Distance from line to line coordinate method. Distance from point to line in space - theory, examples, solutions. Point projection coordinates and distance
Formula for calculating the distance from a point to a straight line on a plane
If the equation of the straight line Ax + By + C \u003d 0 is given, then the distance from the point M (M x, M y) to the straight line can be found using the following formula
Examples of tasks for calculating the distance from a point to a straight line on a plane
Example 1.
Find the distance between the line 3x + 4y - 6 \u003d 0 and the point M (-1, 3).
Decision. Substitute in the formula the coefficients of the line and the coordinates of the point
Answer: the distance from a point to a straight line is 0.6.
equation of a plane passing through points perpendicular to a vector General equation of a plane
A nonzero vector perpendicular to a given plane is called normal vector (or, in short, normal ) for this plane.
Let the coordinate space (in a rectangular coordinate system) be given:
a) point 
;
b) a nonzero vector (Figure 4.8, a).
It is required to draw up an equation of a plane passing through a point 
perpendicular to vector End of proof.
Let us now consider various types of equations of a straight line on a plane.
1) General equation of the planeP .
It follows from the derivation of the equation that simultaneously A, B and C not equal to 0 (explain why).
The point belongs to the plane P only if its coordinates satisfy the plane equation. Depending on the odds A, B, C and Dplane 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,
- the plane is parallel to the axis X,
X,
- the plane is parallel to the axis Y,
- the plane is not parallel to the axis Y,
- the plane is 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). Consider now 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 the end of the last vector are respectively at the points that belong to the plane P... Therefore, the vector is perpendicular to the plane P... Distance from point to plane P, the general equation of which is 
determined by the formula 
The proof of this formula is completely analogous to the proof of the formula for the distance between a point and a line (see Fig. 2). 
Figure: 2. To the derivation of the formula for the distance between a plane and a straight line.
Indeed, the distance d between a straight line and a plane is
where is a point lying on a plane. Hence, as in lecture No. 11, the above formula is obtained. Two planes are parallel if their normal vectors are parallel. From this we obtain the condition of 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 of perpendicularity of two planes if their general equations are known
Angle f between two planes is equal to the angle between their normal vectors (see Fig. 3) and can, therefore, be calculated by the formula 
Determination of the angle between the planes.
(11)
Distance from point to plane and ways to find 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 is located from point to plane: maybe it lies in some convenient plane, is the height in some convenient (or not very) triangle, or maybe this perpendicular is generally the height in some pyramid.
After this first and most difficult stage, the task breaks down into several specific planimetric tasks (perhaps in different planes).
In the algebraic way 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.
Consider the application of the analyzed methods for finding the distance from a given point to a given straight line on a plane when solving an example.
Find the distance from a point to a straight line:
First, let's solve the problem in the first way.
In the condition of the problem, we are given a general equation of the straight line a of the form:
Let's find the general equation of the straight line b, which passes through a given point perpendicular to the straight line:
Since line b is perpendicular to line a, the direction vector of line b is the normal vector of a given line:
that is, the direction vector of the straight line b has coordinates. Now we can write the canonical equation of the straight line b on the plane, since we know the coordinates of the point M 1 through which the straight line b passes, and the coordinates of the direction vector of the straight line b:
From the obtained canonical equation of the straight line b, we pass to the general equation of the straight line:
Now we will find the coordinates of the point of intersection of the straight lines a and b (denote it by H 1) by solving the system of equations composed of the general equations of the straight lines a and b (if necessary, refer to the article solving systems linear equations):
Thus, the point H 1 has coordinates.
It remains to calculate the required distance from point M 1 to line a as the distance between points and:
The second way to solve the problem.
We obtain the normal equation of the given line. To do this, we calculate the value of the normalizing factor and multiply by it both sides of the original general equation of the straight line:
(we talked about this in the section on reducing the general equation of a straight line to normal form).
The normalizing factor is
then the normal equation of the straight line has the form:
Now we take the expression on the left side of the resulting normal equation of the straight line, and calculate its value at:
The required distance from a given point to a given straight line:
equally absolute value the obtained value, that is, five ().
distance from point to line:
Obviously, the advantage of the method for finding the distance from a point to a straight line on a plane, based on the use of the normal equation of a straight line, is a relatively smaller amount of computational work. In turn, the first method of finding the distance from a point to a straight line is intuitive and is distinguished by consistency and consistency.
A rectangular coordinate system Oxy is fixed on the plane, a point and a straight line are specified:
Find the distance from a given point to a given straight line.
First way.
You can go from a given equation of a straight line with a slope to the general equation of this straight line and act in the same way as in the above example.
But you can do otherwise.
We know that the product of the slopes of perpendicular lines is 1 (see the article perpendicular lines, perpendicular lines). Therefore, the slope of a straight line that is perpendicular to a given straight line:
is equal to 2. Then the equation of a line perpendicular to a given line and passing through a point has the form:
Now let's find the coordinates of the point H 1 - the intersection points of the lines:
Thus, the required distance from a point to a straight line:
is equal to the distance between points and:
Second way.
Let us pass from a given equation of a straight line with a slope to a normal equation of this straight line:
the normalizing factor is:
therefore, the normal equation of a given line has the form:
Now we calculate the required distance from a point to a line:
Calculate the distance from a point to a straight line:
and to the straight line:
We get the normal equation of the line:
Now let's calculate the distance from a point to a line:
The normalizing factor for the straight-line equation:
is equal to 1. Then the normal equation of this line has the form:
Now we can calculate the distance from a point to a line:
it is equal.
Answer: and 5.
In conclusion, we separately consider how the distance from a given point of the plane to the coordinate lines Ox and Oy is found.
In the rectangular coordinate system Oxy, the coordinate line Oy is given by the incomplete general equation of the line x \u003d 0, and the coordinate line Ox is given by the equation y \u003d 0. These equations are normal equations lines Oy and Ox, therefore, the distance from a point to these lines is calculated by the formulas:
respectively.
Figure 5
A rectangular coordinate system Oxy is introduced on the plane. Find the distances from the point to the coordinate lines.
The distance from a given point М 1 to the coordinate line Ox (it is given by the equation y \u003d 0) is equal to the modulus of the ordinate of point М 1, that is,.
The distance from a given point M 1 to the coordinate line Oy (it corresponds to the equation x \u003d 0) is equal to the absolute value of the abscissa of point M 1:.
Answer: the distance from point М 1 to line Ox is 6, and the distance from a given point to coordinate line Oy is equal to.
Let a rectangular coordinate system be fixed in three-dimensional space Oxyz, given point, straight a and it is required to find the distance from the point AND to straight a.
We will show two ways to calculate the distance from a point to a straight line in space. In the first case, finding the distance from the point M 1 to straight a is reduced to finding the distance from a point M 1 to the point H 1 where H 1 - the base of the perpendicular dropped from the point M 1 on a straight line a... In the second case, the distance from the point to the plane will be found as the height of the parallelogram.
So let's get started.
The first way to find the distance from point to line a in space.
Since by definition the distance from the point M 1
to straight a Is the length of the perpendicular M 1
H 1
, then, having determined the coordinates of the point H 1
, we will be able to calculate the required distance as the distance between points 
and 
according to the formula .
Thus, the problem is reduced to finding the coordinates of the base of the perpendicular constructed from the point M 1 to straight a... This is easy enough: point H 1 Is the intersection point of the straight line a with a plane passing through the point M 1 perpendicular to straight line a.
Hence, algorithm for determining the distance from a point 
to straighta
in spaceis this:
The second method allows you to find the distance from a point to a straight line a in space.
Since in the problem statement we are given a straight line a, then we can define its direction vector 
and coordinates of some point M 3
lying on a straight line a... Then 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).
Set aside vectors 
and from point M 3
and build 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 straight a... We will find it.
On the one hand, the area of \u200b\u200bthe parallelogram (we denote it S) can be found in terms of the vector product of vectors 
and by the formula 
... On the other hand, the area of \u200b\u200ba parallelogram is equal to the product of the length of its side by the height, that is, 
where 
- vector length 
equal to the length of the side of the parallelogram under consideration. Therefore, the distance from a given point M 1
to a given straight line a can be found from equality 
as 
.
So, to find the distance from a point 
to straighta
in space you need
Solving problems on finding the distance from a given point to a given straight line in space.
Let's consider the solution of an example.
Example.
Find the distance from the point 
to straight 
.
Decision.
First way.
Let us write the equation of the plane passing through the point M 1 perpendicular to a given straight line:
Find the coordinates of the point H 1
- points of intersection of the plane and a given straight line. To do this, we perform 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 straight line as the distance between points 
and:.
Second way.
The numbers in the denominators of fractions in the canonical equations of the straight line represent the corresponding coordinates of the direction vector of this straight line, that is, 
- directing vector of a straight line 
... Let's calculate its length: 
.
Obviously, the straight line 
goes through the point 
, then the vector starting at the point 
and end at point 
there is 
... Find the vector product of vectors 
and 
:
then the length of this cross 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:
Mutual arrangement of straight lines in space
OoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooTherefore, we will proceed to the first section, I hope that by the end of the article I will maintain a cheerful frame of mind.
The relative position of two straight lines
The case when the audience sings along with the chorus. Two straight lines can:
1) match;
2) be parallel:;
3) or intersect at a single point:.
Help for Dummies : please remember the mathematical sign of the intersection, it will be very common. The notation indicates that a straight line intersects a straight line at a point.
How to determine the relative position of two straight lines?
Let's start with the first case:
Two straight lines coincide if and only if their corresponding coefficients are proportional, that is, there is such a number "lambda" that the equalities
Consider straight lines and compose three equations from the corresponding coefficients:. It follows from each equation that, therefore, these lines coincide.
Indeed, if all the coefficients of the equation 
multiply by –1 (change signs), and reduce all the coefficients of the equation by 2, then you get the same equation:.
The second case, when the lines are parallel:
Two straight lines are parallel if and only if their coefficients for the variables are proportional: 
but.
As an example, consider two lines. We check the proportionality of the corresponding coefficients for the variables: 
However, it is quite clear that.
And the third case, when the lines intersect:
Two straight lines intersect if and only if their coefficients for variables are NOT proportional, that is, there is NOT such a lambda value that the equalities are satisfied 
So, for straight lines we will compose the system: 
From the first equation it follows that, and from the second equation:, therefore, 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 considered. By the way, it is very similar to the algorithm for checking vectors for collinearity, which we considered in the lesson The concept of linear (non) dependence of vectors. Vector basis... But there is a more civilized packaging:
Example 1
Find out the relative position of the straight lines: 
Decision based on the study of direction vectors of straight lines:
a) From the equations we find the direction vectors of the straight lines: 
.
, so the vectors are not collinear and the lines intersect.
Just in case, I'll put a stone with pointers at the crossroads:
The rest jump over the stone and follow on, straight to Kashchei the Immortal \u003d)
b) Find the direction vectors of straight lines: 
Lines have the same direction vector, which means that they are either parallel or coincide. There is no need to count the determinant here.
It is obvious that the coefficients for the unknowns are proportional, while.
Let us find out whether the equality is true: 
Thus,
c) Find the direction vectors of straight lines: 
Let us calculate the determinant composed of the coordinates of these vectors: 
hence the direction vectors are collinear. Lines are either parallel or coincide.
The coefficient of proportionality "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 if 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) how to solve the problem considered orally literally in a matter of seconds. In this regard, I see no reason to propose anything for an independent solution, it is better to lay another important brick in the geometric foundation:
How to build a straight line parallel to a given one?
For ignorance of this simplest task, the Nightingale the Robber severely punishes.
Example 2
The straight line is given by the equation. Equate a parallel line that goes through a point.
Decision: Let's denote the unknown direct letter. What does the condition say about her? The straight line goes through the point. And if the straight lines are parallel, then it is obvious that the directing vector of the straight line "tse" is also suitable for constructing the straight line "de".
We take out the direction vector from the equation:
Answer:
The geometry of the example looks simple: 
Analytical verification consists of the following steps:
1) Check that the straight lines have the same direction vector (if the equation of the straight line is not simplified properly, then the vectors will be collinear).
2) Check if the point satisfies the obtained equation.
Analytical review is in most cases easy to do orally. Look at the two equations and many of you will quickly figure out the parallelism of straight lines without any drawing.
Examples for self-solution today will be creative. Because you still have to compete with Baba Yaga, and she, you know, is a lover of all kinds of riddles.
Example 3
Make an equation of a straight line passing through a point parallel to a straight line if
There is a rational and not very rational solution. The shortest way is at the end of the lesson.
We have worked a little with parallel lines and will return to them later. The case of coinciding straight lines is of little interest, so consider a problem that is well known to you from school curriculum:
How to find the intersection point of two lines?
If straight 
intersect at a point, then its coordinates are the solution systems of linear equations 
How to find the point of intersection of lines? Solve the system.
So much for you geometric meaning of a system of two linear equations in two unknowns Are two intersecting (most often) straight lines on a plane.
Example 4
Find the point of intersection of lines
Decision: There are two ways of solving - graphical and analytical.
The graphical way is to simply draw the data lines and find out the intersection point directly from the drawing: 
Here's our point:. To check it, you should substitute its coordinates in each equation of the straight line, they should fit both there and there. In other words, the coordinates of a point are the solution of the system. Basically, we looked at a graphical way to solve systems of linear equations with two equations, two unknowns.
The graphical method, of course, is not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide so, the point is that it will take time to get a correct and EXACT drawing. In addition, some straight lines are not so easy to construct, and the point of intersection itself may be located somewhere in the thirty 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. Visit the lesson to build relevant skills. How to solve a system of equations?
Answer:
The check is trivial - the coordinates of the intersection point must satisfy every equation in the system.
Example 5
Find the intersection point of the lines if they intersect.
This is an example for a do-it-yourself solution. It is convenient to divide the task into several stages. Analysis of the condition suggests what is needed:
1) Make the equation of the straight line.
2) Make the equation of the straight line.
3) Find out the relative position of the straight lines.
4) If the lines intersect, then find the intersection point.
The development of an algorithm of actions is typical for many geometric problems, and I will repeatedly focus on this.
Complete solution and the answer at the end of the tutorial:
A pair of shoes has not yet been worn down, as we got to the second section of the lesson:
Perpendicular straight lines. Distance from point to 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 build a line perpendicular to a given one?
Example 6
The straight line is given by the equation. Equate a perpendicular line through a point.
Decision: By condition it is known that. It would be nice to find the direction vector of the straight line. Since the lines are perpendicular, the trick is simple:
From the equation "remove" the normal vector:, which will be the direction vector of the straight line.
Let us compose the equation of a straight line by 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) Take out the direction vectors from the equations 
and with the help dot product of vectors we come to the conclusion that the straight lines are indeed perpendicular:.
By the way, you can use normal vectors, it's even easier.
2) Check if the point satisfies the obtained equation 
.
The check, again, is easy to do orally.
Example 7
Find the point of intersection of perpendicular lines if the equation is known 
and point.
This is an example for a do-it-yourself solution. There are several actions in the task, so it is convenient to draw up the solution point by point.
Our exciting journey continues:
Distance from point to line
Before us is a straight strip of the river and our task is to reach it by the shortest way. There are no obstacles, and the most optimal route will be movement along the perpendicular. That is, the distance from a point to a straight line is the length of a perpendicular line.
Distance in geometry is traditionally denoted by the 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 point to line 
Decision: all you need is to carefully substitute the numbers into the formula and perform the calculations:
Answer: 
Let's execute the drawing: 
The distance from the point to the line found is exactly the length of the red line. If you draw up a drawing on checkered paper on a scale of 1 unit. \u003d 1 cm (2 cells), then the distance can be measured with an ordinary ruler.
Consider another task for the same blueprint:
The task is to find the coordinates of a point that is symmetrical to a point with respect to a straight line 
... I propose to perform the actions 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 detailed in this lesson.
3) The point is the midpoint of the line segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the midpoint we find.
It will not be superfluous to check that the distance is also 2.2 units.
Difficulties here may arise in calculations, but in the tower, a micro calculator helps out great, allowing you to count ordinary fractions. Repeatedly advised, will advise and again.
How to find the distance between two parallel lines?
Example 9
Find the distance between two parallel lines
This is another example for an independent solution. Let me give you a little hint: there are infinitely many ways to solve it. Debriefing at the end of the lesson, but better try to guess for yourself, I think your ingenuity was dispersed quite well.
Angle between two straight lines
Every angle is a jamb: 
In geometry, the angle between two straight lines is taken as the SMALLEST 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 straight lines. And his "green" neighbor is considered as such, or oppositely oriented "Crimson" corner.
If the straight lines are perpendicular, then any of the 4 angles can be taken as the angle between them.
How do angles differ? Orientation. First, the direction of the corner "scrolling" is of fundamental importance. Second, a negatively oriented angle is written with a minus sign, for example, if.
Why did I tell this? It seems that you can get by with the usual concept of an angle. The fact is that in the formulas by which we will find the angles, you can easily get 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 straight lines
Decision and Method one
Consider two straight lines given by equations in general form: 
If straight not perpendicularthen oriented the angle between them can be calculated using the formula: 
Let's pay close attention to the denominator - this is exactly scalar product direction vectors of straight lines:
If, then the denominator of the formula vanishes, and the vectors will be orthogonal and the straight lines are perpendicular. That is why a reservation was made about the non-perpendicularity of the straight lines in the formulation.
Based on the above, it is convenient to arrange a solution in two steps:
1) Calculate the scalar product of the direction vectors of straight lines:
, therefore, the straight lines are not perpendicular.
2) The angle between the straight lines is found by the formula:
Through inverse function the corner itself is easy to find. 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 the approximate value (preferably both in degrees and in radians), calculated using a calculator.
Well, minus, so minus, that's okay. Here's a geometric illustration: 
It is not surprising that the angle turned out to have a negative orientation, because in the problem statement, the first number is a straight line and the "twisting" of the angle began with it.
If you really want to get a positive angle, you need to swap the straight lines, that is, take the coefficients from the second equation 
, and the coefficients are taken from the first equation. In short, you must start with a straight line 
.
Coordinate method (distance between point and plane, between straight lines)
Distance between point and plane.
Distance between point and line.
Distance between two straight lines.
The first thing that is useful to know is how to find the distance from a point to a plane:
Values \u200b\u200bA, B, C, D - plane coefficients
x, y, z - point coordinates
A task. Find the distance between the point A \u003d (3; 7; −2) and the plane 4x + 3y + 13z - 20 \u003d 0.
Everything is given, you can immediately substitute the values \u200b\u200binto the equation:
A task. Find the distance from the point K \u003d (1; −2; 7) to the straight line passing through the points V \u003d (8; 6; −13) and T \u003d (−1; −6; 7).
- Find the vector of a straight line.
- We calculate the vector passing through the desired point and any point on the line.
- We set the matrix and find the determinant by the two obtained vectors in the 1st and 2nd points.
- We get the distance when square root from the sum of the squares of the coefficients of the matrix, we divide by the length of the vector that defines the straight line(I think it's not clear, so let's move on to a specific example).
1) TV \u003d (8 - (- 1); 6 - (- 6); -13-7) \u003d (9; 12; −20)
2) The vector is found through the points K and T, although the same could be through K and V or any other point on this line.
TK \u003d (1 - (- 1); −2 - (- 6); 7-7) \u003d (2; 4; 0)
3) We get an m matrix without coefficient D (here it is not needed for the solution):
4) The plane turned out with the coefficients A \u003d 80, B \u003d 40, C \u003d 12,
x, y, z - coordinates of the straight line vector, in this case - vector TV has coordinates (9; 12; −20)
A task. Find the distance between the straight line passing through the points Е \u003d (1; 0; −2), G \u003d (2; 2; −1), and the straight line passing through the points M \u003d (4; −1; 4), L \u003d ( −2; 3; 0).
- We set the vectors of both lines.
- Find the vector by taking one point from each line.
- We write down a matrix of 3 vectors (two lines from the 1st item, one line from the 2nd) and find its numerical determinant.
- We set a matrix of the first two vectors (in step 1). The first line is set as x, y, z.
- The distance is obtained when we divide the resulting value from point 3 modulo by the square root of the sum of squares of point 4.
Let's move on to the numbers.
