Formula for calculating the distance from a point to a line on a plane
If the equation of the line Ax + By + C = 0 is given, then the distance from the point M(M x , M y) to the line can be found using the following formula
Examples of problems for calculating the distance from a point to a line on a plane
Example 1.
Find the distance between the line 3x + 4y - 6 = 0 and the point M(-1, 3).
Solution. Let's substitute the coefficients of the line and the coordinates of the point into the formula
Answer: the distance from the point to the line is 0.6.
equation of a plane passing through points perpendicular to a vectorGeneral equation of a plane
A nonzero vector perpendicular to a given plane is called normal vector (or, in short, normal ) for this plane.
Let in coordinate space (in rectangular system coordinates) are given:
a) point 
;
b) non-zero vector (Fig. 4.8, a).
You need to create an equation for a plane passing through a point 
perpendicular to the vector End of proof.
Let's now consider Various types equations of a straight line on a plane.
1) General equation of the planeP .
From the derivation of the equation it follows that at the same time A, B And C are not equal to 0 (explain why).
The point belongs to the plane P only if its coordinates satisfy the equation of the plane. Depending on the odds A, B, C And D plane P occupies one position or another:
- the plane passes through the origin of the coordinate system, - the plane does not pass through the origin of the coordinate system,
- plane parallel to the axis X,
X,
- plane parallel to the axis Y,
- the plane is not parallel to the axis Y,
- plane parallel to the axis Z,
- the plane is not parallel to the axis Z.
Prove these statements yourself.
Equation (6) is easily derived from equation (5). Indeed, let the point lie on the plane P. Then its coordinates satisfy the equation. Subtracting equation (7) from equation (5) and grouping the terms, we obtain equation (6). Let us now consider two vectors with coordinates respectively. From formula (6) it follows that their scalar product is equal to zero. Therefore, the vector is perpendicular to the vector. The beginning and end of the last vector are located, respectively, at points that belong to the plane P. Therefore, the vector is perpendicular to the plane P. Distance from point to plane P, whose general equation 
determined by the formula 
The proof of this formula is completely similar to the proof of the formula for the distance between a point and a line (see Fig. 2). 
Rice. 2. To derive the formula for the distance between a plane and a straight line.
Indeed, the distance d between a straight line and a plane is equal
where is a point lying on the plane. From here, as in lecture No. 11, the above formula is obtained. Two planes are parallel if their normal vectors are parallel. From here we obtain the condition for parallelism of two planes 
- coefficients of general equations of planes. Two planes are perpendicular if their normal vectors are perpendicular, hence we obtain the condition for the perpendicularity of two planes if their general equations are known
Corner f between two planes is equal to the angle between their normal vectors (see Fig. 3) and can therefore be calculated using the formula 
Determining the angle between planes.
(11)
Distance from a point to a plane and methods for finding it
Distance from point to 
plane– the length of the perpendicular dropped from a point onto this plane. There are at least two ways to find the distance from a point to a plane: geometric And algebraic.
With the geometric method You must first understand how the perpendicular from a point to a plane is located: maybe it lies in some convenient plane, is a height in some convenient (or not so convenient) triangle, or maybe this perpendicular is generally a height in some pyramid.
After this first and most complex stage, the problem breaks down into several specific planimetric problems (perhaps in different planes).
With the algebraic method in order to find the distance from a point to a plane, you need to enter a coordinate system, find the coordinates of the point and the equation of the plane, and then apply the formula for the distance from a point to a plane.
This article talks about the topic « distance from a point to a line », Discusses the definition of the distance from a point to a line with illustrated examples using the coordinate method. Each theory block at the end has shown examples of solving similar problems.
The distance from a point to a line is found by determining the distance from point to point. Let's take a closer look.
Let there be a line a and a point M 1 that does not belong to the given line. Through it we draw a straight line b, located perpendicular to the straight line a. Let's take the point of intersection of the lines as H 1. We obtain that M 1 H 1 is a perpendicular that was lowered from point M 1 to straight line a.
Definition 1
Distance from point M 1 to straight line a is called the distance between points M 1 and H 1.
There are definitions that include the length of the perpendicular.
Definition 2
Distance from point to line is the length of the perpendicular drawn from a given point to a given line.
The definitions are equivalent. Consider the figure below.
It is known that the distance from a point to a line is the smallest of all possible. Let's look at this with an example.
If we take a point Q lying on a straight line a, which does not coincide with the point M 1, then we obtain that the segment M 1 Q is called an inclined segment, lowered from M 1 to a straight line a. It is necessary to indicate that the perpendicular from point M 1 is less than any other inclined line drawn from the point to the straight line.
To prove this, consider the triangle M 1 Q 1 H 1, where M 1 Q 1 is the hypotenuse. It is known that its length is always greater than the length of any of the legs. This means we have that M 1 H 1< M 1 Q . Рассмотрим рисунок, приведенный ниже.
The initial data for finding from a point to a line allows you to use several solution methods: through the Pythagorean theorem, determination of sine, cosine, tangent of an angle and others. Most tasks of this type are solved at school during geometry lessons.
When, when finding the distance from a point to a line, a rectangular coordinate system can be introduced, then the coordinate method is used. In this paragraph we will consider the main two methods of finding the required distance from given point.
The first method involves searching for the distance as a perpendicular drawn from M 1 to straight line a. The second method uses the normal equation of straight line a to find the required distance.
If there is a point on the plane with coordinates M 1 (x 1 , y 1), located in a rectangular coordinate system, straight line a, and you need to find the distance M 1 H 1, you can make the calculation in two ways. Let's look at them.
First way
If there are coordinates of point H 1 equal to x 2, y 2, then the distance from the point to the line is calculated using the coordinates from the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2.
Now let's move on to finding the coordinates of point H 1.
It is known that a straight line in O x y corresponds to the equation of a straight line on the plane. Let's take the method of defining a straight line a by writing a general equation of a straight line or an equation with an angular coefficient. We compose the equation of a straight line that passes through point M 1 perpendicular to a given straight line a. Let's denote the straight line by the letter b. H 1 is the point of intersection of lines a and b, which means to determine the coordinates you need to use the article in which we're talking about about the coordinates of the points of intersection of two lines.
It can be seen that the algorithm for finding the distance from a given point M 1 (x 1, y 1) to straight line a is carried out according to the points:
Definition 3
- finding the general equation of a straight line a, having the form A 1 x + B 1 y + C 1 = 0, or an equation with an angular coefficient, having the form y = k 1 x + b 1;
- obtaining a general equation of line b, having the form A 2 x + B 2 y + C 2 = 0 or an equation with an angular coefficient y = k 2 x + b 2, if line b intersects point M 1 and is perpendicular to a given line a;
- determination of the coordinates x 2, y 2 of the point H 1, which is the intersection point of a and b, for this purpose the system is solved linear equations A 1 x + B 1 y + C 1 = 0 A 2 x + B 2 y + C 2 = 0 or y = k 1 x + b 1 y = k 2 x + b 2 ;
- calculating the required distance from a point to a line using the formula M 1 H 1 = (x 2 - x 1) 2 + (y 2 - y 1) 2.
Second way
The theorem can help answer the question of finding the distance from a given point to a given straight line on a plane.
Theorem
The rectangular coordinate system has O x y has a point M 1 (x 1, y 1), from which a straight line is drawn to the plane, given by the normal equation of the plane, having the form cos α x + cos β y - p = 0, equal to The absolute value obtained on the left side of the normal equation of the line, calculated at x = x 1, y = y 1, means that M 1 H 1 = cos α · x 1 + cos β · y 1 - p.
Proof
Line a corresponds to the normal equation of the plane, having the form cos α x + cos β y - p = 0, then n → = (cos α, cos β) is considered the normal vector of line a at a distance from the origin to line a with p units . It is necessary to display all the data in the figure, add a point with coordinates M 1 (x 1, y 1), where the radius vector of the point M 1 - O M 1 → = (x 1, y 1). It is necessary to draw a straight line from a point to a straight line, which we denote as M 1 H 1 . It is necessary to show the projections M 2 and H 2 of the points M 1 and H 2 onto a straight line passing through the point O with a direction vector of the form n → = (cos α, cos β), and denote the numerical projection of the vector as O M 1 → = (x 1, y 1) to the direction n → = (cos α , cos β) as n p n → O M 1 → .
The variations depend on the location of the M1 point itself. Let's look at the figure below.
We fix the results using the formula M 1 H 1 = n p n → O M → 1 - p. Then we bring the equality to this form M 1 H 1 = cos α · x 1 + cos β · y 1 - p in order to obtain n p n → O M → 1 = cos α · x 1 + cos β · y 1 .
The scalar product of vectors results in a transformed formula of the form n → , O M → 1 = n → · n p n → O M 1 → = 1 · n p n → O M 1 → = n p n → O M 1 → , which is a product in coordinate form of the form n → , O M 1 → = cos α · x 1 + cos β · y 1 . This means that we get that n p n → O M 1 → = cos α · x 1 + cos β · y 1 . It follows that M 1 H 1 = n p n → O M 1 → - p = cos α · x 1 + cos β · y 1 - p. The theorem has been proven.
We find that to find the distance from point M 1 (x 1 , y 1) to straight line a on the plane, you need to perform several actions:
Definition 4
- obtaining the normal equation of the straight line a cos α · x + cos β · y - p = 0, provided that it is not in the task;
- calculation of the expression cos α · x 1 + cos β · y 1 - p, where the resulting value takes M 1 H 1.
Let's apply these methods to solve problems with finding the distance from a point to a plane.
Example 1
Find the distance from the point with coordinates M 1 (- 1, 2) to the straight line 4 x - 3 y + 35 = 0.
Solution
Let's use the first method to solve.
To do this, it is necessary to find the general equation of the line b, which passes through a given point M 1 (- 1, 2), perpendicular to the line 4 x - 3 y + 35 = 0. From the condition it is clear that line b is perpendicular to line a, then its direction vector has coordinates equal to (4, - 3). Thus, we have the opportunity to write down the canonical equation of line b on the plane, since there are coordinates of the point M 1, which belongs to line b. Let's determine the coordinates of the directing vector of the straight line b. We get that x - (- 1) 4 = y - 2 - 3 ⇔ x + 1 4 = y - 2 - 3. The resulting canonical equation must be converted to a general one. Then we get that
x + 1 4 = y - 2 - 3 ⇔ - 3 · (x + 1) = 4 · (y - 2) ⇔ 3 x + 4 y - 5 = 0
Let us find the coordinates of the points of intersection of the lines, which we will take as the designation H 1. The transformations look like this:
4 x - 3 y + 35 = 0 3 x + 4 y - 5 = 0 ⇔ x = 3 4 y - 35 4 3 x + 4 y - 5 = 0 ⇔ x = 3 4 y - 35 4 3 3 4 y - 35 4 + 4 y - 5 = 0 ⇔ ⇔ x = 3 4 y - 35 4 y = 5 ⇔ x = 3 4 5 - 35 4 y = 5 ⇔ x = - 5 y = 5
From what was written above, we have that the coordinates of point H 1 are equal to (- 5; 5).
It is necessary to calculate the distance from point M 1 to straight line a. We have that the coordinates of the points M 1 (- 1, 2) and H 1 (- 5, 5), then we substitute them into the formula to find the distance and get that
M 1 H 1 = (- 5 - (- 1) 2 + (5 - 2) 2 = 25 = 5
Second solution.
In order to solve in another way, it is necessary to obtain the normal equation of the line. We calculate the value of the normalizing factor and multiply both sides of the equation 4 x - 3 y + 35 = 0. From here we get that the normalizing factor is equal to - 1 4 2 + (- 3) 2 = - 1 5, and the normal equation will be of the form - 1 5 4 x - 3 y + 35 = - 1 5 0 ⇔ - 4 5 x + 3 5 y - 7 = 0 .
According to the calculation algorithm, it is necessary to obtain the normal equation of the line and calculate it with the values x = - 1, y = 2. Then we get that
4 5 · - 1 + 3 5 · 2 - 7 = - 5
From this we obtain that the distance from point M 1 (- 1, 2) to the given straight line 4 x - 3 y + 35 = 0 has the value - 5 = 5.
Answer: 5 .
It is clear that in this method It is important to use the normal equation of a line, since this method is the shortest. But the first method is convenient because it is consistent and logical, although it has more calculation points.
Example 2
On the plane there is a rectangular coordinate system O x y with point M 1 (8, 0) and straight line y = 1 2 x + 1. Find the distance from a given point to a straight line.
Solution
The first method involves reducing a given equation with an angular coefficient to a general equation. To simplify, you can do it differently.
If the product of the angular coefficients of perpendicular lines has a value of - 1, then the angular coefficient of a line perpendicular to a given one y = 1 2 x + 1 has a value of 2. Now we get the equation of a line passing through a point with coordinates M 1 (8, 0). We have that y - 0 = - 2 · (x - 8) ⇔ y = - 2 x + 16 .
We proceed to finding the coordinates of point H 1, that is, the intersection points y = - 2 x + 16 and y = 1 2 x + 1. We compose a system of equations and get:
y = 1 2 x + 1 y = - 2 x + 16 ⇔ y = 1 2 x + 1 1 2 x + 1 = - 2 x + 16 ⇔ y = 1 2 x + 1 x = 6 ⇔ ⇔ y = 1 2 · 6 + 1 x = 6 = y = 4 x = 6 ⇒ H 1 (6, 4)
It follows that the distance from the point with coordinates M 1 (8, 0) to the straight line y = 1 2 x + 1 is equal to the distance from the start point and end point with coordinates M 1 (8, 0) and H 1 (6, 4) . Let's calculate and find that M 1 H 1 = 6 - 8 2 + (4 - 0) 2 20 = 2 5.
The solution in the second way is to move from an equation with a coefficient to its normal form. That is, we get y = 1 2 x + 1 ⇔ 1 2 x - y + 1 = 0, then the value of the normalizing factor will be - 1 1 2 2 + (- 1) 2 = - 2 5. It follows that the normal equation of the line takes the form - 2 5 1 2 x - y + 1 = - 2 5 0 ⇔ - 1 5 x + 2 5 y - 2 5 = 0. Let's carry out the calculation from the point M 1 8, 0 to a line of the form - 1 5 x + 2 5 y - 2 5 = 0. We get:
M 1 H 1 = - 1 5 8 + 2 5 0 - 2 5 = - 10 5 = 2 5
Answer: 2 5 .
Example 3
It is necessary to calculate the distance from the point with coordinates M 1 (- 2, 4) to the lines 2 x - 3 = 0 and y + 1 = 0.
Solution
We get the equation normal looking straight line 2 x - 3 = 0:
2 x - 3 = 0 ⇔ 1 2 2 x - 3 = 1 2 0 ⇔ x - 3 2 = 0
Then we proceed to calculating the distance from the point M 1 - 2, 4 to the straight line x - 3 2 = 0. We get:
M 1 H 1 = - 2 - 3 2 = 3 1 2
The equation of the straight line y + 1 = 0 has a normalizing factor with a value equal to -1. This means that the equation will take the form - y - 1 = 0. We proceed to calculating the distance from the point M 1 (- 2, 4) to the straight line - y - 1 = 0. We find that it is equal to - 4 - 1 = 5.
Answer: 3 1 2 and 5.
Let's take a closer look at finding the distance from a given point on the plane to the coordinate axes O x and O y.
In a rectangular coordinate system, the O axis y has an equation of a straight line, which is incomplete and has the form x = 0, and O x - y = 0. The equations are normal for the coordinate axes, then it is necessary to find the distance from the point with coordinates M 1 x 1, y 1 to the lines. This is done based on the formulas M 1 H 1 = x 1 and M 1 H 1 = y 1. Let's look at the figure below.
Example 4
Find the distance from the point M 1 (6, - 7) to the coordinate lines located in the O x y plane.
Solution
Since the equation y = 0 refers to the straight line O x, you can find the distance from M 1 with given coordinates to this straight line using the formula. We get that 6 = 6.
Since the equation x = 0 refers to the straight line O y, you can find the distance from M 1 to this straight line using the formula. Then we get that - 7 = 7.
Answer: the distance from M 1 to O x has a value of 6, and from M 1 to O y has a value of 7.
When in three-dimensional space we have a point with coordinates M 1 (x 1, y 1, z 1), it is necessary to find the distance from point A to straight line a.
Let's consider two methods that allow you to calculate the distance from a point to a straight line a located in space. The first case considers the distance from point M 1 to a line, where a point on the line is called H 1 and is the base of a perpendicular drawn from point M 1 to line a. The second case suggests that the points of this plane must be sought as the height of the parallelogram.
First way
From the definition we have that the distance from point M 1 located on straight line a is the length of the perpendicular M 1 H 1, then we obtain that with the found coordinates of point H 1, then we find the distance between M 1 (x 1, y 1, z 1 ) and H 1 (x 1 , y 1 , z 1) , based on the formula M 1 H 1 = x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2.
We find that the whole solution goes towards finding the coordinates of the base of the perpendicular drawn from M 1 to the straight line a. This is done as follows: H 1 is the point where straight line a intersects with the plane that passes through the given point.
This means that the algorithm for determining the distance from point M 1 (x 1, y 1, z 1) to line a in space implies several points:
Definition 5
- drawing up the equation of the plane χ as an equation of the plane passing through a given point located perpendicular to the line;
- determination of the coordinates (x 2, y 2, z 2) belonging to the point H 1, which is the intersection point of straight line a and plane χ;
- calculating the distance from a point to a line using the formula M 1 H 1 = x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2.
Second way
From the condition we have a straight line a, then we can determine the direction vector a → = a x, a y, a z with coordinates x 3, y 3, z 3 and a certain point M 3 belonging to straight a. If you have the coordinates of the points M 1 (x 1, y 1) and M 3 x 3, y 3, z 3, you can calculate M 3 M 1 →:
M 3 M 1 → = (x 1 - x 3, y 1 - y 3, z 1 - z 3)
We should set aside the vectors a → = a x , a y , a z and M 3 M 1 → = x 1 - x 3 , y 1 - y 3 , z 1 - z 3 from point M 3 , connect them and get a parallelogram figure. M 1 H 1 is the height of the parallelogram.
Let's look at the figure below.
We have that the height M 1 H 1 is the required distance, then it is necessary to find it using the formula. That is, we are looking for M 1 H 1.
Let us denote the area of the parallelogram by the letter S, found by the formula using the vector a → = (a x, a y, a z) and M 3 M 1 → = x 1 - x 3. y 1 - y 3, z 1 - z 3. The area formula is S = a → × M 3 M 1 → . Also, the area of the figure is equal to the product of the lengths of its sides and the height, we get that S = a → · M 1 H 1 with a → = a x 2 + a y 2 + a z 2, which is the length of the vector a → = (a x, a y, a z), which is equal to the side of the parallelogram. This means that M 1 H 1 is the distance from the point to the line. It is found using the formula M 1 H 1 = a → × M 3 M 1 → a → .
To find the distance from a point with coordinates M 1 (x 1, y 1, z 1) to a straight line a in space, you need to perform several steps of the algorithm:
Definition 6
- determination of the direction vector of the straight line a - a → = (a x, a y, a z);
- calculating the length of the direction vector a → = a x 2 + a y 2 + a z 2 ;
- obtaining coordinates x 3 , y 3 , z 3 belonging to point M 3 located on straight line a;
- calculating the coordinates of the vector M 3 M 1 → ;
- finding the vector product of vectors a → (a x , a y , a z) and M 3 M 1 → = x 1 - x 3 , y 1 - y 3 , z 1 - z 3 as a → × M 3 M 1 → = i → j → k → a x a y a z x 1 - x 3 y 1 - y 3 z 1 - z 3 to obtain the length using the formula a → × M 3 M 1 → ;
- calculating the distance from a point to a line M 1 H 1 = a → × M 3 M 1 → a → .
Solving problems of finding the distance from a given point to a given line in space
Example 5Find the distance from the point with coordinates M 1 2, - 4, - 1 to the line x + 1 2 = y - 1 = z + 5 5.
Solution
The first method begins with writing the equation of the plane χ passing through M 1 and perpendicular to a given point. We get an expression like:
2 (x - 2) - 1 (y - (- 4)) + 5 (z - (- 1)) = 0 ⇔ 2 x - y + 5 z - 3 = 0
It is necessary to find the coordinates of the point H 1, which is the point of intersection with the χ plane to the line specified by the condition. You should move from the canonical view to the intersecting one. Then we get a system of equations of the form:
x + 1 2 = y - 1 = z + 5 5 ⇔ - 1 · (x + 1) = 2 · y 5 · (x + 1) = 2 · (z + 5) 5 · y = - 1 · (z + 5) ⇔ x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 5 y + z + 5 = 0 ⇔ x + 2 y + 1 = 0 5 x - 2 z - 5 = 0
It is necessary to calculate the system x + 2 y + 1 = 0 5 x - 2 z - 5 = 0 2 x - y + 5 z - 3 = 0 ⇔ x + 2 y = - 1 5 x - 2 z = 5 2 x - y + 5 z = 3 by Cramer’s method, then we get that:
∆ = 1 2 0 5 0 - 2 2 - 1 5 = - 60 ∆ x = - 1 2 0 5 0 - 2 3 - 1 5 = - 60 ⇔ x = ∆ x ∆ = - 60 - 60 = 1 ∆ y = 1 - 1 0 5 5 2 2 3 5 = 60 ⇒ y = ∆ y ∆ = 60 - 60 = - 1 ∆ z = 1 2 - 1 5 0 5 2 - 1 3 = 0 ⇒ z = ∆ z ∆ = 0 - 60 = 0
From here we have that H 1 (1, - 1, 0).
M 1 H 1 = 1 - 2 2 + - 1 - - 4 2 + 0 - - 1 2 = 11
The second method must begin by searching for coordinates in the canonical equation. To do this, you need to pay attention to the denominators of the fraction. Then a → = 2, - 1, 5 is the direction vector of the line x + 1 2 = y - 1 = z + 5 5. It is necessary to calculate the length using the formula a → = 2 2 + (- 1) 2 + 5 2 = 30.
It is clear that the straight line x + 1 2 = y - 1 = z + 5 5 intersects the point M 3 (- 1 , 0 , - 5), hence we have that the vector with the origin M 3 (- 1 , 0 , - 5) and its end at the point M 1 2, - 4, - 1 is M 3 M 1 → = 3, - 4, 4. Find the vector product a → = (2, - 1, 5) and M 3 M 1 → = (3, - 4, 4).
We get an expression of the form a → × M 3 M 1 → = i → j → k → 2 - 1 5 3 - 4 4 = - 4 i → + 15 j → - 8 k → + 20 i → - 8 · j → = 16 · i → + 7 · j → - 5 · k →
we find that the length of the vector product is equal to a → × M 3 M 1 → = 16 2 + 7 2 + - 5 2 = 330.
We have all the data to use the formula for calculating the distance from a point for a straight line, so let’s apply it and get:
M 1 H 1 = a → × M 3 M 1 → a → = 330 30 = 11
Answer: 11 .
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Let a rectangular coordinate system be fixed in three-dimensional space Oxyz, given point , straight line a and you need to find the distance from the point A to a straight line a.
We will show two methods that allow you to calculate the distance from a point to a line in space. In the first case, finding the distance from a point M 1 to a straight line a comes down to finding the distance from the point M 1 to the point H 1 , Where H 1 - the base of a perpendicular dropped from a point M 1 directly a. In the second case, we will find the distance from the point to the plane as the height of the parallelogram.
So let's get started.
The first way to find the distance from a point to a line a in space.
Since by definition the distance from a point M 1
to a straight line a is the length of the perpendicular M 1
H 1
, then, having determined the coordinates of the point H 1
, we can calculate the required distance as the distance between points 
And 
according to the formula .
Thus, the problem comes down to finding the coordinates of the base of the perpendicular constructed from the point M 1 to a straight line a. This is quite simple to do: period H 1 is the point of intersection of the line a with a plane passing through a point M 1 perpendicular to the line a.
Hence, algorithm that allows you to determine the distance from a point 
to a straight linea
in space, is:
The second method allows you to find the distance from a point to a line a in space.
Since in the problem statement we are given a straight line a, then we can determine its direction vector 
and the coordinates of some point M 3
, lying on the straight line a. Then, according to the coordinates of the points and 
we can calculate the coordinates of a vector: (if necessary, refer to the article coordinates of a vector through the coordinates of its start and end points).
Let's put the vectors aside 
and from the point M 3
and construct a parallelogram on them. In this parallelogram we draw the height M 1
H 1
.
Obviously the height M 1 H 1 of the constructed parallelogram is equal to the required distance from the point M 1 to a straight line a. Let's find it.
On one side, the area of the parallelogram (let us denote it S) can be found through the vector product of vectors 
and according to the formula 
. On the other hand, the area of a parallelogram is equal to the product of the length of its side and its height, that is, 
, Where 
- vector length 
,
equal to length sides of the parallelogram in question. Therefore, the distance from a given point M 1
to a given straight line a can be found from the equality 
How 
.
So, to find the distance from a point 
to a straight linea
in space needed
Solving problems of finding the distance from a given point to a given line in space.
Let's look at the example solution.
Example.
Find the distance from the point 
to a straight line 
.
Solution.
First way.
Let's write the equation of the plane passing through the point M 1 perpendicular to a given line:
Find the coordinates of the point H 1
- points of intersection of the plane and a given straight line. To do this, let's make the transition from canonical equations straight line to the equations of two intersecting planes 
after which we solve the system of linear equations 
Cramer's method: 
Thus, .
It remains to calculate the required distance from a point to a line as the distance between points 
And : .
Second way.
The numbers in the denominators of fractions in the canonical equations of a line represent the corresponding coordinates of the direction vector of this line, that is, 
- direct vector 
. Let's calculate its length: 
.
Obviously straight 
passes through a point 
, then a vector with origin at point 
and end at a point 
There is 
. Let's find the vector product of vectors 
And 
:
then the length of this vector product is 
.
Now we have all the data to use the formula to calculate the distance from a given point to a given plane: 
.
Answer:
The relative position of lines in space
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 the relative position of the lines: 
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 terms are zero, so:
The resulting value satisfies this equation(any number generally 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 example geometry 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 in 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.
The graphical 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 look for the intersection point 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) Write down the equation of the 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
We have a straight strip of river in front of us 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 by the Greek letter “rho”, 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 count 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 the equations in 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 scalar 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 scalar product directing vectors of straight 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 your 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 
.
The ability to find the distance between different geometric objects is important when calculating the surface area of shapes and their volumes. In this article we will consider the question of how to find the distance from a point to a line in space and on a plane.
Mathematical description of a line
To understand how to find the distance from a point to a line, you need to understand the question of the mathematical definition of these geometric objects.
Everything is simple with a point; it is described by a set of coordinates, the number of which corresponds to the dimension of space. For example, on a plane these are two coordinates, in three-dimensional space - three.
As for a one-dimensional object - a straight line, several types of equations are used to describe it. Let's consider only two of them.
The first type is called a vector equation. Below are expressions for lines in three-dimensional and two-dimensional space:
(x; y; z) = (x 0 ; y 0 ; z 0) + α × (a; b; c);
(x; y) = (x 0 ; y 0) + α × (a; b)
In these expressions, coordinates with zero indices describe the point through which a given line passes, the set of coordinates (a; b; c) and (a; b) are the so-called direction vectors for the corresponding line, α is a parameter that can take any actual value.
The vector equation is convenient in the sense that it explicitly contains the direction vector of the line, the coordinates of which can be used when solving problems of parallelism or perpendicularity of various geometric objects, for example, two straight lines.
The second type of equation that we will consider for a line is called general. In space this type is given general equations two planes. On a plane it has the following shape:
A × x + B × y + C = 0
When plotting a graph, it is often written as a dependence on X/Y, that is:
y = -A / B × x +(-C / B)
Here the free term -C / B corresponds to the coordinate of the intersection of the line with the y-axis, and the coefficient -A / B is associated with the angle of inclination of the line to the x-axis.
The concept of the distance between a line and a point
Having dealt with the equations, you can directly move on to answering the question of how to find the distance from a point to a straight line. In the 7th grade, schools begin to consider this issue by determining the appropriate value.
The distance between a line and a point is the length of the segment perpendicular to this line, which is omitted from the point in question. The figure below shows a straight line r and a point A. The segment perpendicular to the straight line r is shown in blue. Its length is the required distance.
The two-dimensional case is depicted here, however this definition distances are also valid for a three-dimensional problem.
Required formulas
Depending on the form in which the equation of a line is written and in what space the problem is solved, two basic formulas can be given that answer the question of how to find the distance between a line and a point.
Let us denote the known point by the symbol P 2 . If the equation of a straight line is given in vector form, then for d the distance between the objects under consideration the formula is valid:
d = || / |v¯|
That is, to determine d, you should calculate the modulus of the vector product of the guide for the straight line vector v¯ and the vector P 1 P 2 ¯, the beginning of which lies at an arbitrary point P 1 on the straight line, and the end is at the point P 2 , then divide this modulus by the length v ¯. This formula is universal for flat and three-dimensional space.
If the problem is considered on a plane in the xy coordinate system and the equation of the line is given in general form, then the following formula allows you to find the distance from the line to the point as follows:
Straight line: A × x + B × y + C = 0;
Point: P 2 (x 2; y 2; z 2);
Distance: d = |A × x 2 + B × y 2 + C| / √(A 2 + B 2)
The above formula is quite simple, but its use is limited by the conditions noted above.
Coordinates of the projection of a point onto a straight line and distance
You can also answer the question of how to find the distance from a point to a line in another way that does not involve memorizing the given formulas. This method involves determining a point on a line that is the projection of the original point.
Suppose that there is a point M and a line r. The projection onto r of a point M corresponds to a certain point M 1 . The distance from M to r is equal to the length of the vector MM 1 ¯.
How to find the coordinates of M 1? Very simple. It is enough to remember that the line vector v¯ will be perpendicular to MM 1 ¯, that is, their scalar product must be equal to zero. Adding to this condition the fact that the coordinates M 1 must satisfy the equation of the line r, we obtain a system of simple linear equations. As a result of its solution, the coordinates of the projection of point M onto r are obtained.
The technique described in this paragraph for finding the distance from a line to a point can be used for a plane and for space, however, its use requires knowledge of the vector equation for the line.
Plane problem
Now it's time to show how to use the presented mathematical apparatus to solve real problems. Suppose that a point M(-4; 5) is given on the plane. It is necessary to find the distance from point M to a straight line, which is described by a general equation:
3 × (-4) + 6 = -6 ≠ 5
That is, M does not lie on a line.
Since the equation of a straight line is not given in general form, we reduce it to such a form in order to be able to use the corresponding formula, we have:
y = 3 × x + 6 =>
3 × x - y + 6 = 0
Now you can substitute known numbers into the formula for d:
d = |A × x 2 + B × y 2 + C| / √(A 2 +B 2) =
= |3 × (-4) -1 × 5+6| / √(3 2 +(-1) 2) = 11 / √10 ≈ 3.48
Problem in space
Now let's consider the case in space. Let the straight line be described by the following equation:
(x; y; z) = (1; -1; 0) + α × (3; -2; 1)
What is the distance from it to the point M(0; 2; -3)?
Just as in the previous case, let us check whether M belongs to the given line. To do this, we substitute the coordinates into the equation and rewrite it explicitly:
x = 0 = 1 + 3 × α => α = -1/3;
y = 2 = -1 -2 × α => α = -3/2;
Since received different parametersα, then M does not lie on this line. Let us now calculate the distance from it to the straight line.
To use the formula for d, take an arbitrary point on a line, for example P(1; -1; 0), then:
Let's calculate the vector product between PM¯ and the line v¯. We get:
= [(-1; 3; -3) * (3; -2; 1)] = (-3; -8; -7)
Now we substitute the modules of the found vector and the vector v¯ into the formula for d, we get:
d = √(9 + 64 + 49) / √(9 + 4 + 1) ≈ 2.95
This answer could be obtained using the technique described above, which involves solving a system of linear equations. In this and the previous problems, the calculated values of the distance from a straight line to a point are presented in units of the corresponding coordinate system.
