Distance from a point to a line in coordinates. The simplest problems with a straight line on a plane. Mutual arrangement of lines. Angle between lines
155*. Determine the actual size of the segment AB of a straight line in general position (Fig. 153, a).
Decision. As you know, the projection of a straight line segment on any plane is equal to the segment itself (taking into account the scale of the drawing), if it is parallel to this plane
(Fig. 153, b). It follows from this that by converting the drawing it is necessary to achieve the parallelism of this segment pl. V or pl. H or supplement the system V, H with another plane perpendicular to the square. V or to pl. H and at the same time parallel to the given segment.
On fig. 153, c shows the introduction of an additional plane S, perpendicular to the square. H and parallel to the given segment AB.
The projection a s b s is equal to the natural value of the segment AB.
On fig. 153, d shows another method: the segment AB is rotated around a straight line passing through point B and perpendicular to the square. H, to a position parallel
sq. V. In this case, point B remains in place, and point A occupies a new position A 1 . Horizon in new position. projection a 1 b || x axis. The projection a "1 b" is equal to the natural value of the segment AB.
156. Pyramid SABCD is given (Fig. 154). Determine the natural size of the pyramid edges AS and CS using the method of changing the projection planes, and the edges BS and DS using the rotation method, and take the axis of rotation perpendicular to the square. H.
157*. Determine the distance from point A to the straight line BC (Fig. 155, a).
Decision. The distance from a point to a line is measured by a segment of a perpendicular drawn from a point to a line.
If the line is perpendicular to any plane (Fig. 155.6), then the distance from the point to the line is measured by the distance between the projection of the point and the projection point of the line on this plane. If a straight line occupies a general position in the V, H system, then in order to determine the distance from a point to a straight line by changing the projection planes, two more additional planes must be introduced into the V, H system.
First (Fig. 155, c) we enter the square. S, parallel to the segment BC (the new axis S/H is parallel to the projection bс), and we construct the projections b s c s and a s . Then (Fig. 155, d) we introduce another square. T perpendicular to line BC (new T/S axis perpendicular to b s c s). We build projections of a straight line and a point - with t (b t) and a t. The distance between points a t and c t (b t) is equal to the distance l from point A to the line BC.
On fig. 155e, the same task is accomplished by the rotation method in its form, which is called the parallel movement method. First, the line BC and point A, keeping their mutual position unchanged, turn around some (not indicated in the drawing) line perpendicular to the square. H, so that the straight line BC is parallel to the square. V. This is equivalent to moving points A, B, C in planes parallel to the square. H. At the same time, the horizon. the projection of a given system (BC + A) does not change either in magnitude or in configuration, only its position relative to the x-axis changes. Set up a horizon. the projection of the straight line BC parallel to the x axis (position b 1 c 1) and determine the projection a 1, setting aside c 1 1 1 \u003d c-1 and a 1 1 1 \u003d a-1, and a 1 1 1 ⊥ c 1 1 1. Drawing straight lines b "b" 1, a "a" 1, c "c" 1 parallel to the x axis, we find the front on them. projections b "1, a" 1, c "1. Next, we move the points B 1, C 1 and A 1 in planes parallel to square V (also without changing their relative position), so as to get B 2 C 2 ⊥ square H. In this case, the projection of the straight line to the front will be perpendicular to the axis x, b 2 c "2 = b" 1 c "1, and to construct the projection a" 2, you need to take b "2 2" 2 = b "1 2" 1 , draw 2 "a" 2 ⊥ b "2 c" 2 and postpone a "2 2" 2 \u003d a "1 2" 1. Now, after spending c 1 c 2 and a 1 a 2 || x 1, we get the projections b 2 s 2 and a 2 and the desired distance l from point A to line BC. You can determine the distance from A to BC by turning the plane defined by point A and line BC around the horizontal of this plane to the position T || pl. H (Fig. 155 , e).
In the plane given by point A and the straight line BC, we draw a horizontal line A-1 (Fig. 155, g) and rotate point B around it. Point B moves to square. R (given in the drawing following R h), perpendicular to A-1; at point O is the center of rotation of point B. Now we determine the natural value of the radius of rotation of VO, (Fig. 155, c). In the required position, i.e. when pl. T defined by point A and line BC will become || sq. H, point B will turn out on R h at a distance Ob 1 from point O (there may be another position on the same track R h, but on the other side of O). Point b 1 is the horizon. the projection of point B after moving it to position B 1 in space, when the plane defined by point A and the straight line BC has taken position T.
Having drawn (Fig. 155, and) the straight line b 1 1, we get the horizon. projection of the straight line BC, already located || sq. H is in the same plane as A. In this position, the distance from a to b 1 1 is equal to the desired distance l. The plane P, in which the given elements lie, can be combined with the square. H (Fig. 155, j), turning the square. P around her horizon. trace. Having passed from setting the plane by the point A and the line BC to setting the lines BC and A-1 (Fig. 155, l), we find the traces of these lines and draw traces P ϑ and P h through them. We are building (Fig. 155, m) combined with the square. H position front. trace - P ϑ0 .
Draw the horizon through point a. frontal projection; the combined frontal passes through point 2 on the trace Р h parallel to Р ϑ0 . Point A 0 - combined with pl. H is the position of point A. Similarly, we find the point B 0 . Direct sun in combined with pl. H position passes through point B 0 and point m (horizontal trace of a straight line).
The distance from the point A 0 to the straight line B 0 C 0 is equal to the desired distance l.
It is possible to carry out the indicated construction by finding only one trace P h (Fig. 155, n and o). The whole construction is similar to turning around the horizontal (see Fig. 155, f, c, i): the trace P h is one of the horizontal lines of the square. R.
Of the methods for converting a drawing given to solve this problem, the method of rotation around a horizontal or frontal is preferable.
158. Pyramid SABC is given (Fig. 156). Determine distances:
a) from top B of the base to its side AC by the method of parallel movement;
b) from the top S of the pyramid to the sides BC and AB of the base by means of rotation around the horizontal;
c) from the top S to the side AC of the base by changing the projection planes.
159. Given a prism (Fig. 157). Determine distances:
a) between the edges AD and CF by changing the projection planes;
b) between ribs BE and CF by rotation around the front;
c) between the edges AD and BE by the method of parallel movement.
160. Determine the actual size of the quadrilateral ABCD (Fig. 158) by combining with the square. N. Use only the horizontal trace of the plane.
161*. Determine the distance between the intersecting lines AB and CD (Fig. 159, a) and construct projections of the common perpendicular to them.
Decision. The distance between the crossing lines is measured by the segment (MN) of the perpendicular to both lines (Fig. 159, b). Obviously, if one of the lines is placed perpendicular to any square. T then
the segment MN of the perpendicular to both lines will be parallel to the square. Its projection on this plane will display the required distance. Projection of the right angle of the maenad MN n AB on the square. T also turns out to be a right angle between m t n t and a t b t , since one of the sides of the right angle AMN, namely MN. parallel to square. T.
On fig. 159, c and d, the desired distance l is determined by the method of changing the projection planes. First, we introduce an additional square. projections S, perpendicular to the square. H and parallel to the straight line CD (Fig. 159, c). Then we introduce another additional square. T, perpendicular to the square. S and perpendicular to the same line CD (Fig. 159, d). Now you can build a projection of the common perpendicular by drawing m t n t from the point c t (d t) perpendicular to the projection a t b t . Points m t and n t are projections of the points of intersection of this perpendicular with lines AB and CD. From the point m t (Fig. 159, e) we find m s on a s b s: the projection m s n s should be parallel to the T / S axis. Further, from m s and n s we find m and n on ab and cd, and from them m "and n" on a "b" and c "d".
On fig. 159, in shows the solution to this problem by the method of parallel movements. First, we put the straight line CD parallel to the square. V: projection c 1 d 1 || X. Next, we move the lines CD and AB from positions C 1 D 1 and A 1 B 1 to positions C 2 B 2 and A 2 B 2 so that C 2 D 2 is perpendicular to H: projection c "2 d" 2 ⊥ x. The segment of the desired perpendicular is located || sq. H, and, therefore, m 2 n 2 expresses the desired distance l between AB and CD. We find the position of the projections m "2, and n" 2 on a "2 b" 2 and c "2 d" 2, then the projections and m 1 and m "1, n 1 and n" 1, finally, the projections m "and n ", m and n.
162. Pyramid SABC is given (Fig. 160). Determine the distance between the edge SB and the side AC of the base of the pyramid and construct projections of the common perpendicular to SB and AC, using the method of changing projection planes.
163. Pyramid SABC is given (Fig. 161). Determine the distance between the edge SH and the side BC of the base of the pyramid and construct projections of the common perpendicular to SX and BC using the parallel displacement method.
164*. Determine the distance from point A to the plane in cases where the plane is given: a) by the triangle BCD (Fig. 162, a); b) traces (Fig. 162, b).
Decision. As you know, the distance from a point to a plane is measured by the magnitude of the perpendicular drawn from the point to the plane. This distance is projected onto any square. life-size projections, if the given plane is perpendicular to the square. projections (Fig. 162, c). This situation can be achieved by converting the drawing, for example, by changing the square. projections. Let's introduce the square. S (Fig. 16ts, d), perpendicular to the square. triangle BCD. To do this, we spend in the square. triangle horizontal B-1 and position the axis of projections S perpendicular to the projection b-1 horizontal. We build projections of a point and a plane - a s and a segment c s d s . The distance from a s to c s d s is equal to the desired distance l of the point to the plane.
On rio. 162, d the method of parallel movement is applied. We move the entire system until the B-1 horizontal of the plane becomes perpendicular to the V plane: the projection b 1 1 1 must be perpendicular to the x-axis. In this position, the plane of the triangle will become front-projecting, and the distance l from point A to it will turn out to be square. V without distortion.
On fig. 162b the plane is given by traces. We introduce (Fig. 162, e) an additional square. S, perpendicular to the square. P: the S/H axis is perpendicular to P h . The rest is clear from the drawing. On fig. 162, well the problem is solved with the help of one displacement: pl. P goes into position P 1, that is, it becomes front-projecting. Track. P 1h is perpendicular to the x-axis. We build a front in this position of the plane. the trace of the horizontal is the point n "1, n 1. The trace P 1ϑ will pass through P 1x and n 1. The distance from a" 1 to P 1ϑ is equal to the desired distance l.
165. Pyramid SABC is given (see fig. 160). Determine the distance from point A to the face SBC of the pyramid using the parallel displacement method.
166. Pyramid SABC is given (see fig. 161). Determine the height of the pyramid using the parallel displacement method.
167*. Determine the distance between the intersecting lines AB and CD (see Fig. 159, a) as the distance between parallel planes drawn through these lines.
Decision. On fig. 163, and planes P and Q are shown parallel to each other, of which pl. Q is drawn through CD parallel to AB, and pl. P - through AB parallel to the square. Q. The distance between such planes is considered to be the distance between the skew lines AB and CD. However, you can restrict yourself to building only one plane, for example Q, parallel to AB, and then determine the distance at least from point A to this plane.
On fig. 163c shows plane Q through CD parallel to AB; in projections held with "e" || a"b" and se || ab. Using the method of changing square. projections (Fig. 163, c), we introduce an additional square. S, perpendicular to the square. V and at the same time
perpendicular to square. Q. To draw the S / V axis, we take the frontal D-1 in this plane. Now we draw S / V perpendicular to d "1" (Fig. 163, c). Pl. Q will be displayed on the square. S as a straight line with s d s . The rest is clear from the drawing.
168. Pyramid SABC is given (see Fig. 160). Determine the distance between the edges SC and AB. Apply: 1) method of changing the area. projections, 2) a method of parallel movement.
169*. Determine the distance between parallel planes, one of which is given by straight lines AB and AC, and the other by straight lines DE and DF (Fig. 164, a). Also perform construction for the case when the planes are given by traces (Fig. 164, b).
Decision. The distance (Fig. 164, c) between parallel planes can be determined by drawing a perpendicular from any point of one plane to another plane. On fig. 164, g introduced an additional square. S perpendicular to the square. H and to both given planes. The S.H axis is perpendicular to the horizon. projection of a horizontal line drawn in one of the planes. We build a projection of this plane and points In another plane on Sq. 5. The distance of the point d s to the line l s a s is equal to the desired distance between parallel planes.
On fig. 164, d another construction is given (according to the method of parallel movement). In order for the plane expressed by the intersecting lines AB and AC to be perpendicular to the square. V, horizon. we set the horizontal projection of this plane perpendicular to the x-axis: 1 1 2 1 ⊥ x. Distance between front. the projection d "1 of the point D and the straight line a" 1 2 "1 (frontal projection of the plane) is equal to the desired distance between the planes.
On fig. 164, e shows the introduction of an additional square. S, perpendicular to pl.H and to the given planes P and Q (the S/H axis is perpendicular to the traces P h and Q h). We construct traces Р s , and Q s . The distance between them (see Fig. 164, c) is equal to the desired distance l between the planes P and Q.
On fig. 164, g shows the movement of the planes P 1 n Q 1, to the position P 1 and Q 1 when the horizon. the traces turn out to be perpendicular to the x-axis. Distance between new front. traces P 1ϑ and Q 1ϑ is equal to the required distance l.
170. Given a parallelepiped ABCDEFGH (Fig. 165). Determine the distances: a) between the bases of the parallelepiped - l 1; b) between faces ABFE and DCGH - l 2 ; c) between the ADHE and BCGF-l 3 faces.
This article talks about the topic « distance from point to line », definitions of the distance from a point to a line are considered with illustrated examples by the method of coordinates. Each block of theory at the end has shown examples of solving similar problems.
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The distance from a point to a line is found by determining the distance from a point to a point. Let's consider in more detail.
Let there be a line a and a point M 1 not belonging to the given line. Draw a line through it blocated perpendicular to the line a. Take the point of intersection of the lines as H 1. We get that M 1 H 1 is a perpendicular, which was lowered from the point M 1 to the line a.
Definition 1
Distance from point M 1 to straight line a called the distance between the points M 1 and H 1 .
There are records of the definition with the figure of 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 straight line is the smallest of all possible. Let's look at this with an example.
If we take the point Q lying on the line a, not coinciding with the point M 1, then we get that the segment M 1 Q is called oblique, lowered from M 1 to the line a. It is necessary to indicate that the perpendicular from the point M 1 is less than any other oblique 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. Hence, we have that M 1 H 1< M 1 Q . Рассмотрим рисунок, приведенный ниже.
The initial data for finding from a point to a straight line allow using several solution methods: through the Pythagorean theorem, definitions of sine, cosine, tangent of an angle, and others. Most tasks of this type are solved at school in geometry lessons.
When, when finding the distance from a point to a line, it is possible to enter a rectangular coordinate system, then the coordinate method is used. In this paragraph, we consider the main two methods for finding the desired distance from a given point.
The first method involves finding the distance as a perpendicular drawn from M 1 to the line a. The second method uses the normal equation of the 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, a straight line a, and you need to find the distance M 1 H 1, you can calculate in two ways. Let's consider them.
First way
If there are coordinates of the point H 1 equal to x 2, y 2, then the distance from the point to the line is calculated from 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 the point H 1.
It is known that a straight line in O x y corresponds to the equation of a straight line in a plane. Let's take a way to define a straight line a through writing a general equation of a straight line or an equation with a slope. We compose the equation of a straight line that passes through the point M 1 perpendicular to a given line a. Let's denote the line by beech b . H 1 is the point of intersection of lines a and b, so to determine the coordinates, you must use the article, which deals with 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 the straight line a is carried out according to the points:
Definition 3
- finding the general equation of the straight line a , having the form A 1 x + B 1 y + C 1 \u003d 0, or an equation with a slope coefficient, having the form y \u003d k 1 x + b 1;
- obtaining the general equation of the line b, which has the form A 2 x + B 2 y + C 2 \u003d 0 or an equation with a slope y \u003d k 2 x + b 2 if the line b intersects the point M 1 and is perpendicular to the 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, the system of linear equations is solved 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 ;
- calculation of the required distance from a point to a straight 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 line on a plane.
Theorem
A rectangular coordinate system has O x y has a point M 1 (x 1, y 1), from which a straight line is drawn a to the plane, given by the normal equation of the plane, having the form cos α x + cos β y - p \u003d 0, equal to modulo the value obtained on the left side of the normal straight line equation, calculated at x = x 1, y = y 1, means that M 1 H 1 = cos α · x 1 + cos β · y 1 - p.
Proof
The line a corresponds to the normal equation of the plane, which has the form cos α x + cos β y - p = 0, then n → = (cos α , cos β) is considered a normal vector of the line a at a distance from the origin to the line a with p units . It is necessary to depict 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 will denote by M 1 H 1 . It is necessary to show the projections M 2 and H 2 of points M 1 and H 2 onto a straight line passing through the point O with a directing vector of the form n → = (cos α , cos β) , and we 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 → .
Variations depend on the location of the point M 1 itself. Consider 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 . Hence, we obtain 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 get that to find the distance from the point M 1 (x 1, y 1) to the straight line a on the plane, several actions must be performed:
Definition 4
- obtaining the normal equation of the 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 line 4 x - 3 y + 35 = 0 .
Decision
Let's use the first method to solve.
To do this, you need 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 . It can be seen from the condition that the line b is perpendicular to the line a, then its direction vector has coordinates equal to (4, - 3) . Thus, we have the opportunity to write the canonical equation of the line b on the plane, since there are coordinates of the point M 1, belongs to the 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's 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 the above, we have that the coordinates of the point H 1 are (- 5; 5) .
It is necessary to calculate the distance from the point M 1 to the straight line a. We have that the coordinates of the points M 1 (- 1, 2) and H 1 (- 5, 5), then we substitute into the formula for finding the distance and we get that
M 1 H 1 \u003d (- 5 - (- 1) 2 + (5 - 2) 2 \u003d 25 \u003d 5
The second solution.
In order to solve in another way, it is necessary to obtain the normal equation of a straight 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 - 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 a straight line and calculate it with the values x = - 1 , y = 2 . Then we get that
4 5 - 1 + 3 5 2 - 7 = - 5
From here we get that the distance from the point M 1 (- 1 , 2) to the given straight line 4 x - 3 y + 35 = 0 has the value - 5 = 5 .
Answer: 5 .
It can be seen that in this method it is important to use the normal equation of a straight line, since this method is the shortest. But the first method is convenient in that 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 a point M 1 (8, 0) and a straight line y = 1 2 x + 1. Find the distance from a given point to a straight line.
Decision
The solution in the first way implies the reduction of a given equation with a slope coefficient to a general equation. To simplify, you can do it differently.
If the product of the slopes of the perpendicular lines is - 1 , then the slope of the line perpendicular to the given y = 1 2 x + 1 is 2 . Now we get the equation of a straight 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 the point H 1, that is, the intersection points y \u003d - 2 x + 16 and y \u003d 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 \u003d 6 \u003d y \u003d 4 x \u003d 6 ⇒ H 1 (6, 4)
It follows that the distance from the point with coordinates M 1 (8 , 0) to the 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 get that M 1 H 1 = 6 - 8 2 + (4 - 0) 2 20 = 2 5 .
The solution in the second way is to pass from the equation with a coefficient to its normal form. That is, we get y \u003d 1 2 x + 1 ⇔ 1 2 x - y + 1 \u003d 0, then the value of the normalizing factor will be - 1 1 2 2 + (- 1) 2 \u003d - 2 5. It follows that the normal equation of a straight 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 calculate from the point M 1 8 , 0 to a straight line of the form - 1 5 x + 2 5 y - 2 5 = 0 . We get:
M 1 H 1 \u003d - 1 5 8 + 2 5 0 - 2 5 \u003d - 10 5 \u003d 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 straight lines 2 x - 3 = 0 and y + 1 = 0 .
Decision
We get the equation of the normal form of the 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 calculate 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 straight line equation y + 1 = 0 has a normalizing factor with a value of -1. This means that the equation will take the form - y - 1 = 0 . We proceed to calculate the distance from the point M 1 (- 2 , 4) to the straight line - y - 1 = 0 . We get that it equals - 4 - 1 = 5.
Answer: 3 1 2 and 5 .
Let us consider in detail the determination of the distance from a given point of the plane to the coordinate axes O x and O y.
In a rectangular coordinate system, the axis O y has an equation of a straight line, which is incomplete and has the form x \u003d 0, and O x - y \u003d 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 straight lines. This is done based on the formulas M 1 H 1 = x 1 and M 1 H 1 = y 1 . Consider 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.
Decision
Since the equation y \u003d 0 refers to the line O x, you can find the distance from M 1 with given coordinates to this line using the formula. We get that 6 = 6 .
Since the equation x \u003d 0 refers to the line O y, you can find the distance from M 1 to this 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 the point A to the line a.
Consider two ways 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 the point M 1 to the line, where the point on the line is called H 1 and is the base of the perpendicular drawn from the point M 1 to the 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 the point M 1 located on the straight line a is the length of the perpendicular M 1 H 1, then we get that with the found coordinates of the 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 get that the whole solution goes to finding the coordinates of the base of the perpendicular drawn from M 1 to the line a. This is done as follows: H 1 is the point where the line a intersects with the plane that passes through the given point.
This means that the algorithm for determining the distance from the point M 1 (x 1, y 1, z 1) to the straight line a of 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 perpendicular to the line;
- determination of the coordinates (x 2 , y 2 , z 2) belonging to the point H 1 which is the point of intersection of the line a and the plane χ ;
- calculation of 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 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 the line a. Given the coordinates of the points M 1 (x 1 , y 1) and M 3 x 3 , y 3 , z 3 , M 3 M 1 → can be calculated:
M 3 M 1 → = (x 1 - x 3, y 1 - y 3, z 1 - z 3)
It is necessary to postpone the vectors a → \u003d a x, a y, a z and M 3 M 1 → \u003d x 1 - x 3, y 1 - y 3, z 1 - z 3 from the point M 3, connect and get a parallelogram figure. M 1 H 1 is the height of the parallelogram.
Consider the figure below.
We have that the height M 1 H 1 is the desired distance, then you need to find it using the formula. That is, we are looking for M 1 H 1 .
Denote the area of the parallelogram by the letter S, is 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 has the form S = a → × M 3 M 1 → . Also, the area of \u200b\u200bthe figure is equal to the product of the lengths of its sides and the height, we get that S \u003d a → M 1 H 1 with a → \u003d a x 2 + a y 2 + a z 2, which is the length of the vector a → \u003d (a x, a y, a z) , which is equal to the side of the parallelogram. Hence, M 1 H 1 is the distance from the point to the line. It is found by 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 points of the algorithm:
Definition 6
- determination of the direction vector of the straight line a - a → = (a x , a y , a z) ;
- calculation of the length of the direction vector a → = a x 2 + a y 2 + a z 2 ;
- obtaining the coordinates x 3 , y 3 , z 3 belonging to the point M 3 located on the line a;
- calculation of the coordinates of the vector M 3 M 1 → ;
- finding the cross 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 according to the formula a → × M 3 M 1 → ;
- calculation of the distance from a point to a line M 1 H 1 = a → × M 3 M 1 → a → .
Solving problems on finding the distance from a given point to a given straight 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 .
Decision
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 plane χ to the straight line given by the condition. It is necessary to move from the canonical form 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
Hence we have that H 1 (1, - 1, 0) .
M 1 H 1 \u003d 1 - 2 2 + - 1 - - 4 2 + 0 - - 1 2 \u003d 11
The second method must be started by searching for coordinates in the canonical equation. To do this, 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 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 get that the length of the cross product is 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 we apply it and get:
M 1 H 1 = a → × M 3 M 1 → a → = 330 30 = 11
Answer: 11 .
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
To calculate the distance from a given point M to a line L, different methods can be used. For example, if we take an arbitrary point M 0 on the line L, then we can define orthogonal projection of the vector M 0 M onto the direction of the normal vector of the straight line. This projection, up to a sign, is the required distance.
Another way to calculate the distance from a point to a line is to use normal equation of a straight line. Let the line L be given by the normal equation (4.23). If the point M(x; y) does not lie on the line L, then the orthogonal projection pr n OM radius-vector point M to the direction of the unit normal vector n of the straight line L is equal to the scalar product of the vectors OM and n, i.e. x cosφ + y sinφ. The same projection is equal to the sum of the distance p from the origin to the straight line and some value δ (Fig. 4.10). The value of δ in absolute value is equal to the distance from the point M to the straight line. In this case, δ > 0 if the points M and O are on opposite sides of the straight line, and δ is the deviation of the point M from the straight line.
The deviation δ for the point M(x; y) from the line L is calculated as the difference between the projection pr n OM and the distance p from the origin to the line (see Fig. 4.10), i.e. δ \u003d x cosφ + y sinφ - p.
Using this formula, one can also obtain the distance p(M, L) from the point M(x; y) to the line L given by the normal equation: p(M, L) = |δ | = |x cosφ + y sinφ - p|.
2 Two adjacent angles add up to 180°
Given the above conversion procedure general equation of a straight line into its normal equation, we obtain a formula for the distance from the point M(x; y) to the line L, given by its general equation:
Example 4.8. Let's find the general equations for the height AH, the median AM and the bisector AD of the triangle ABC coming out of the vertex A. The coordinates of the vertices of the triangle A(-1;-3), B(7; 3), C(1;7) are known.
First of all, let's clarify the condition of the example: the indicated equations mean the equations of the lines L AH, L AM and L AD, on which the height AH, the median AM and the bisector AD of the specified triangle are located, respectively (Fig. 4.11).
To find the equation of the straight line L AM , we use the fact that the median divides the opposite side of the triangle in half. Having found the coordinates (x 1; y 1) of the middle of the side BC x 1 = (7 + 1)/2 = 4, y 1 = (3 + 7)/2 = 5, we write the equation for L AM in the form equation of a straight line passing through two points,(x + 1)/(4 + 1) = (y + 3)/(5 + 3). After transformations, we obtain the general equation of the median 8x - 5y - 7 \u003d 0./p>
To find the equation for the height L AH, we use the fact that the height is perpendicular to the opposite side of the triangle. Therefore, the vector BC is perpendicular to the height AH and can be chosen as the normal vector of the line L AH . The equation of this line is obtained from (4.15) by substituting the coordinates of the point A and the normal vector of the line L AH:
(-6)(x + 1) + 4(y + 3) = 0.
After transformations, we obtain the general equation for the height 3x - 2y - 3 = 0.
To find the equation of the bisector L AD , we use the fact that the bisector AD belongs to the set of those points N(x; y) that are equidistant from the lines L AB and L AC . The equation of this set has the form
P(N, L AB) = P(N, L AC), (4.28)
and it defines two lines passing through the point A and dividing the angles between the lines L AB and L AC in half. Using the equation of a straight line passing through two points, we find the general equations of the lines L AB and L AC:
L AB: (x + 1)/(7 + 1) = (y + 3)/(3 + 3), L AC: (x + 1)/(1 + 1) = (y + 3)/(7 + 3)
After transformations, we obtain L AB: 3x - 4y - 9 \u003d 0, L AC: 5x - y + 2 \u003d 0. Equation (4.28) using formula (4.27) to calculate the distance from a point to a straight line, we write in the form
Let's transform it by expanding the modules:
As a result, we obtain the general equations of two lines
(3 ± 25/√26)x + (-4 ± 5/√26)y + (-9 ± 10/√26) = 0
To choose the bisector equation from them, we take into account that the vertices B and C of the triangle are located on opposite sides of the desired line and therefore substituting their coordinates into the left side of the general equation of the line L AD should give values with different signs. We choose the equation corresponding to the upper sign, i.e.
(3 - 25/√26)x + (-4 + 5/√26)y + (-9 - 10/√26) = 0
Substituting the coordinates of point B into the left side of this equation gives a negative value because
(3 - 25/√26)7 + (-4 + 5/√26)3 + (-9 - 10/√26) = 21 - 12 - 9 + (-175 + 15 - 10)/√26 = -170/√26
and the same sign is obtained for the coordinates of the point C, since
(3 - 25/√26)1 + (-4 + 5/√26)7 + (-9 - 10/√26) = 3 - 28 - 9 + (-25 + 35 - 10)/√26 = -34
Therefore, vertices B and C are located on the same side of the straight line with the chosen equation, and therefore the equation of the bisector is
(3 + 25/√26)x + (-4 - 5/√26)y + (-9 + 10/√26) = 0.
First level
Coordinates and vectors. Comprehensive Guide (2019)
In this article, you and I will begin a discussion of one "magic wand" that will allow you to reduce many problems in geometry to simple arithmetic. This “wand” can make your life much easier, especially when you feel insecure in building spatial figures, sections, etc. All this requires a certain imagination and practical skills. The method, which we will begin to consider here, will allow you to abstract almost completely from all kinds of geometric constructions and reasoning. The method is called "coordinate method". In this article, we will consider the following questions:
- Coordinate plane
- Points and vectors on the plane
- Building a vector from two points
- Vector length (distance between two points)
- Midpoint coordinates
- Dot product of vectors
- Angle between two vectors
I think you already guessed why the coordinate method is called that? It is true that it got such a name, since it does not operate with geometric objects, but with their numerical characteristics (coordinates). And the transformation itself, which makes it possible to move from geometry to algebra, consists in introducing a coordinate system. If the original figure was flat, then the coordinates are two-dimensional, and if the figure is three-dimensional, then the coordinates are three-dimensional. In this article, we will consider only the two-dimensional case. And the main purpose of the article is to teach you how to use some basic techniques of the coordinate method (they sometimes turn out to be useful when solving problems in planimetry in part B of the Unified State Examination). The following two sections on this topic are devoted to the discussion of methods for solving problems C2 (the problem of stereometry).
Where would it be logical to start discussing the coordinate method? Probably with the concept of a coordinate system. Remember when you first met her. It seems to me that in the 7th grade, when you learned about the existence of a linear function, for example. Let me remind you that you built it point by point. Do you remember? You chose an arbitrary number, substituted it into the formula and calculated in this way. For example, if, then, if, then, etc. What did you get as a result? And you received points with coordinates: and. Then you drew a “cross” (coordinate system), chose a scale on it (how many cells you will have as a single segment) and marked the points you received on it, which you then connected with a straight line, the resulting line is the graph of the function.
There are a few things that need to be explained to you in a little more detail:
1. You choose a single segment for reasons of convenience, so that everything fits nicely and compactly in the picture
2. It is assumed that the axis goes from left to right, and the axis goes from bottom to top
3. They intersect at a right angle, and the point of their intersection is called the origin. It is marked with a letter.
4. In the record of the coordinate of a point, for example, on the left in brackets is the coordinate of the point along the axis, and on the right, along the axis. In particular, simply means that the point
5. In order to set any point on the coordinate axis, you need to specify its coordinates (2 numbers)
6. For any point lying on the axis,
7. For any point lying on the axis,
8. The axis is called the x-axis
9. The axis is called the y-axis
Now let's take the next step with you: mark two points. Connect these two points with a line. And let's put the arrow as if we were drawing a segment from point to point: that is, we will make our segment directed!
Remember what another name for a directed segment is? That's right, it's called a vector!
Thus, if we connect a dot to a dot, and the beginning will be point A, and the end will be point B, then we get a vector. You also did this construction in the 8th grade, remember?
It turns out that vectors, like points, can be denoted by two numbers: these numbers are called the coordinates of the vector. Question: do you think it is enough for us to know the coordinates of the beginning and end of the vector to find its coordinates? It turns out that yes! And it's very easy to do:
Thus, since in the vector the point is the beginning, and the end, the vector has the following coordinates:
For example, if, then the coordinates of the vector
Now let's do the opposite, find the coordinates of the vector. What do we need to change for this? Yes, you need to swap the beginning and end: now the beginning of the vector will be at a point, and the end at a point. Then:
Look closely, what is the difference between vectors and? Their only difference is the signs in the coordinates. They are opposite. This fact is written like this:
Sometimes, if it is not specifically stated which point is the beginning of the vector, and which is the end, then the vectors are denoted not by two capital letters, but by one lower case, for example:, etc.
Now a little practice and find the coordinates of the following vectors:
Examination:
Now solve the problem a little more difficult:
A vector torus with on-cha-scrap at a point has co-or-di-on-you. Find-di-te abs-cis-su points.
All the same is quite prosaic: Let be the coordinates of the point. Then
I compiled the system by determining what the coordinates of a vector are. Then the point has coordinates. We are interested in the abscissa. Then
Answer:
What else can you do with vectors? Yes, almost everything is the same as with ordinary numbers (except that you cannot divide, but you can multiply in two ways, one of which we will discuss here a little later)
- Vectors can be stacked with each other
- Vectors can be subtracted from each other
- Vectors can be multiplied (or divided) by an arbitrary non-zero number
- Vectors can be multiplied with each other
All these operations have a quite visual geometric representation. For example, the triangle (or parallelogram) rule for addition and subtraction:
A vector stretches or shrinks or changes direction when multiplied or divided by a number:
However, here we will be interested in the question of what happens to the coordinates.
1. When adding (subtracting) two vectors, we add (subtract) their coordinates element by element. I.e:
2. When multiplying (dividing) a vector by a number, all its coordinates are multiplied (divided) by this number:
For example:
· Find-di-the sum of ko-or-di-nat century-to-ra.
Let's first find the coordinates of each of the vectors. Both of them have the same origin - the origin point. Their ends are different. Then, . Now we calculate the coordinates of the vector Then the sum of the coordinates of the resulting vector is equal to.
Answer:
Now solve the following problem yourself:
· Find the sum of the coordinates of the vector
We check:
Let's now consider the following problem: we have two points on the coordinate plane. How to find the distance between them? Let the first point be, and the second. Let's denote the distance between them as . Let's make the following drawing for clarity:
What I've done? I, firstly, connected the points and, and also drew a line parallel to the axis from the point, and drew a line parallel to the axis from the point. Did they intersect at a point, forming a wonderful figure? Why is she wonderful? Yes, you and I almost know everything about a right triangle. Well, the Pythagorean theorem, for sure. The desired segment is the hypotenuse of this triangle, and the segments are the legs. What are the coordinates of the point? Yes, they are easy to find from the picture: Since the segments are parallel to the axes and, respectively, their lengths are easy to find: if we denote the lengths of the segments, respectively, through, then
Now let's use the Pythagorean theorem. We know the lengths of the legs, we will find the hypotenuse:
Thus, the distance between two points is the root sum of the squared differences from the coordinates. Or - the distance between two points is the length of the segment connecting them. It is easy to see that the distance between the points does not depend on the direction. Then:
From this we draw three conclusions:
Let's practice a bit on calculating the distance between two points:
For example, if, then the distance between and is
Or let's go differently: find the coordinates of the vector
And find the length of the vector:
As you can see, it's the same!
Now practice a little on your own:
Task: find the distance between the given points:
We check:
Here are a couple more problems for the same formula, though they sound a little different:
1. Find-di-te the square of the length of the eyelid-to-ra.
2. Nai-di-te square of eyelid length-to-ra
I'm guessing you can handle them easily? We check:
1. And this is for attentiveness) We have already found the coordinates of the vectors before: . Then the vector has coordinates. The square of its length will be:
2. Find the coordinates of the vector
Then the square of its length is
Nothing complicated, right? Simple arithmetic, nothing more.
The following puzzles cannot be unambiguously classified, they are rather for general erudition and the ability to draw simple pictures.
1. Find-di-those sine of the angle on-clo-on-from-cut, connect-one-n-th-th point, with the abscissa axis.
and
How are we going to do it here? You need to find the sine of the angle between and the axis. And where can we look for the sine? That's right, in a right triangle. So what do we need to do? Build this triangle!
Since the coordinates of the point and, then the segment is equal, and the segment. We need to find the sine of the angle. Let me remind you that the sine is the ratio of the opposite leg to the hypotenuse, then
What are we left to do? Find the hypotenuse. You can do it in two ways: using the Pythagorean theorem (the legs are known!) or using the formula for the distance between two points (actually the same as the first method!). I will go the second way:
Answer:
The next task will seem even easier to you. She - on the coordinates of the point.
Task 2. From the point, the per-pen-di-ku-lar is lowered onto the abs-ciss axis. Nai-di-te abs-cis-su os-no-va-niya per-pen-di-ku-la-ra.
Let's make a drawing:
The base of the perpendicular is the point at which it intersects the x-axis (axis) for me this is a point. The figure shows that it has coordinates: . We are interested in the abscissa - that is, the "X" component. She is equal.
Answer: .
Task 3. Under the conditions of the previous problem, find the sum of the distances from the point to the coordinate axes.
The task is generally elementary if you know what the distance from a point to the axes is. You know? I hope, but still I remind you:
So, in my drawing, located a little higher, I have already depicted one such perpendicular? What axis is it? to the axis. And what is its length then? She is equal. Now draw a perpendicular to the axis yourself and find its length. It will be equal, right? Then their sum is equal.
Answer: .
Task 4. In the conditions of task 2, find the ordinate of a point symmetrical to a point about the x-axis.
I think you intuitively understand what symmetry is? Very many objects have it: many buildings, tables, planes, many geometric shapes: a ball, a cylinder, a square, a rhombus, etc. Roughly speaking, symmetry can be understood as follows: a figure consists of two (or more) identical halves. This symmetry is called axial. What then is an axis? This is exactly the line along which the figure can, relatively speaking, be “cut” into identical halves (in this picture, the axis of symmetry is straight):
Now let's get back to our task. We know that we are looking for a point that is symmetric about the axis. Then this axis is the axis of symmetry. So, we need to mark a point so that the axis cuts the segment into two equal parts. Try to mark such a point yourself. Now compare with my solution:
Did you do the same? Well! At the found point, we are interested in the ordinate. She is equal
Answer:
Now tell me, after thinking for a second, what will be the abscissa of the point symmetrical to point A about the y-axis? What is your answer? Correct answer: .
In general, the rule can be written like this:
A point symmetrical to a point about the x-axis has the coordinates:
A point symmetrical to a point about the y-axis has coordinates:
Well, now it's really scary. task: Find the coordinates of a point that is symmetrical to a point, relative to the origin. You first think for yourself, and then look at my drawing!
Answer:
Now parallelogram problem:
Task 5: The points are ver-shi-na-mi-pa-ral-le-lo-gram-ma. Find-dee-te or-dee-on-tu points.
You can solve this problem in two ways: logic and the coordinate method. I will first apply the coordinate method, and then I will tell you how you can decide otherwise.
It is quite clear that the abscissa of the point is equal. (it lies on the perpendicular drawn from the point to the x-axis). We need to find the ordinate. Let's take advantage of the fact that our figure is a parallelogram, which means that. Find the length of the segment using the formula for the distance between two points:
We lower the perpendicular connecting the point with the axis. The point of intersection is denoted by a letter.
The length of the segment is equal. (find the problem yourself, where we discussed this moment), then we will find the length of the segment using the Pythagorean theorem:
The length of the segment is exactly the same as its ordinate.
Answer: .
Another solution (I'll just provide a picture that illustrates it)
Solution progress:
1. Spend
2. Find point coordinates and length
3. Prove that.
Another one cut length problem:
The points are-la-yut-xia top-shi-on-mi tri-angle-no-ka. Find the length of his midline, par-ral-lel-noy.
Do you remember what the middle line of a triangle is? Then for you this task is elementary. If you do not remember, then I will remind you: the middle line of a triangle is a line that connects the midpoints of opposite sides. It is parallel to the base and equal to half of it.
The base is a segment. We had to look for its length earlier, it is equal. Then the length of the midline is half as long and equal.
Answer: .
Comment: This problem can be solved in another way, which we will turn to a little later.
In the meantime, here are a few tasks for you, practice on them, they are quite simple, but they help to “fill your hand” using the coordinate method!
1. The points appear-la-yut-xia top-shi-on-mi tra-pe-tion. Find the length of its midline.
2. Points and yav-la-yut-xia ver-shi-na-mi pa-ral-le-lo-gram-ma. Find-dee-te or-dee-on-tu points.
3. Find the length from the cut, connect the second point and
4. Find-di-te the area for-the-red-shen-noy fi-gu-ry on the ko-or-di-nat-noy plane.
5. A circle centered at na-cha-le ko-or-di-nat passes through a point. Find-de-te her ra-di-mustache.
6. Nai-di-te ra-di-us circle-no-sti, describe-san-noy near the right-angle-no-ka, the tops-shi-ny of something-ro-go have co-or -di-na-you co-from-reply-but
Solutions:
1. It is known that the midline of a trapezoid is equal to half the sum of its bases. The base is equal, but the base. Then
Answer:
2. The easiest way to solve this problem is to notice that (parallelogram rule). Calculate the coordinates of the vectors and is not difficult: . When adding vectors, the coordinates are added. Then has coordinates. The point has the same coordinates, since the beginning of the vector is a point with coordinates. We are interested in the ordinate. She is equal.
Answer:
3. We act immediately according to the formula for the distance between two points:
Answer:
4. Look at the picture and say, between which two figures is the shaded area “squeezed”? It is sandwiched between two squares. Then the area of the desired figure is equal to the area of the large square minus the area of the small one. The side of the small square is a segment connecting the points and its length is
Then the area of the small square is
We do the same with a large square: its side is a segment connecting the points and its length is equal to
Then the area of the large square is
The area of the desired figure is found by the formula:
Answer:
5. If the circle has the origin as its center and passes through a point, then its radius will be exactly equal to the length of the segment (make a drawing and you will understand why this is obvious). Find the length of this segment:
Answer:
6. It is known that the radius of a circle circumscribed about a rectangle is equal to half of its diagonal. Let's find the length of any of the two diagonals (after all, in a rectangle they are equal!)
Answer:
Well, did you manage everything? It wasn't that hard to figure it out, was it? There is only one rule here - to be able to make a visual picture and simply “read” all the data from it.
We have very little left. There are literally two more points that I would like to discuss.
Let's try to solve this simple problem. Let two points and be given. Find the coordinates of the middle of the segment. The solution to this problem is as follows: let the point be the desired middle, then it has coordinates:
I.e: coordinates of the middle of the segment = arithmetic mean of the corresponding coordinates of the ends of the segment.
This rule is very simple and usually does not cause difficulties for students. Let's see in what problems and how it is used:
1. Find-di-te or-di-na-tu se-re-di-us from-cut, connect-nya-yu-th-th point and
2. The points are yav-la-yut-xia ver-shi-na-mi-che-you-reh-coal-no-ka. Find-di-te or-di-na-tu points of re-re-se-che-niya of his dia-go-on-lei.
3. Find-di-te abs-cis-su of the center of the circle, describe-san-noy near the rectangle-no-ka, the tops-shi-we have something-ro-go co-or-di-na-you co-from-vet-stvenno-but.
Solutions:
1. The first task is just a classic. We act immediately by determining the midpoint of the segment. She has coordinates. The ordinate is equal.
Answer:
2. It is easy to see that the given quadrilateral is a parallelogram (even a rhombus!). You can prove it yourself by calculating the lengths of the sides and comparing them with each other. What do I know about a parallelogram? Its diagonals are bisected by the intersection point! Aha! So the point of intersection of the diagonals is what? This is the middle of any of the diagonals! I will choose, in particular, the diagonal. Then the point has coordinates. The ordinate of the point is equal to.
Answer:
3. What is the center of the circle circumscribed about the rectangle? It coincides with the point of intersection of its diagonals. What do you know about the diagonals of a rectangle? They are equal and the intersection point is divided in half. The task has been reduced to the previous one. Take, for example, the diagonal. Then if is the center of the circumscribed circle, then is the middle. I'm looking for coordinates: The abscissa is equal.
Answer:
Now practice a little on your own, I will only give the answers to each problem so that you can check yourself.
1. Nai-di-te ra-di-us circle-no-sti, describe-san-noy near the triangle-no-ka, the tops of someone-ro-go have ko-or-di -no misters
2. Find-di-te or-di-na-tu the center of the circle, describe the san-noy near the triangle-no-ka, the tops-shi-we have something-ro-go coordinates
3. What kind of ra-di-y-sa should there be a circle with a center at a point so that it touches the abs-ciss axis?
4. Find-di-te or-di-on-that point of re-re-se-che-ing of the axis and from-cut, connect-nya-yu-th-th point and
Answers:
Did everything work out? I really hope for it! Now - the last push. Now be especially careful. The material that I'm going to explain now is not only relevant to the simple coordinate method problems in Part B, but is also ubiquitous in Problem C2.
Which of my promises have I not yet kept? Remember what operations on vectors I promised to introduce and which ones I eventually introduced? Am I sure I haven't forgotten anything? Forgot! I forgot to explain what multiplication of vectors means.
There are two ways to multiply a vector by a vector. Depending on the chosen method, we will get objects of a different nature:
The vector product is quite tricky. How to do it and why it is needed, we will discuss with you in the next article. And in this we will focus on the scalar product.
There are already two ways that allow us to calculate it:
As you guessed, the result should be the same! So let's look at the first way first:
Dot product through coordinates
Find: - common notation for dot product
The formula for the calculation is as follows:
That is, the dot product = the sum of the products of the coordinates of the vectors!
Example:
Find-dee-te
Decision:
Find the coordinates of each of the vectors:
We calculate the scalar product by the formula:
Answer:
You see, absolutely nothing complicated!
Well, now try it yourself:
Find-di-te scalar-noe pro-from-ve-de-nie century-to-ditch and
Did you manage? Maybe he noticed a little trick? Let's check:
Vector coordinates, as in the previous task! Answer: .
In addition to the coordinate, there is another way to calculate the scalar product, namely, through the lengths of the vectors and the cosine of the angle between them:
Denotes the angle between the vectors and.
That is, the scalar product is equal to the product of the lengths of the vectors and the cosine of the angle between them.
Why do we need this second formula, if we have the first one, which is much simpler, at least there are no cosines in it. And we need it so that from the first and second formulas we can deduce how to find the angle between vectors!
Let Then remember the formula for the length of a vector!
Then if I plug this data into the dot product formula, I get:
But on the other side:
So what have we got? We now have a formula to calculate the angle between two vectors! Sometimes, for brevity, it is also written like this:
That is, the algorithm for calculating the angle between vectors is as follows:
- We calculate the scalar product through the coordinates
- Find the lengths of vectors and multiply them
- Divide the result of point 1 by the result of point 2
Let's practice with examples:
1. Find the angle between the eyelids-to-ra-mi and. Give your answer in degrees.
2. Under the conditions of the previous problem, find the cosine between the vectors
Let's do this: I'll help you solve the first problem, and try to do the second one yourself! I agree? Then let's start!
1. These vectors are our old friends. We have already considered their scalar product and it was equal. Their coordinates are: , . Then we find their lengths:
Then we are looking for the cosine between the vectors:
What is the cosine of the angle? This is the corner.
Answer:
Well, now solve the second problem yourself, and then compare! I'll just give a very short solution:
2. has coordinates, has coordinates.
Let be the angle between the vectors and, then
Answer:
It should be noted that the tasks directly on the vectors and the method of coordinates in part B of the examination paper are quite rare. However, the vast majority of C2 problems can be easily solved by introducing a coordinate system. So you can consider this article as a foundation, on the basis of which we will make quite tricky constructions that we will need to solve complex problems.
COORDINATES AND VECTORS. INTERMEDIATE LEVEL
You and I continue to study the method of coordinates. In the last part, we derived a number of important formulas that allow:
- Find vector coordinates
- Find the length of a vector (alternatively: the distance between two points)
- Add, subtract vectors. Multiply them by a real number
- Find the midpoint of a segment
- Calculate dot product of vectors
- Find the angle between vectors
Of course, the entire coordinate method does not fit into these 6 points. It underlies such a science as analytical geometry, which you will get acquainted with at the university. I just want to build a foundation that will allow you to solve problems in a single state. exam. We figured out the tasks of part B in Now it's time to move to a qualitatively new level! This article will be devoted to a method for solving those C2 problems in which it would be reasonable to switch to the coordinate method. This reasonableness is determined by what needs to be found in the problem, and what figure is given. So, I would use the coordinate method if the questions are:
- Find the angle between two planes
- Find the angle between a line and a plane
- Find the angle between two lines
- Find the distance from a point to a plane
- Find the distance from a point to a line
- Find the distance from a straight line to a plane
- Find the distance between two lines
If the figure given in the condition of the problem is a body of revolution (ball, cylinder, cone ...)
Suitable figures for the coordinate method are:
- cuboid
- Pyramid (triangular, quadrangular, hexagonal)
Also in my experience it is inappropriate to use the coordinate method for:
- Finding the areas of sections
- Calculations of volumes of bodies
However, it should be immediately noted that three “unfavorable” situations for the coordinate method are quite rare in practice. In most tasks, it can become your savior, especially if you are not very strong in three-dimensional constructions (which are sometimes quite intricate).
What are all the figures I have listed above? They are no longer flat, such as a square, triangle, circle, but voluminous! Accordingly, we need to consider not a two-dimensional, but a three-dimensional coordinate system. It is built quite easily: just in addition to the abscissa and ordinates, we will introduce another axis, the applicate axis. The figure schematically shows their relative position:
All of them are mutually perpendicular, intersect at one point, which we will call the origin. The abscissa axis, as before, will be denoted, the ordinate axis - , and the introduced applicate axis - .
If earlier each point on the plane was characterized by two numbers - the abscissa and the ordinate, then each point in space is already described by three numbers - the abscissa, the ordinate, the applicate. For example:
Accordingly, the abscissa of the point is equal, the ordinate is , and the applicate is .
Sometimes the abscissa of a point is also called the projection of the point onto the abscissa axis, the ordinate is the projection of the point onto the y-axis, and the applicate is the projection of the point onto the applicate axis. Accordingly, if a point is given then, a point with coordinates:
called the projection of a point onto a plane
called the projection of a point onto a plane
A natural question arises: are all the formulas derived for the two-dimensional case valid in space? The answer is yes, they are just and have the same appearance. For a small detail. I think you already guessed which one. In all formulas, we will have to add one more term responsible for the applicate axis. Namely.
1. If two points are given: , then:
- Vector coordinates:
- Distance between two points (or vector length)
- The middle of the segment has coordinates
2. If two vectors are given: and, then:
- Their dot product is:
- The cosine of the angle between the vectors is:
However, space is not so simple. As you understand, the addition of one more coordinate introduces a significant variety in the spectrum of figures "living" in this space. And for further narration, I need to introduce some, roughly speaking, "generalization" of the straight line. This "generalization" will be a plane. What do you know about plane? Try to answer the question, what is a plane? It's very difficult to say. However, we all intuitively imagine what it looks like:
Roughly speaking, this is a kind of endless “leaf” thrust into space. "Infinity" should be understood that the plane extends in all directions, that is, its area is equal to infinity. However, this explanation "on the fingers" does not give the slightest idea about the structure of the plane. And we will be interested in it.
Let's remember one of the basic axioms of geometry:
- A straight line passes through two different points on a plane, moreover, only one:
Or its analog in space:
Of course, you remember how to derive the equation of a straight line from two given points, this is not at all difficult: if the first point has coordinates: and the second, then the equation of the straight line will be as follows:
You went through this in 7th grade. In space, the equation of a straight line looks like this: let us have two points with coordinates: , then the equation of a straight line passing through them has the form:
For example, a line passes through points:
How should this be understood? This should be understood as follows: a point lies on a line if its coordinates satisfy the following system:
We will not be very interested in the equation of a straight line, but we need to pay attention to the very important concept of the directing vector of a straight line. - any non-zero vector lying on a given line or parallel to it.
For example, both vectors are direction vectors of a straight line. Let be a point lying on a straight line, and be its directing vector. Then the equation of a straight line can be written in the following form:
Once again, I will not be very interested in the equation of a straight line, but I really need you to remember what a direction vector is! Again: it is ANY non-zero vector lying on a line, or parallel to it.
Withdraw three-point equation of a plane is no longer so trivial, and is usually not covered in a high school course. But in vain! This technique is vital when we resort to the coordinate method to solve complex problems. However, I assume that you are full of desire to learn something new? Moreover, you will be able to impress your teacher at the university when it turns out that you already know how to use the technique that is usually studied in the course of analytic geometry. So let's get started.
The equation of a plane is not too different from the equation of a straight line on a plane, namely, it has the form:
some numbers (not all equal to zero), but variables, for example: etc. As you can see, the equation of a plane is not very different from the equation of a straight line (linear function). However, remember what we argued with you? We said that if we have three points that do not lie on one straight line, then the equation of the plane is uniquely restored from them. But how? I'll try to explain to you.
Since the plane equation is:
And the points belong to this plane, then when substituting the coordinates of each point into the equation of the plane, we should get the correct identity:
Thus, there is a need to solve three equations already with unknowns! Dilemma! However, we can always assume that (for this we need to divide by). Thus, we get three equations with three unknowns:
However, we will not solve such a system, but write out the cryptic expression that follows from it:
Equation of a plane passing through three given points
\[\left| (\begin(array)(*(20)(c))(x - (x_0))&((x_1) - (x_0))&((x_2) - (x_0))\\(y - (y_0) )&((y_1) - (y_0))&((y_2) - (y_0))\\(z - (z_0))&((z_1) - (z_0))&((z_2) - (z_0)) \end(array)) \right| = 0\]
Stop! What else is this? Some very unusual module! However, the object that you see in front of you has nothing to do with the module. This object is called a third-order determinant. From now on, when you deal with the method of coordinates on a plane, you will often come across these very determinants. What is a third order determinant? Oddly enough, it's just a number. It remains to understand what specific number we will compare with the determinant.
Let's first write the third-order determinant in more general view:
Where are some numbers. Moreover, by the first index we mean the row number, and by the index - the column number. For example, it means that the given number is at the intersection of the second row and the third column. Let's pose the following question: how exactly are we going to calculate such a determinant? That is, what specific number will we compare it with? For the determinant of precisely the third order, there is a heuristic (visual) triangle rule, it looks like this:
- The product of the elements of the main diagonal (from upper left to lower right) the product of the elements that form the first triangle "perpendicular" to the main diagonal the product of the elements that form the second triangle "perpendicular" to the main diagonal
- The product of the elements of the secondary diagonal (from the upper right to the lower left) the product of the elements that form the first triangle "perpendicular" to the secondary diagonal the product of the elements that form the second triangle "perpendicular" to the secondary diagonal
- Then the determinant is equal to the difference between the values obtained at the step and
If we write all this in numbers, then we get the following expression:
However, you don’t need to memorize the calculation method in this form, it’s enough to just keep the triangles in your head and the very idea of \u200b\u200bwhat is added to what and what is then subtracted from what).
Let's illustrate the triangle method with an example:
1. Calculate the determinant:
Let's figure out what we add and what we subtract:
Terms that come with a "plus":
This is the main diagonal: the product of the elements is
The first triangle, "perpendicular to the main diagonal: the product of the elements is
The second triangle, "perpendicular to the main diagonal: the product of the elements is
We add three numbers:
Terms that come with a "minus"
This is a side diagonal: the product of the elements is
The first triangle, "perpendicular to the secondary diagonal: the product of the elements is
The second triangle, "perpendicular to the secondary diagonal: the product of the elements is
We add three numbers:
All that remains to be done is to subtract from the sum of the plus terms the sum of the minus terms:
Thus,
As you can see, there is nothing complicated and supernatural in the calculation of third-order determinants. It is simply important to remember about triangles and not to make arithmetic mistakes. Now try to calculate yourself:
We check:
- The first triangle perpendicular to the main diagonal:
- The second triangle perpendicular to the main diagonal:
- The sum of the plus terms:
- First triangle perpendicular to the side diagonal:
- The second triangle, perpendicular to the side diagonal:
- The sum of terms with a minus:
- Sum of plus terms minus sum of minus terms:
Here's a couple more determinants for you, calculate their values yourself and compare with the answers:
Answers:
Well, did everything match? Great, then you can move on! If there are difficulties, then my advice is this: on the Internet there are a bunch of programs for calculating the determinant online. All you need is to come up with your own determinant, calculate it yourself, and then compare it with what the program calculates. And so on until the results start to match. I'm sure this moment will not be long in coming!
Now let's return to the determinant that I wrote out when I talked about the equation of a plane passing through three given points:
All you have to do is calculate its value directly (using the triangle method) and set the result equal to zero. Naturally, since they are variables, you will get some expression that depends on them. It is this expression that will be the equation of a plane passing through three given points that do not lie on one straight line!
Let's illustrate this with a simple example:
1. Construct the equation of the plane passing through the points
We compose a determinant for these three points:
Simplifying:
Now we calculate it directly according to the rule of triangles:
\[(\left| (\begin(array)(*(20)(c))(x + 3)&2&6\\(y - 2)&0&1\\(z + 1)&5&0\end(array)) \ right| = \left((x + 3) \right) \cdot 0 \cdot 0 + 2 \cdot 1 \cdot \left((z + 1) \right) + \left((y - 2) \right) \cdot 5 \cdot 6 - )\]
Thus, the equation of the plane passing through the points is:
Now try to solve one problem yourself, and then we will discuss it:
2. Find the equation of the plane passing through the points
Well, let's discuss the solution now:
We make a determinant:
And calculate its value:
Then the equation of the plane has the form:
Or, reducing by, we get:
Now two tasks for self-control:
- Construct the equation of a plane passing through three points:
Answers:
Did everything match? Again, if there are certain difficulties, then my advice is this: you take three points from your head (with a high degree of probability they will not lie on one straight line), build a plane on them. And then check yourself online. For example, on the site:
However, with the help of determinants, we will construct not only the equation of the plane. Remember, I told you that for vectors, not only the dot product is defined. There is also a vector, as well as a mixed product. And if the scalar product of two vectors will be a number, then the vector product of two vectors will be a vector, and this vector will be perpendicular to the given ones:
Moreover, its modulus will be equal to the area of the parallelogram built on the vectors and. We will need this vector to calculate the distance from a point to a line. How can we calculate the cross product of vectors and if their coordinates are given? The determinant of the third order again comes to our aid. However, before I move on to the algorithm for calculating the cross product, I have to make a small lyrical digression.
This digression concerns the basis vectors.
Schematically they are shown in the figure:
Why do you think they are called basic? The fact is that :
Or in the picture:
The validity of this formula is obvious, because:
vector product
Now I can start introducing the cross product:
The vector product of two vectors is a vector that is calculated according to the following rule:
Now let's give some examples of calculating the cross product:
Example 1: Find the cross product of vectors:
Solution: I make a determinant:
And I calculate it:
Now, from writing through basis vectors, I will return to the usual vector notation:
Thus:
Now try.
Ready? We check:
And traditionally two tasks to control:
- Find the cross product of the following vectors:
- Find the cross product of the following vectors:
Answers:
Mixed product of three vectors
The last construction I need is the mixed product of three vectors. It, like a scalar, is a number. There are two ways to calculate it. - through the determinant, - through the mixed product.
Namely, let's say we have three vectors:
Then the mixed product of three vectors, denoted by can be calculated as:
1. - that is, the mixed product is the scalar product of a vector and the vector product of two other vectors
For example, the mixed product of three vectors is:
Try to calculate it yourself using the vector product and make sure that the results match!
And again - two examples for an independent decision:
Answers:
Choice of coordinate system
Well, now we have all the necessary foundation of knowledge to solve complex stereometric problems in geometry. However, before proceeding directly to the examples and algorithms for solving them, I believe that it will be useful to dwell on the following question: how exactly choose a coordinate system for a particular figure. After all, it is the choice of the relative position of the coordinate system and the figure in space that will ultimately determine how cumbersome the calculations will be.
I remind you that in this section we are considering the following shapes:
- cuboid
- Straight prism (triangular, hexagonal…)
- Pyramid (triangular, quadrangular)
- Tetrahedron (same as triangular pyramid)
For a cuboid or cube, I recommend the following construction:
That is, I will place the figure “in the corner”. The cube and the box are very good figures. For them, you can always easily find the coordinates of its vertices. For example, if (as shown in the picture)
then the vertex coordinates are:
Of course, you don’t need to remember this, but remembering how best to position a cube or a rectangular box is desirable.
straight prism
Prism is a more harmful figure. You can arrange it in space in different ways. However, I think the following is the best option:
Triangular prism:
That is, we put one of the sides of the triangle entirely on the axis, and one of the vertices coincides with the origin.
Hexagonal prism:
That is, one of the vertices coincides with the origin, and one of the sides lies on the axis.
Quadrangular and hexagonal pyramid:
A situation similar to a cube: we combine two sides of the base with the coordinate axes, we combine one of the vertices with the origin. The only small difficulty will be to calculate the coordinates of the point.
For a hexagonal pyramid - the same as for a hexagonal prism. The main task will again be in finding the coordinates of the vertex.
Tetrahedron (triangular pyramid)
The situation is very similar to the one I gave for the triangular prism: one vertex coincides with the origin, one side lies on the coordinate axis.
Well, now you and I are finally close to starting to solve problems. From what I said at the very beginning of the article, you could draw the following conclusion: most C2 problems fall into 2 categories: problems for the angle and problems for the distance. First, we will consider problems for finding an angle. They, in turn, are divided into the following categories (as the complexity increases):
Problems for finding corners
- Finding the angle between two lines
- Finding the angle between two planes
Let's consider these problems sequentially: let's start by finding the angle between two straight lines. Come on, remember, have you and I solved similar examples before? You remember, because we already had something similar ... We were looking for an angle between two vectors. I remind you, if two vectors are given: and, then the angle between them is found from the relation:
Now we have a goal - finding the angle between two straight lines. Let's turn to the "flat picture":
How many angles do we get when two lines intersect? Already things. True, only two of them are not equal, while others are vertical to them (and therefore coincide with them). So what angle should we consider the angle between two straight lines: or? Here the rule is: the angle between two straight lines is always no more than degrees. That is, from two angles, we will always choose the angle with the smallest degree measure. That is, in this picture, the angle between the two lines is equal. In order not to bother with finding the smallest of the two angles every time, cunning mathematicians suggested using the module. Thus, the angle between two straight lines is determined by the formula:
You, as an attentive reader, should have had a question: where, in fact, do we get these very numbers that we need to calculate the cosine of an angle? Answer: we will take them from the direction vectors of the lines! Thus, the algorithm for finding the angle between two lines is as follows:
- We apply formula 1.
Or in more detail:
- We are looking for the coordinates of the direction vector of the first straight line
- We are looking for the coordinates of the direction vector of the second line
- Calculate the modulus of their scalar product
- We are looking for the length of the first vector
- We are looking for the length of the second vector
- Multiply the results of point 4 by the results of point 5
- We divide the result of point 3 by the result of point 6. We get the cosine of the angle between the lines
- If this result allows us to calculate the angle exactly, we look for it
- Otherwise, we write through the arccosine
Well, now is the time to move on to the tasks: I will demonstrate the solution of the first two in detail, I will present the solution of another one in brief, and I will only give answers to the last two tasks, you must do all the calculations for them yourself.
Tasks:
1. In the right tet-ra-ed-re, find-di-te the angle between you-so-that tet-ra-ed-ra and the me-di-a-noy bo-ko-how side.
2. In the right-forward six-coal-pi-ra-mi-de, the hundred-ro-na-os-no-va-niya are somehow equal, and the side ribs are equal, find the angle between the straight lines and.
3. The lengths of all edges of the right-handed four-you-rech-coal-noy pi-ra-mi-dy are equal to each other. Find the angle between the straight lines and if from-re-zok - you-so-that given pi-ra-mi-dy, the point is se-re-di-on her bo-ko- th rib
4. On the edge of the cube from-me-che-to a point so that Find-di-te the angle between the straight lines and
5. Point - se-re-di-on the edges of the cube Nai-di-te the angle between the straight lines and.
It is no coincidence that I placed the tasks in this order. While you have not yet had time to begin to navigate the coordinate method, I myself will analyze the most “problematic” figures, and I will leave you to deal with the simplest cube! Gradually you have to learn how to work with all the figures, I will increase the complexity of the tasks from topic to topic.
Let's start solving problems:
1. Draw a tetrahedron, place it in the coordinate system as I suggested earlier. Since the tetrahedron is regular, then all its faces (including the base) are regular triangles. Since we are not given the length of the side, I can take it equal. I think you understand that the angle will not really depend on how much our tetrahedron will be "stretched" ?. I will also draw the height and median in the tetrahedron. Along the way, I will draw its base (it will also come in handy for us).
I need to find the angle between and. What do we know? We only know the coordinate of the point. So, we need to find more coordinates of the points. Now we think: a point is a point of intersection of heights (or bisectors or medians) of a triangle. A dot is an elevated point. The point is the midpoint of the segment. Then finally we need to find: the coordinates of the points: .
Let's start with the simplest: point coordinates. Look at the figure: It is clear that the applicate of a point is equal to zero (the point lies on a plane). Its ordinate is equal (because it is the median). It is more difficult to find its abscissa. However, this is easily done on the basis of the Pythagorean theorem: Consider a triangle. Its hypotenuse is equal, and one of the legs is equal Then:
Finally we have:
Now let's find the coordinates of the point. It is clear that its applicate is again equal to zero, and its ordinate is the same as that of a point, that is. Let's find its abscissa. This is done rather trivially if one remembers that the heights of an equilateral triangle are divided by the intersection point in the proportion counting from the top. Since:, then the desired abscissa of the point, equal to the length of the segment, is equal to:. Thus, the coordinates of the point are:
Let's find the coordinates of the point. It is clear that its abscissa and ordinate coincide with the abscissa and ordinate of the point. And the applique is equal to the length of the segment. - this is one of the legs of the triangle. The hypotenuse of a triangle is a segment - a leg. It is searched for the reasons that I highlighted in bold:
The point is the midpoint of the segment. Then we need to remember the formula for the coordinates of the middle of the segment:
That's it, now we can look for the coordinates of the direction vectors:
Well, everything is ready: we substitute all the data into the formula:
Thus,
Answer:
You should not be afraid of such "terrible" answers: for problems C2 this is a common practice. I would rather be surprised by the "beautiful" answer in this part. Also, as you noted, I practically did not resort to anything other than the Pythagorean theorem and the property of the heights of an equilateral triangle. That is, to solve the stereometric problem, I used the very minimum of stereometry. The gain in this is partially "extinguished" by rather cumbersome calculations. But they are quite algorithmic!
2. Draw a regular hexagonal pyramid along with the coordinate system, as well as its base:
We need to find the angle between the lines and. Thus, our task is reduced to finding the coordinates of points: . We will find the coordinates of the last three from the small drawing, and we will find the coordinate of the vertex through the coordinate of the point. Lots of work, but gotta get started!
a) Coordinate: it is clear that its applicate and ordinate are zero. Let's find the abscissa. To do this, consider a right triangle. Alas, in it we only know the hypotenuse, which is equal to. We will try to find the leg (because it is clear that twice the length of the leg will give us the abscissa of the point). How can we look for it? Let's remember what kind of figure we have at the base of the pyramid? This is a regular hexagon. What does it mean? This means that all sides and all angles are equal. We need to find one such corner. Any ideas? There are a lot of ideas, but there is a formula:
The sum of the angles of a regular n-gon is .
Thus, the sum of the angles of a regular hexagon is degrees. Then each of the angles is equal to:
Let's look at the picture again. It is clear that the segment is the bisector of the angle. Then the angle is degrees. Then:
Then where.
So it has coordinates
b) Now we can easily find the coordinate of the point: .
c) Find the coordinates of the point. Since its abscissa coincides with the length of the segment, it is equal. Finding the ordinate is also not very difficult: if we connect the points and and denote the point of intersection of the line, say for. (do it yourself simple construction). Then Thus, the ordinate of point B is equal to the sum of the lengths of the segments. Let's look at the triangle again. Then
Then since Then the point has coordinates
d) Now find the coordinates of the point. Consider a rectangle and prove that Thus, the coordinates of the point are:
e) It remains to find the coordinates of the vertex. It is clear that its abscissa and ordinate coincide with the abscissa and ordinate of the point. Let's find an app. Since then. Consider a right triangle. By the condition of the problem, the lateral edge. This is the hypotenuse of my triangle. Then the height of the pyramid is the leg.
Then the point has coordinates:
That's it, I have the coordinates of all points of interest to me. I am looking for the coordinates of the directing vectors of the straight lines:
We are looking for the angle between these vectors:
Answer:
Again, when solving this problem, I did not use any sophisticated tricks, except for the formula for the sum of the angles of a regular n-gon, as well as the definition of the cosine and sine of a right triangle.
3. Since we are again not given the lengths of the edges in the pyramid, I will consider them equal to one. Thus, since ALL edges, and not just the side ones, are equal to each other, then at the base of the pyramid and me lies a square, and the side faces are regular triangles. Let's depict such a pyramid, as well as its base on a plane, marking all the data given in the text of the problem:
We are looking for the angle between and. I will make very brief calculations when I am looking for the coordinates of points. You will need to "decrypt" them:
b) - the middle of the segment. Her coordinates:
c) I will find the length of the segment using the Pythagorean theorem in a triangle. I will find by the Pythagorean theorem in a triangle.
Coordinates:
d) - the middle of the segment. Its coordinates are
e) Vector coordinates
f) Vector coordinates
g) Looking for an angle:
The cube is the simplest figure. I'm sure you can figure it out on your own. The answers to problems 4 and 5 are as follows:
Finding the angle between a line and a plane
Well, the time for simple puzzles is over! Now the examples will be even more difficult. To find the angle between a line and a plane, we will proceed as follows:
- Using three points, we build the equation of the plane
,
using a third order determinant. - By two points we are looking for the coordinates of the directing vector of the straight line:
- We apply the formula to calculate the angle between a straight line and a plane:
As you can see, this formula is very similar to the one we used to find the angles between two lines. The structure of the right side is just the same, and on the left we are now looking for a sine, and not a cosine, as before. Well, one nasty action was added - the search for the equation of the plane.
Let's not shelve solving examples:
1. Os-no-va-ni-em straight-my prize-we are-la-et-xia equal-but-poor-ren-ny triangle-nick you-with-that prize-we are equal. Find the angle between the straight line and the plane
2. In a rectangular pa-ral-le-le-pi-pe-de from the West Nai-di-te the angle between the straight line and the plane
3. In the right-handed six-coal prism, all edges are equal. Find the angle between the straight line and the plane.
4. In the right triangular pi-ra-mi-de with the os-but-va-ni-em from the west of the rib Nai-di-te angle, ob-ra-zo-van -ny plane of the os-no-va-niya and straight-my, passing through the se-re-di-na of the ribs and
5. The lengths of all edges of the right quadrangular pi-ra-mi-dy with the top are equal to each other. Find the angle between the straight line and the plane, if the point is se-re-di-on the bo-ko-in-th edge of the pi-ra-mi-dy.
Again, I will solve the first two problems in detail, the third - briefly, and I leave the last two for you to solve on your own. In addition, you already had to deal with triangular and quadrangular pyramids, but not yet with prisms.
Solutions:
1. Draw a prism, as well as its base. Let's combine it with the coordinate system and mark all the data that are given in the problem statement:
I apologize for some non-observance of proportions, but for solving the problem this, in fact, is not so important. The plane is just the "back wall" of my prism. It is enough to simply guess that the equation of such a plane has the form:
However, this can also be shown directly:
We choose arbitrary three points on this plane: for example, .
Let's make the equation of the plane:
Exercise for you: calculate this determinant yourself. Did you succeed? Then the equation of the plane has the form:
Or simply
Thus,
To solve the example, I need to find the coordinates of the directing vector of the straight line. Since the point coincided with the origin, the coordinates of the vector will simply coincide with the coordinates of the point. To do this, we first find the coordinates of the point.
To do this, consider a triangle. Let's draw a height (it is also a median and a bisector) from the top. Since, then the ordinate of the point is equal. In order to find the abscissa of this point, we need to calculate the length of the segment. By the Pythagorean theorem we have:
Then the point has coordinates:
A dot is a "raised" on a dot:
Then the coordinates of the vector:
Answer:
As you can see, there is nothing fundamentally difficult in solving such problems. In fact, the “straightness” of a figure such as a prism simplifies the process a little more. Now let's move on to the next example:
2. We draw a parallelepiped, draw a plane and a straight line in it, and also separately draw its lower base:
First, we find the equation of the plane: The coordinates of the three points lying in it:
(the first two coordinates are obtained in an obvious way, and you can easily find the last coordinate from the picture from the point). Then we compose the equation of the plane:
We calculate:
We are looking for the coordinates of the direction vector: It is clear that its coordinates coincide with the coordinates of the point, isn't it? How to find coordinates? These are the coordinates of the point, raised along the applicate axis by one! . Then we are looking for the desired angle:
Answer:
3. Draw a regular hexagonal pyramid, and then draw a plane and a straight line in it.
Here it is even problematic to draw a plane, not to mention the solution of this problem, but the coordinate method does not care! It is in its versatility that its main advantage lies!
The plane passes through three points: . We are looking for their coordinates:
one) . Display the coordinates for the last two points yourself. You will need to solve the problem with a hexagonal pyramid for this!
2) We build the equation of the plane:
We are looking for the coordinates of the vector: . (See triangular pyramid problem again!)
3) We are looking for an angle:
Answer:
As you can see, there is nothing supernaturally difficult in these tasks. You just need to be very careful with the roots. To the last two problems, I will give only answers:
As you can see, the technique for solving problems is the same everywhere: the main task is to find the coordinates of the vertices and substitute them into some formulas. It remains for us to consider one more class of problems for calculating angles, namely:
Calculating angles between two planes
The solution algorithm will be as follows:
- For three points we are looking for the equation of the first plane:
- For the other three points, we are looking for the equation of the second plane:
- We apply the formula:
As you can see, the formula is very similar to the previous two, with the help of which we were looking for angles between straight lines and between a straight line and a plane. So remembering this one will not be difficult for you. Let's jump right into the problem:
1. A hundred-ro-on the basis of the right triangular prism is equal, and the dia-go-nal of the side face is equal. Find the angle between the plane and the plane of the base of the prize.
2. In the right-forward four-you-re-coal-noy pi-ra-mi-de, all the edges of someone are equal, find the sine of the angle between the plane and the plane Ko-Stu, passing through the point of per-pen-di-ku-lyar-but straight-my.
3. In a regular four-coal prism, the sides of the os-no-va-nia are equal, and the side edges are equal. On the edge from-me-che-to the point so that. Find the angle between the planes and
4. In the right quadrangular prism, the sides of the bases are equal, and the side edges are equal. On the edge from-me-che-to a point so that Find the angle between the planes and.
5. In the cube, find the co-si-nus of the angle between the planes and
Problem solutions:
1. I draw a regular (at the base - an equilateral triangle) triangular prism and mark on it the planes that appear in the condition of the problem:
We need to find the equations of two planes: The base equation is obtained trivially: you can make the corresponding determinant for three points, but I will make the equation right away:
Now let's find the equation The point has coordinates The point - Since - the median and the height of the triangle, it is easy to find by the Pythagorean theorem in a triangle. Then the point has coordinates: Find the applicate of the point To do this, consider a right triangle
Then we get the following coordinates: We compose the equation of the plane.
We calculate the angle between the planes:
Answer:
2. Making a drawing:
The most difficult thing is to understand what kind of mysterious plane it is, passing through a point perpendicularly. Well, the main thing is what is it? The main thing is attentiveness! Indeed, the line is perpendicular. The line is also perpendicular. Then the plane passing through these two lines will be perpendicular to the line, and, by the way, will pass through the point. This plane also passes through the top of the pyramid. Then the desired plane - And the plane is already given to us. We are looking for coordinates of points.
We find the coordinate of the point through the point. It is easy to deduce from a small drawing that the coordinates of the point will be as follows: What is now left to find in order to find the coordinates of the top of the pyramid? Still need to calculate its height. This is done using the same Pythagorean theorem: first, prove that (trivially from small triangles forming a square at the base). Since by condition, we have:
Now everything is ready: vertex coordinates:
We compose the equation of the plane:
You are already an expert in calculating determinants. Easily you will receive:
Or otherwise (if we multiply both parts by the root of two)
Now let's find the equation of the plane:
(You didn’t forget how we get the equation of the plane, right? If you don’t understand where this minus one came from, then go back to the definition of the equation of the plane! It just always turned out before that that my plane belonged to the origin!)
We calculate the determinant:
(You may notice that the equation of the plane coincided with the equation of the straight line passing through the points and! Think why!)
Now we calculate the angle:
We need to find the sine:
Answer:
3. A tricky question: what is a rectangular prism, what do you think? It's just a well-known parallelepiped to you! Drawing right away! You can even not separately depict the base, there is little use from it here:
The plane, as we noted earlier, is written as an equation:
Now we make a plane
We immediately compose the equation of the plane:
Looking for an angle
Now the answers to the last two problems:
Well, now is the time to take a break, because you and I are great and have done a great job!
Coordinates and vectors. Advanced level
In this article, we will discuss with you another class of problems that can be solved using the coordinate method: distance problems. Namely, we will consider the following cases:
- Calculating the distance between skew lines.
I have ordered the given tasks as their complexity increases. The easiest is to find point to plane distance and the hardest part is finding distance between intersecting lines. Although, of course, nothing is impossible! Let's not procrastinate and immediately proceed to the consideration of the first class of problems:
Calculating the distance from a point to a plane
What do we need to solve this problem?
1. Point coordinates
So, as soon as we get all the necessary data, we apply the formula:
You should already know how we build the equation of the plane from the previous problems that I analyzed in the last part. Let's get down to business right away. The scheme is as follows: 1, 2 - I help you decide, and in some detail, 3, 4 - only the answer, you make the decision yourself and compare. Started!
Tasks:
1. Given a cube. The edge length of the cube is Find-di-te distance from se-re-di-ny from cut to flat
2. Given the right-vil-naya four-you-rekh-coal-naya pi-ra-mi-da Bo-ko-voe edge hundred-ro-on the os-no-va-nia is equal. Find-di-those distances from a point to a plane where - se-re-di-on the edges.
3. In the right triangular pi-ra-mi-de with os-but-va-ni-em, the other edge is equal, and one hundred-ro-on os-no-va- niya is equal. Find-di-those distances from the top to the plane.
4. In the right-handed six-coal prism, all edges are equal. Find-di-those distances from a point to a plane.
Solutions:
1. Draw a cube with single edges, build a segment and a plane, denote the middle of the segment by the letter
.
First, let's start with an easy one: find the coordinates of a point. Since then (remember the coordinates of the middle of the segment!)
Now we compose the equation of the plane on three points
\[\left| (\begin(array)(*(20)(c))x&0&1\\y&1&0\\z&1&1\end(array)) \right| = 0\]
Now I can start finding the distance:
2. We start again with a drawing, on which we mark all the data!
For a pyramid, it would be useful to draw its base separately.
Even the fact that I draw like a chicken paw will not prevent us from easily solving this problem!
Now it's easy to find the coordinates of a point
Since the coordinates of the point
2. Since the coordinates of the point a are the middle of the segment, then
We can easily find the coordinates of two more points on the plane. We compose the equation of the plane and simplify it:
\[\left| (\left| (\begin(array)(*(20)(c))x&1&(\frac(3)(2))\\y&0&(\frac(3)(2))\\z&0&(\frac( (\sqrt 3 ))(2))\end(array)) \right|) \right| = 0\]
Since the point has coordinates: , then we calculate the distance:
Answer (very rare!):
Well, did you understand? It seems to me that everything here is just as technical as in the examples that we considered with you in the previous part. So I am sure that if you have mastered that material, then it will not be difficult for you to solve the remaining two problems. I'll just give you the answers:
Calculating the Distance from a Line to a Plane
In fact, there is nothing new here. How can a line and a plane be located relative to each other? They have all the possibilities: to intersect, or a straight line is parallel to the plane. What do you think is the distance from the line to the plane with which the given line intersects? It seems to me that it is clear that such a distance is equal to zero. Uninteresting case.
The second case is trickier: here the distance is already non-zero. However, since the line is parallel to the plane, then each point of the line is equidistant from this plane:
Thus:
And this means that my task has been reduced to the previous one: we are looking for the coordinates of any point on the line, we are looking for the equation of the plane, we calculate the distance from the point to the plane. In fact, such tasks in the exam are extremely rare. I managed to find only one problem, and the data in it was such that the coordinate method was not very applicable to it!
Now let's move on to another, much more important class of problems:
Calculating the Distance of a Point to a Line
What will we need?
1. The coordinates of the point from which we are looking for the distance:
2. Coordinates of any point lying on a straight line
3. Direction vector coordinates of the straight line
What formula do we use?
What does the denominator of this fraction mean to you and so it should be clear: this is the length of the directing vector of the straight line. Here is a very tricky numerator! The expression means the module (length) of the vector product of vectors and How to calculate the vector product, we studied in the previous part of the work. Refresh your knowledge, it will be very useful to us now!
Thus, the algorithm for solving problems will be as follows:
1. We are looking for the coordinates of the point from which we are looking for the distance:
2. We are looking for the coordinates of any point on the line to which we are looking for the distance:
3. Building a vector
4. We build the directing vector of a straight line
5. Calculate the cross product
6. We are looking for the length of the resulting vector:
7. Calculate the distance:
We have a lot of work, and the examples will be quite complex! So now focus all your attention!
1. Dana is a right-handed triangular pi-ra-mi-da with a vertex. One hundred-ro-on the os-no-va-niya pi-ra-mi-dy is equal, you-so-ta is equal. Find-di-those distances from the se-re-di-ny of the bo-ko-th edge to the straight line, where the points and are the se-re-di-ny of the ribs and co-from- vet-stven-but.
2. The lengths of the ribs and the right-angle-no-para-ral-le-le-pi-pe-da are equal, respectively, and Find-di-te distance from top-shi-ny to straight-my
3. In the right six-coal prism, all the edges of a swarm are equal find-di-those distance from a point to a straight line
Solutions:
1. We make a neat drawing, on which we mark all the data:
We have a lot of work for you! I would first like to describe in words what we will look for and in what order:
1. Coordinates of points and
2. Point coordinates
3. Coordinates of points and
4. Coordinates of vectors and
5. Their cross product
6. Vector length
7. The length of the vector product
8. Distance from to
Well, we have a lot of work to do! Let's roll up our sleeves!
1. To find the coordinates of the height of the pyramid, we need to know the coordinates of the point. Its applicate is zero, and the ordinate is equal to its abscissa. Finally, we got the coordinates:
Point coordinates
2. - middle of the segment
3. - the middle of the segment
midpoint
4.Coordinates
Vector coordinates
5. Calculate the vector product:
6. The length of the vector: the easiest way is to replace that the segment is the middle line of the triangle, which means it is equal to half the base. So that.
7. We consider the length of the vector product:
8. Finally, find the distance:
Phew, that's all! Honestly, I'll tell you: solving this problem by traditional methods (through constructions) would be much faster. But here I reduced everything to a ready-made algorithm! I think that the solution algorithm is clear to you? Therefore, I will ask you to solve the remaining two problems on your own. Compare answers?
Again, I repeat: it is easier (faster) to solve these problems through constructions, rather than resorting to the coordinate method. I demonstrated this way of solving only to show you a universal method that allows you to "do not complete anything."
Finally, consider the last class of problems:
Calculating the distance between skew lines
Here the algorithm for solving problems will be similar to the previous one. What we have:
3. Any vector connecting the points of the first and second lines:
How do we find the distance between lines?
The formula is:
The numerator is the module of the mixed product (we introduced it in the previous part), and the denominator - as in the previous formula (the module of the vector product of the directing vectors of the lines, the distance between which we are looking for).
I will remind you that
then the distance formula can be rewritten as:
Divide this determinant by the determinant! Although, to be honest, I'm not in the mood for jokes here! This formula, in fact, is very cumbersome and leads to rather complicated calculations. If I were you, I would only use it as a last resort!
Let's try to solve a few problems using the above method:
1. In the right triangular prism, all the edges are somehow equal, find the distance between the straight lines and.
2. Given a right-fore-shaped triangular prism, all the edges of the os-no-va-niya of someone are equal to Se-che-tion, passing through the other rib and se-re-di-nu ribs are yav-la-et-sya square-ra-tom. Find-di-te dis-sto-I-nie between straight-we-mi and
I decide the first, and based on it, you decide the second!
1. I draw a prism and mark the lines and
Point C coordinates: then
Point coordinates
Vector coordinates
Point coordinates
Vector coordinates
Vector coordinates
\[\left((B,\overrightarrow (A(A_1)) \overrightarrow (B(C_1)) ) \right) = \left| (\begin(array)(*(20)(l))(\begin(array)(*(20)(c))0&1&0\end(array))\\(\begin(array)(*(20) (c))0&0&1\end(array))\\(\begin(array)(*(20)(c))(\frac((\sqrt 3 ))(2))&( - \frac(1) (2))&1\end(array))\end(array)) \right| = \frac((\sqrt 3 ))(2)\]
We consider the cross product between the vectors and
\[\overrightarrow (A(A_1)) \cdot \overrightarrow (B(C_1)) = \left| \begin(array)(l)\begin(array)(*(20)(c))(\overrightarrow i )&(\overrightarrow j )&(\overrightarrow k )\end(array)\\\begin(array )(*(20)(c))0&0&1\end(array)\\\begin(array)(*(20)(c))(\frac((\sqrt 3 ))(2))&( - \ frac(1)(2))&1\end(array)\end(array) \right| - \frac((\sqrt 3 ))(2)\overrightarrow k + \frac(1)(2)\overrightarrow i \]
Now we consider its length:
Answer:
Now try to carefully complete the second task. The answer to it will be:.
Coordinates and vectors. Brief description and basic formulas
A vector is a directed segment. - the beginning of the vector, - the end of the vector.
The vector is denoted by or.
Absolute value vector - the length of the segment representing the vector. Designated as.
Vector coordinates:
,
where are the ends of the vector \displaystyle a .
Sum of vectors: .
The product of vectors:
Dot product of vectors:
Formula for calculating the distance from a point to a line in 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 tasks for calculating the distance from a point to a line in a plane
Example 1
Find the distance between the line 3x + 4y - 6 = 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 line is 0.6.
equation of a plane passing through points perpendicular to a vectorGeneral equation of a plane
A non-zero vector perpendicular to a given plane is called normal vector (or, in short, normal ) for this plane.
Let in the coordinate space (in a rectangular coordinate system) given:
a) dot 
;
b) a non-zero vector (Fig. 4.8, a).
It is required to write an equation for a plane passing through a point 
perpendicular to the vector End of proof.
Let us now consider various types of equations of a straight line in 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 not equal to 0 (explain why).
Point belongs to the plane P only if its coordinates satisfy the equation of the plane. Depending on the coefficients 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,
- 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. It follows from formula (6) 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 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 is 
is 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 the derivation of the formula for the distance between a plane and a straight line.
Indeed, the distance d between a line and a plane is
where is a point lying on a 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 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
Injection f between two planes is equal to the angle between their normal vectors (see Fig. 3) and can therefore be calculated from the formula 
Determining the angle between planes.
(11)
Distance from a point to a plane and how to find it
Distance from point to 
plane is the length of the perpendicular dropped from a point to 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 first need to understand how the perpendicular is located from a point to a plane: maybe it lies in some convenient plane, it is a height in some convenient (or not so) triangle, or maybe this perpendicular is generally a height in some pyramid.
After this first and most difficult stage, the problem breaks down into several specific planimetric problems (perhaps in different planes).
With 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 the point to the plane.