find the angle φ between lines CA and DB.
Let CA and DB be the direction vectors of straight lines CA and DB.
Let's take point M as the origin of coordinates. By the condition of the equation we have A (4; 0; 0), B(0; 0; 3), C(0; 6; 0), D (2; 0; 0). Therefore \(\overrightarrow(CA)\) = (4; - 6;0), \(\overrightarrow(DB)\)= (-2; 0; 3). Let's use formula (1):
$$ cos\phi=\frac(|4\cdot (-2)+(-6)\cdot 0+0\cdot 3|)(\sqrt(16+36+0)\sqrt(4+0+9 )) $$
Using the cosine table, we find that the angle between straight lines CA and DB is approximately 72°.
It will be useful for every student who is preparing for the Unified State Exam in mathematics to repeat the topic “Finding an angle between straight lines.” As statistics show, when passing the certification test, tasks in this section of stereometry cause difficulties for large quantities students. At the same time, tasks that require finding the angle between straight lines are found in the Unified State Examination of both basic and profile level. This means that everyone should be able to solve them.
Highlights
There are 4 types of relative positions of lines in space. They can coincide, intersect, be parallel or intersecting. The angle between them can be acute or straight.
To find the angle between lines in the Unified State Exam or, for example, in solving, schoolchildren in Moscow and other cities can use several ways to solve problems in this section of stereometry. You can complete the task using classical constructions. To do this, it is worth learning the basic axioms and theorems of stereometry. The student needs to be able to reason logically and create drawings in order to bring the task to a planimetric problem.
You can also use the vector coordinate method using simple formulas, rules and algorithms. The main thing in this case is to perform all calculations correctly. Hone your skills in solving problems in stereometry and other areas school course will help you educational project"Shkolkovo".
Oh-oh-oh-oh-oh... well, it’s tough, as if he was reading out a sentence to himself =) However, relaxation will help later, especially since today I bought the appropriate accessories. Therefore, let's proceed to the first section, I hope that by the end of the article I will maintain a cheerful mood.
The relative position of two straight lines
This is the case when the audience sings along in chorus. Two straight lines can:
1) match;
2) be parallel: ;
3) or intersect at a single point: .
Help for dummies : Please remember the mathematical intersection sign, it will appear very often. The notation means that the line intersects with the line at point .
How to determine the relative position of two lines?
Let's start with the first case:
Two lines coincide if and only if their corresponding coefficients are proportional, that is, there is a number “lambda” such that the equalities are satisfied
Let's consider the straight lines and create three equations from the corresponding coefficients: . From each equation it follows that, therefore, these lines coincide.
Indeed, if all the coefficients of the equation 
multiply by –1 (change signs), and reduce all coefficients of the equation by 2, you get the same equation: .
The second case, when the lines are parallel:
Two lines are parallel if and only if their coefficients of the variables are proportional: 
, But.
As an example, consider two straight lines. We check the proportionality of the corresponding coefficients for the variables: 
However, it is quite obvious that.
And the third case, when the lines intersect:
Two lines intersect if and only if their coefficients of the variables are NOT proportional, that is, there is NO such value of “lambda” that the equalities are satisfied 
So, for straight lines we will create a system: 
From the first equation it follows that , and from the second equation: , which means the system is inconsistent(no solutions). Thus, the coefficients of the variables are not proportional.
Conclusion: lines intersect
IN practical problems you can use the solution scheme just discussed. By the way, it is very reminiscent of the algorithm for checking vectors for collinearity, which we looked at in class The concept of linear (in)dependence of vectors. Basis of vectors. But there is a more civilized packaging:
Example 1
Find out relative position direct: 
Solution based on the study of directing vectors of straight lines:
a) From the equations we find the direction vectors of the lines: 
.
, which means that the vectors are not collinear and the lines intersect.
Just in case, I’ll put a stone with signs at the crossroads:
The rest jump over the stone and follow further, straight to Kashchei the Immortal =)
b) Find the direction vectors of the lines: 
The lines have the same direction vector, which means they are either parallel or coincident. There is no need to count the determinant here.
It is obvious that the coefficients of the unknowns are proportional, and .
Let's find out whether the equality is true: 
Thus,
c) Find the direction vectors of the lines: 
Let's calculate the determinant made up of the coordinates of these vectors: 
, therefore, the direction vectors are collinear. The lines are either parallel or coincident.
The proportionality coefficient “lambda” is easy to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: 
.
Now let's find out whether the equality is true. Both free members zero, therefore:
The resulting value satisfies this equation (any number in general satisfies it).
Thus, the lines coincide.
Answer:
Very soon you will learn (or even have already learned) to solve the problem discussed verbally literally in a matter of seconds. In this regard, I see no point in offering anything for independent decision, it’s better to lay another important brick in the geometric foundation:
How to construct a line parallel to a given one?
For ignorance of this simplest task Nightingale the Robber severely punishes.
Example 2
The straight line is given by the equation. Write an equation for a parallel line that passes through the point.
Solution: Let's denote the unknown line by the letter . What does the condition say about her? The straight line passes through the point. And if the lines are parallel, then it is obvious that the direction vector of the straight line “tse” is also suitable for constructing the straight line “de”.
We take the direction vector out of the equation:
Answer:
The geometry of the example looks simple: 
Analytical testing consists of the following steps:
1) We check that the lines have the same direction vector (if the equation of the line is not simplified properly, then the vectors will be collinear).
2) Check whether the point satisfies the resulting equation.
In most cases, analytical testing can be easily performed orally. Look at the two equations, and many of you will quickly determine the parallelism of the lines without any drawing.
Examples for independent solutions today will be creative. Because you will still have to compete with Baba Yaga, and she, you know, is a lover of all sorts of riddles.
Example 3
Write an equation for a line passing through a point parallel to the line if
There is a rational and not so rational way to solve it. The shortest way is at the end of the lesson.
We worked a little with parallel lines and will return to them later. The case of coinciding lines is of little interest, so let’s consider a problem that is familiar to you from school curriculum:
How to find the point of intersection of two lines?
If straight 
intersect at point , then its coordinates are the solution systems of linear equations 
How to find the point of intersection of lines? Solve the system.
Here you go geometric meaning systems of two linear equations with two unknowns- these are two intersecting (most often) lines on a plane.
Example 4
Find the point of intersection of lines
Solution: There are two ways to solve - graphical and analytical.
Graphic method is to simply draw the given lines and find out the intersection point directly from the drawing: 
Here is our point: . To check, you should substitute its coordinates into each equation of the line; they should fit both there and there. In other words, the coordinates of a point are a solution to the system. Essentially, we looked at a graphical solution systems of linear equations with two equations, two unknowns.
The graphical method is, of course, not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide this way, the point is that it will take time to create a correct and ACCURATE drawing. In addition, some straight lines are not so easy to construct, and the point of intersection itself may be located somewhere in the thirtieth kingdom outside the notebook sheet.
Therefore, it is more expedient to search for the intersection point using the analytical method. Let's solve the system: 
To solve the system, the method of term-by-term addition of equations was used. To develop relevant skills, take a lesson How to solve a system of equations?
Answer:
The check is trivial - the coordinates of the intersection point must satisfy each equation of the system.
Example 5
Find the point of intersection of the lines if they intersect.
This is an example for you to solve on your own. It is convenient to split the task into several stages. Analysis of the condition suggests that it is necessary:
1) Write down the equation of the straight line.
2) Create an equation of a straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.
Development of an action algorithm is typical for many geometric problems, and I will repeatedly focus on this.
Full solution and answer at the end of the lesson:
Not even a pair of shoes were worn out before we got to the second section of the lesson:
Perpendicular lines. Distance from a point to a line.
Angle between straight lines
Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to this one, and now the hut on chicken legs will turn 90 degrees:
How to construct a line perpendicular to a given one?
Example 6
The straight line is given by the equation. Write an equation perpendicular to the line passing through the point.
Solution: By condition it is known that . It would be nice to find the directing vector of the line. Since the lines are perpendicular, the trick is simple:
From the equation we “remove” the normal vector: , which will be the directing vector of the straight line.
Let's compose the equation of a straight line using a point and a direction vector:
Answer: 
Let's expand the geometric sketch:
Hmmm... Orange sky, orange sea, orange camel.
Analytical verification of the solution:
1) We take out the direction vectors from the equations 
and with the help scalar product of vectors we come to the conclusion that the lines are indeed perpendicular: .
By the way, you can use normal vectors, it's even easier.
2) Check whether the point satisfies the resulting equation 
.
The test, again, is easy to perform orally.
Example 7
Find the point of intersection of perpendicular lines if the equation is known 
and period.
This is an example for you to solve on your own. There are several actions in the problem, so it is convenient to formulate the solution point by point.
Our exciting journey continues:
Distance from point to line
In front of us is a straight strip of the river and our task is to get to it by the shortest route. There are no obstacles, and the most optimal route will be to move along the perpendicular. That is, the distance from a point to a line is the length of the perpendicular segment.
Distance in geometry is traditionally denoted Greek letter“ro”, for example: – the distance from the point “em” to the straight line “de”.
Distance from point to line 
expressed by the formula
Example 8
Find the distance from a point to a line 
Solution: all you need to do is carefully substitute the numbers into the formula and carry out the calculations:
Answer: 
Let's make the drawing: 
The found distance from the point to the line is exactly the length of the red segment. If you draw up a drawing on checkered paper on a scale of 1 unit. = 1 cm (2 cells), then the distance can be measured with an ordinary ruler.
Let's consider another task based on the same drawing:
The task is to find the coordinates of a point that is symmetrical to the point relative to the straight line 
. I suggest performing the steps yourself, but I will outline the solution algorithm with intermediate results:
1) Find a line that is perpendicular to the line.
2) Find the point of intersection of the lines: 
.
Both actions are discussed in detail in this lesson.
3) The point is the midpoint of the segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the midpoint of a segment we find .
It would be a good idea to check that the distance is also 2.2 units.
Difficulties may arise in calculations here, but a microcalculator is a great help in the tower, allowing you to calculate common fractions. I have advised you many times and will recommend you again.
How to find the distance between two parallel lines?
Example 9
Find the distance between two parallel lines
This is another example for you to decide on your own. I’ll give you a little hint: there are infinitely many ways to solve this. Debriefing at the end of the lesson, but it’s better to try to guess for yourself, I think your ingenuity was well developed.
Angle between two straight lines
Every corner is a jamb: 
In geometry, the angle between two straight lines is taken to be the SMALLER angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered the angle between intersecting lines. And his “green” neighbor or oppositely oriented"raspberry" corner.
If the lines are perpendicular, then any of the 4 angles can be taken as the angle between them.
How are the angles different? Orientation. Firstly, the direction in which the angle is “scrolled” is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example if .
Why did I tell you this? It seems that we can get by with the usual concept of an angle. The fact is that the formulas by which we will find angles can easily result in a negative result, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing, for a negative angle, be sure to indicate its orientation with an arrow (clockwise).
How to find the angle between two straight lines? There are two working formulas:
Example 10
Find the angle between lines
Solution And Method one
Consider two straight lines, given by equations V general view:
If straight not perpendicular, That oriented The angle between them can be calculated using the formula: 
Let us pay close attention to the denominator - this is exactly dot product directing vectors of straight lines:
If , then the denominator of the formula becomes zero, and the vectors will be orthogonal and the lines will be perpendicular. That is why a reservation was made about the non-perpendicularity of straight lines in the formulation.
Based on the above, it is convenient to formalize the solution in two steps:
1) Let's calculate the scalar product of the direction vectors of the lines:
, which means the lines are not perpendicular.
2) Find the angle between straight lines using the formula:
By using inverse function It's easy to find the corner itself. In this case, we use the oddness of the arctangent (see. Graphs and properties of elementary functions):
Answer: 
In the answer we indicate the exact value, as well as an approximate value (preferably in both degrees and radians), calculated using a calculator.
Well, minus, minus, no big deal. Here is a geometric illustration: 
It is not surprising that the angle turned out to be of a negative orientation, because in the problem statement the first number is a straight line and the “unscrewing” of the angle began precisely with it.
If you really want to get a positive angle, you need to swap the lines, that is, take the coefficients from the second equation 
, and take the coefficients from the first equation. In short, you need to start with a direct 
.
A. Let two straight lines be given. These straight lines, as indicated in Chapter 1, form various positive and negative angles, which can be either acute or obtuse. Knowing one of these angles, we can easily find any other.
By the way, for all these angles the numerical value of the tangent is the same, the difference can only be in the sign
Equations of lines. The numbers are the projections of the direction vectors of the first and second straight lines. The angle between these vectors is equal to one of the angles formed by straight lines. Therefore, the problem comes down to determining the angle between the vectors. We get
For simplicity, we can agree that the angle between two straight lines is an acute positive angle (as, for example, in Fig. 53).
Then the tangent of this angle will always be positive. Thus, if there is a minus sign on the right side of formula (1), then we must discard it, i.e., save only the absolute value.
Example. Determine the angle between straight lines
According to formula (1) we have
With. If it is indicated which of the sides of the angle is its beginning and which is its end, then, always counting the direction of the angle counterclockwise, we can extract something more from formula (1). As can be easily seen from Fig. 53, the sign obtained on the right side of formula (1) will indicate what kind of angle - acute or obtuse - the second straight line forms with the first.
(Indeed, from Fig. 53 we see that the angle between the first and second direction vectors is either equal to the desired angle between the straight lines, or differs from it by ±180°.)
d. If the lines are parallel, then their direction vectors are parallel. Applying the condition of parallelism of two vectors, we get!
This is a necessary and sufficient condition for the parallelism of two lines.
Example. Direct
are parallel because
e. If the lines are perpendicular then their direction vectors are also perpendicular. Applying the condition of perpendicularity of two vectors, we obtain the condition of perpendicularity of two straight lines, namely
Example. Direct
are perpendicular due to the fact that
In connection with the conditions of parallelism and perpendicularity, we will solve the following two problems.
f. Draw a line through a point parallel to the given line
The solution is carried out like this. Since the desired line is parallel to this one, then for its direction vector we can take the same one as that of the given line, i.e., a vector with projections A and B. And then the equation of the desired line will be written in the form (§ 1)
Example. Equation of a line passing through the point (1; 3) parallel to the line
there will be next!
g. Draw a line through a point perpendicular to the given line
Here it is no longer suitable to take the vector with projections A and as the guiding vector, but it is necessary to take the vector perpendicular to it. The projections of this vector must therefore be chosen according to the condition of perpendicularity of both vectors, i.e. according to the condition
This condition can be fulfilled in countless ways, since here is one equation with two unknowns. But the easiest way is to take or Then the equation of the desired line will be written in the form
Example. Equation of a line passing through the point (-7; 2) in a perpendicular line
there will be the following (according to the second formula)!
h. In the case when the lines are given by equations of the form
rewriting these equations differently, we have
