Distance from point c to straight line AB. Determines the distance from a point to a straight line. The relative position of two straight lines

This article talks about the topic « distance from point to line », the determination of the distance from a point to a straight line with illustrated examples by the method of coordinates is considered. Each block of the theory at the end has shown examples of solving similar problems.

The distance from point to line is found through the definition of the distance from point to point. Let's take a closer look.

Let there be a line a and a point M 1 that does not belong to a given line. Draw line b through it, which is perpendicular to line a. The point of intersection of the lines is taken as H 1. We get that M 1 H 1 is the perpendicular, which was lowered from the point M 1 to the line a.

Definition 1

Distance from point М 1 to line a called the distance between points M 1 and H 1.

There are definition records 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 straight 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 an example.

If we take a point Q lying on the straight line a, not coinciding with the point M 1, then we get that the segment M 1 Q is called inclined, dropped from M 1 to the line a. It is necessary to indicate that the perpendicular from point M 1 is less than any other inclined line drawn from point to line.

To prove this, consider a 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. 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 allows you to use several solution methods: through the Pythagorean theorem, determining the 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 straight line, you can enter a rectangular coordinate system, then the coordinate method is used. In this paragraph, we will 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 straight line a. In the second method, the normal equation of the straight line a is used to find the desired distance.

If there is a point on the plane with coordinates M 1 (x 1, y 1) located in a rectangular coordinate system, straight line a, and you need to find the distance M 1 H 1, you can calculate in two ways. Let's consider them.

The 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 straight line is calculated by the coordinates from the formula M 1 H 1 \u003d (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 on a plane. Let's take a way of specifying a straight line a through writing the 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 the given straight line a. The straight line will be denoted by beech b. H 1 is the point of intersection of lines a and b, which means that 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 a straight line a is carried out according to 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, having the form y \u003d k 1 x + b 1;
• obtaining a general equation of line b, having 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 line b intersects point M 1 and is perpendicular to a given line a;
• determination of the coordinates x 2, y 2 of the point H 1, which is the intersection point of a and b, for this the system is solved linear equations A 1 x + B 1 y + C 1 \u003d 0 A 2 x + B 2 y + C 2 \u003d 0 or y \u003d k 1 x + b 1 y \u003d k 2 x + b 2;
• calculating the required distance from a point to a straight line using the formula M 1 H 1 \u003d (x 2 - x 1) 2 + (y 2 - y 1) 2.

Second way

The theorem can help answer the question of finding the distance from a given point to a given straight line on a plane.

Theorem

The rectangular coordinate system has O x y has a point M 1 (x 1, y 1), from which a straight line a is drawn to the plane, given by the normal equation of the plane, which has the form cos α x + cos β y - p \u003d 0, equal to to the modulus of the value obtained on the left-hand side of the normal equation of the line, calculated at x \u003d x 1, y \u003d y 1, which means that M 1 H 1 \u003d cos α x 1 + cos β y 1 - p.

Evidence

Line a corresponds to the normal equation of the plane, which has the form cos α x + cos β y - p \u003d 0, then n → \u003d (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 display all the data in the figure, add a point with coordinates M 1 (x 1, y 1), where the radius vector of the point M 1 - O M 1 → \u003d (x 1, y 1). It is necessary to draw a straight line from a point to a straight line, which we denote by M 1 H 1. It is necessary to show the projections М 2 and Н 2 of points М 1 and Н 2 onto a straight line passing through point O with a direction vector of the form n → \u003d (cos α, cos β), and the numerical projection of the vector is denoted as OM 1 → \u003d (x 1, y 1) to the direction n → \u003d (cos α, cos β) as npn → OM 1 →.

Variations depend on the location of the point M 1 itself. Consider in the figure below.

We fix the results using the formula M 1 H 1 \u003d n p n → O M → 1 - p. Then we reduce the equality to this form M 1 H 1 \u003d cos α x 1 + cos β y 1 - p in order to obtain n p n → O M → 1 \u003d cos α x 1 + cos β y 1.

The scalar product of vectors as a result gives a transformed formula of the form n →, OM → 1 \u003d n → npn → OM 1 → \u003d 1 npn → OM 1 → \u003d npn → OM 1 →, which is a product in coordinate form of the form n →, OM 1 → \u003d cos α x 1 + cos β y 1. Hence, we obtain that n p n → O M 1 → \u003d cos α x 1 + cos β y 1. It follows that M 1 H 1 \u003d n p n → O M 1 → - p \u003d cos α x 1 + cos β y 1 - p. The theorem is proved.

We get that to find the distance from the point M 1 (x 1, y 1) to the straight line a on the plane, you need to perform several actions:

Definition 4

• obtaining the normal equation of the straight line a cos α x + cos β y - p \u003d 0, provided that it is not in the task;
• calculation of the expression cos α · x 1 + cos β · y 1 - p, where the obtained value takes M 1 H 1.

Let us apply these methods to solving 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 \u003d 0.

Decision

Let's apply the first method to solve.

To do this, it is necessary to find the general equation of the straight line b, which passes through a given point M 1 (- 1, 2), perpendicular to the straight line 4 x - 3 y + 35 \u003d 0. It is seen from the condition that straight line b is perpendicular to straight line a, then its direction vector has coordinates equal to (4, - 3). Thus, we have the opportunity to write the canonical equation of the straight line b on the plane, since there are coordinates of the point M 1 belongs to the straight line b. Determine the coordinates of the direction vector of the straight line b. We get x - (- 1) 4 \u003d y - 2 - 3 ⇔ x + 1 4 \u003d y - 2 - 3. The resulting canonical equation must be transformed to the general one. Then we get that

x + 1 4 \u003d y - 2 - 3 ⇔ - 3 (x + 1) \u003d 4 (y - 2) ⇔ 3 x + 4 y - 5 \u003d 0

Let's find the coordinates of the points of intersection of straight lines, which we will take as the designation H 1. The transformations look like this:

4 x - 3 y + 35 \u003d 0 3 x + 4 y - 5 \u003d 0 ⇔ x \u003d 3 4 y - 35 4 3 x + 4 y - 5 \u003d 0 ⇔ x \u003d 3 4 y - 35 4 3 3 4 y - 35 4 + 4 y - 5 \u003d 0 ⇔ ⇔ x \u003d 3 4 y - 35 4 y \u003d 5 ⇔ x \u003d 3 4 5 - 35 4 y \u003d 5 ⇔ x \u003d - 5 y \u003d 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 point M 1 to line a. We have that the coordinates of the points M 1 (- 1, 2) and H 1 (- 5, 5), then we substitute in 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

Second solution.

In order to solve in another way, it is necessary to obtain the normal equation of the line. Evaluate the normalizing factor and multiply both sides of the equation 4 x - 3 y + 35 \u003d 0. From this we get that the normalizing factor is - 1 4 2 + (- 3) 2 \u003d - 1 5, and the normal equation will be of the form - 1 5 4 x - 3 y + 35 \u003d - 1 5 0 ⇔ - 4 5 x + 3 5 y - 7 \u003d 0.

According to the calculation algorithm, it is necessary to obtain the normal equation of the straight line and calculate it with the values \u200b\u200bx \u003d - 1, y \u003d 2. Then we get that

4 5 - 1 + 3 5 2 - 7 \u003d - 5

Hence, we find that the distance from the point M 1 (- 1, 2) to the given straight line 4 x - 3 y + 35 \u003d 0 has the value - 5 \u003d 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 \u003d 1 2 x + 1. Find the distance from a given point to a straight line.

Decision

The solution in the first way implies bringing the given equation with the slope to the general equation. For simplicity, you can do it differently.

If the product of the slopes of perpendicular lines has a value of - 1, then the slope of the line perpendicular to the given y \u003d 1 2 x + 1 is 2. Now we get the equation of the straight line passing through the point with coordinates M 1 (8, 0). We have that y - 0 \u003d - 2 (x - 8) ⇔ y \u003d - 2 x + 16.

We turn 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 \u003d 1 2 x + 1 y \u003d - 2 x + 16 ⇔ y \u003d 1 2 x + 1 1 2 x + 1 \u003d - 2 x + 16 ⇔ y \u003d 1 2 x + 1 x \u003d 6 ⇔ ⇔ y \u003d 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 straight line y \u003d 1 2 x + 1 is equal to the distance from the start point and the end point with coordinates M 1 (8, 0) and H 1 (6, 4) ... We calculate and get that M 1 H 1 \u003d 6 - 8 2 + (4 - 0) 2 20 \u003d 2 5.

The solution in the second way is to go from an 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 the line takes the form - 2 5 1 2 x - y + 1 \u003d - 2 5 0 ⇔ - 1 5 x + 2 5 y - 2 5 \u003d 0. Let's make a calculation from the point M 1 8, 0 to a straight line of the form - 1 5 x + 2 5 y - 2 5 \u003d 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 \u003d 0 and y + 1 \u003d 0.

Decision

We obtain the equation of the normal form of the straight line 2 x - 3 \u003d 0:

2 x - 3 \u003d 0 ⇔ 1 2 2 x - 3 \u003d 1 2 0 ⇔ x - 3 2 \u003d 0

Then we proceed to calculating the distance from the point M 1 - 2, 4 to the straight line x - 3 2 \u003d 0. We get:

M 1 H 1 \u003d - 2 - 3 2 \u003d 3 1 2

The equation of the straight line y + 1 \u003d 0 has a normalizing factor of -1. This means that the equation will take the form - y - 1 \u003d 0. We proceed to calculating the distance from the point M 1 (- 2, 4) to the straight line - y - 1 \u003d 0. We get that it is equal to - 4 - 1 \u003d 5.

Answer: 3 1 2 and 5.

Consider in detail the finding 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 O y axis has an equation of a straight line, which is incomplete, has the form x \u003d 0, and O x - y \u003d 0. The equations are normal for the coordinate axes, then you need to find the distance from the point with coordinates M 1 x 1, y 1 to straight lines. This is done based on the formulas M 1 H 1 \u003d x 1 and M 1 H 1 \u003d y 1. Consider in the figure below.

Example 4

Find the distance from the point M 1 (6, - 7) to the coordinate lines located in the plane O x y.

Decision

Since the equation y \u003d 0 refers to the straight line O x, you can find the distance from M 1 with the given coordinates to this straight line using the formula. We get that 6 \u003d 6.

Since the equation x \u003d 0 refers to the straight line O y, then you can find the distance from M 1 to this straight line using the formula. Then we get that - 7 \u003d 7.

Answer:the distance from M 1 to O x is 6, and from M 1 to O y is 7.

When in three-dimensional space we have a point with coordinates M 1 (x 1, y 1, z 1), it is necessary to find the distance from point A to line a.

Consider two methods that allow you to calculate the distance from a point to a straight line a located in space. The first case considers the distance from the point M 1 to the straight line, where the point on the straight line is called H 1 and is the base of the perpendicular drawn from the point M 1 to the straight line a. The second case suggests that the points of this plane must be sought as the height of the parallelogram.

The 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 \u003d x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2.

We get that the whole solution goes to find the coordinates of the base of the perpendicular drawn from М 1 to line a. This is done as follows: H 1 is the point where the straight line a intersects with the plane that passes through the given point.

Hence, the algorithm for determining the distance from the point M 1 (x 1, y 1, z 1) to the line a in space implies several points:

Definition 5

• drawing up the equation of the χ plane as the equation of the plane passing through a given point that is perpendicular to the straight line;
• determination of coordinates (x 2, y 2, z 2) belonging to the point H 1, which is the point of intersection of the straight line a and the plane χ;
• calculating the distance from a point to a straight line using the formula M 1 H 1 \u003d x 2 - x 1 2 + y 2 - y 1 2 + z 2 - z 1 2.

Second way

From the condition we have a straight line a, then we can determine the direction vector a → \u003d a x, a y, a z with coordinates x 3, y 3, z 3 and a certain point M 3 belonging to the straight line a. Given the coordinates of points M 1 (x 1, y 1) and M 3 x 3, y 3, z 3, you can calculate M 3 M 1 →:

M 3 M 1 → \u003d (x 1 - x 3, y 1 - y 3, z 1 - z 3)

You should 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 in the figure below.

We have that the height M 1 H 1 is the desired distance, then it is necessary to find it by the formula. That is, we are looking for M 1 H 1.

Let us denote the area of \u200b\u200bthe parallelogram for the letter S, is found by the formula using the vector 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. The area formula is S \u003d a → × M 3 M 1 →. Also, the area of \u200b\u200bthe figure is equal to the product of the lengths of its sides by the height, we get that S \u003d a → M 1 H 1 with a → \u003d ax 2 + ay 2 + az 2, which is the length of the vector a → \u003d (ax, ay, az), which is equal to the side of the parallelogram. Hence, M 1 H 1 is the distance from a point to a line. It is found by the formula M 1 H 1 \u003d a → × M 3 M 1 → a →.

To find the distance from a point with coordinates M 1 (x 1, y 1, z 1) to a straight line a in space, you need to perform several steps of the algorithm:

Definition 6

• determination of the directing vector of the straight line a - a → \u003d (a x, a y, a z);
• calculating the length of the direction vector a → \u003d a x 2 + a y 2 + a z 2;
• obtaining coordinates x 3, y 3, z 3 belonging to the point M 3, located on the straight line a;
• calculation of the coordinates of the vector M 3 M 1 →;
• finding the vector product of vectors a → (ax, ay, az) and M 3 M 1 → \u003d x 1 - x 3, y 1 - y 3, z 1 - z 3 as a → × M 3 M 1 → \u003d i → j → k → axayazx 1 - x 3 y 1 - y 3 z 1 - z 3 to obtain the length by the formula a → × M 3 M 1 →;
• calculating the distance from a point to a straight line M 1 H 1 \u003d a → × M 3 M 1 → a →.

Solving problems on finding the distance from a given point to a given straight line in space

Example 5

Find the distance from the point with coordinates M 1 2, - 4, - 1 to the line x + 1 2 \u003d y - 1 \u003d z + 5 5.

Decision

The first method begins with writing the equation of the plane χ passing through M 1 and perpendicular to set point... We get an expression of the form:

2 (x - 2) - 1 (y - (- 4)) + 5 (z - (- 1)) \u003d 0 ⇔ 2 x - y + 5 z - 3 \u003d 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 line specified by the condition. You should go from the canonical to the intersecting one. Then we get a system of equations of the form:

x + 1 2 \u003d y - 1 \u003d z + 5 5 ⇔ - 1 (x + 1) \u003d 2 y 5 (x + 1) \u003d 2 (z + 5) 5 y \u003d - 1 (z + 5) ⇔ x + 2 y + 1 \u003d 0 5 x - 2 z - 5 \u003d 0 5 y + z + 5 \u003d 0 ⇔ x + 2 y + 1 \u003d 0 5 x - 2 z - 5 \u003d 0

It is necessary to calculate the system x + 2 y + 1 \u003d 0 5 x - 2 z - 5 \u003d 0 2 x - y + 5 z - 3 \u003d 0 ⇔ x + 2 y \u003d - 1 5 x - 2 z \u003d 5 2 x - y + 5 z \u003d 3 according to Cramer's method, then we get that:

∆ \u003d 1 2 0 5 0 - 2 2 - 1 5 \u003d - 60 ∆ x \u003d - 1 2 0 5 0 - 2 3 - 1 5 \u003d - 60 ⇔ x \u003d ∆ x ∆ \u003d - 60 - 60 \u003d 1 ∆ y \u003d 1 - 1 0 5 5 2 2 3 5 \u003d 60 ⇒ y \u003d ∆ y ∆ \u003d 60 - 60 \u003d - 1 ∆ z \u003d 1 2 - 1 5 0 5 2 - 1 3 \u003d 0 ⇒ z \u003d ∆ z ∆ \u003d 0 - 60 \u003d 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 way is to start by looking for coordinates in canonical equation... To do this, you need to pay attention to the denominators of the fraction. Then a → \u003d 2, - 1, 5 is the direction vector of the line x + 1 2 \u003d y - 1 \u003d z + 5 5. It is necessary to calculate the length by the formula a → \u003d 2 2 + (- 1) 2 + 5 2 \u003d 30.

It is clear that the line x + 1 2 \u003d y - 1 \u003d 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 point M 1 2, - 4, - 1 is M 3 M 1 → \u003d 3, - 4, 4. Find the vector product a → \u003d (2, - 1, 5) and M 3 M 1 → \u003d (3, - 4, 4).

We get an expression of the form a → × M 3 M 1 → \u003d i → j → k → 2 - 1 5 3 - 4 4 \u003d - 4 i → + 15 j → - 8 k → + 20 i → - 8 J → \u003d 16 i → + 7 j → - 5 k →

we get that the length of the vector product is a → × M 3 M 1 → \u003d 16 2 + 7 2 + - 5 2 \u003d 330.

We have all the data for using the formula for calculating the distance from a point for a straight line, so we apply it and get:

M 1 H 1 \u003d a → × M 3 M 1 → a → \u003d 330 30 \u003d 11

Answer: 11 .

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Formula for calculating the distance from a point to a straight line on a plane

If the equation of the straight line Ax + By + C \u003d 0 is given, then the distance from the point M (M x, M y) to the straight line can be found using the following formula

Examples of tasks for calculating the distance from a point to a straight line on a plane

Example 1.

Find the distance between the line 3x + 4y - 6 \u003d 0 and the point M (-1, 3).

Decision. Substitute in the formula the coefficients of the line and the coordinates of the point

Answer: the distance from a point to a straight line is 0.6.

equation of a plane passing through points perpendicular to a vector General equation of a plane

A nonzero vector perpendicular to a given plane is called normal vector (or, in short, normal ) for this plane.

Let the coordinate space (in a rectangular coordinate system) be given:

a) point ;

b) a nonzero vector (Figure 4.8, a).

It is required to draw up an equation of a plane passing through a point perpendicular to vector End of proof.

Let us now consider various types of equations of a straight line on a plane.

1) General equation of the planeP .

It follows from the derivation of the equation that simultaneously A, B and C not equal to 0 (explain why).

The point belongs to the plane P only if its coordinates satisfy the plane equation. Depending on the odds A, B, C and Dplane P occupies one position or another:

- the plane passes through the origin of the coordinate system, - the plane does not pass through the origin of the coordinate system,

- the plane is parallel to the axis X,

X,

- the plane is parallel to the axis Y,

- the plane is not parallel to the axis Y,

- the plane is parallel to the axis Z,

- the plane is not parallel to the axis Z.

Prove these statements yourself.

Equation (6) is easily derived from equation (5). Indeed, let the point lie on the plane P... Then its coordinates satisfy the equation Subtracting equation (7) from equation (5) and grouping the terms, we obtain equation (6). Consider now two vectors with coordinates respectively. From formula (6) it follows that their scalar product is equal to zero. Therefore, the vector is perpendicular to the vector.The beginning and the end of the last vector are respectively at the points that belong to the plane P... Therefore, the vector is perpendicular to the plane P... Distance from point to plane P, the general equation of which is determined by the formula The proof of this formula is completely analogous to the proof of the formula for the distance between a point and a line (see Fig. 2).
Figure: 2. To the derivation of the formula for the distance between a plane and a straight line.

Indeed, the distance d between a straight line and a plane is

where is a point lying on a plane. Hence, as in lecture No. 11, the above formula is obtained. Two planes are parallel if their normal vectors are parallel. From this we obtain the condition of parallelism of two planes - coefficients of general equations of planes. Two planes are perpendicular if their normal vectors are perpendicular, hence we obtain the condition of perpendicularity of two planes, if their general equations are known

Angle f between two planes is equal to the angle between their normal vectors (see Fig. 3) and can, therefore, be calculated by the formula
Determination of the angle between the planes.

(11)

Distance from point to plane and ways to find it

Distance from point to plane - the length of the perpendicular dropped from a point onto this plane. There are at least two ways to find the distance from a point to a plane: geometric and algebraic.

With the geometric method you first need to understand how the perpendicular is located from point to plane: maybe it lies in some convenient plane, is the height in some convenient (or not so) triangle, or maybe this perpendicular is generally the height in some pyramid.

After this first and most difficult stage, the task breaks down into several specific planimetric tasks (perhaps in different planes).

In the algebraic way in order to find the distance from a point to a plane, you need to enter a coordinate system, find the coordinates of the point and the equation of the plane, and then apply the formula for the distance from a point to a plane.

OoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooTherefore, we will proceed to the first section, I hope that by the end of the article I will maintain a cheerful frame of mind.

The relative position of two straight lines

The case when the audience sings along with the chorus. Two straight lines can:

1) match;

2) be parallel:;

3) or intersect at a single point:.

Help for Dummies : please remember the mathematical sign of the intersection, it will be very common. The notation indicates that a straight line intersects a straight line at a point.

How to determine the relative position of two straight lines?

Let's start with the first case:

Two straight lines coincide if and only if their corresponding coefficients are proportional, that is, there is such a number "lambda" that the equalities

Consider straight lines and compose three equations from the corresponding coefficients:. It follows from each equation that, therefore, these lines coincide.

Indeed, if all the coefficients of the equation multiply by –1 (change signs), and reduce all the coefficients of the equation by 2, then you get the same equation:.

The second case, when the lines are parallel:

Two straight lines are parallel if and only if their coefficients for the variables are proportional: but.

As an example, consider two lines. We check the proportionality of the corresponding coefficients for the variables:

However, it is quite clear that.

And the third case, when the lines intersect:

Two straight lines intersect if and only if their coefficients for variables are NOT proportional, that is, there is NO such "lambda" value to make the equalities

So, for straight lines we will compose the system:

From the first equation it follows that, and from the second equation:, therefore, the system is inconsistent (no solutions). Thus, the coefficients of the variables are not proportional.

Conclusion: lines intersect

In practical problems, you can use the solution scheme just considered. By the way, it is very similar to the algorithm for checking vectors for collinearity, which we considered in the lesson The concept of linear (non) dependence of vectors. Vector basis... But there is a more civilized packaging:

Example 1

Find out the relative position of the straight lines:

Decision based on the study of direction vectors of straight lines:

a) From the equations we find the direction vectors of the straight lines: .

, so the vectors are not collinear and the lines intersect.

Just in case, I'll put a stone with pointers at the crossroads:

The rest jump over the stone and follow on, straight to Kashchei the Immortal \u003d)

b) Find the direction vectors of straight lines:

Lines have the same direction vector, which means that they are either parallel or coincide. There is no need to count the determinant here.

It is obvious that the coefficients for the unknowns are proportional, while.

Let us find out whether the equality is true:

Thus,

c) Find the direction vectors of straight lines:

Let us calculate the determinant composed of the coordinates of these vectors:
hence the direction vectors are collinear. Lines are either parallel or coincide.

The coefficient of proportionality "lambda" is easy to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: .

Now let's find out if the equality is true. Both free terms are zero, so:

The resulting value satisfies this equation (any number generally satisfies it).

Thus, the lines coincide.

Answer:

Very soon you will learn (or even have already learned) how to solve the problem considered orally literally in a matter of seconds. In this regard, I see no reason to propose anything for an independent solution, it is better to lay another important brick in the geometric foundation:

How to build a straight line parallel to a given one?

For ignorance of this simplest task, the Nightingale the Robber severely punishes.

Example 2

The straight line is given by the equation. Equate a parallel line that goes through a point.

Decision: Let's denote the unknown direct letter. What does the condition say about her? The straight line goes through the point. And if the straight lines are parallel, then it is obvious that the directing vector of the straight line "tse" is also suitable for constructing the straight line "de".

We take out the direction vector from the equation:

Answer:

The geometry of the example looks simple:

Analytical verification consists of the following steps:

1) Check that the straight lines have the same direction vector (if the equation of the straight line is not simplified properly, then the vectors will be collinear).

2) Check if the point satisfies the obtained equation.

Analytical review is in most cases easy to do orally. Look at the two equations and many of you will quickly figure out the parallelism of straight lines without any drawing.

Examples for self-solution today will be creative. Because you still have to compete with Baba Yaga, and she, you know, is a lover of all kinds of riddles.

Example 3

Make an equation of a straight line passing through a point parallel to a straight line if

There is a rational and not very rational solution. The shortest way is at the end of the lesson.

We have worked a little with parallel lines and will return to them later. The case of coinciding straight lines is of little interest, so consider a problem that is well known to you from school curriculum:

How to find the intersection point of two lines?

If straight intersect at a point, then its coordinates are the solution systems of linear equations

How to find the point of intersection of lines? Solve the system.

So much for you geometric meaning of a system of two linear equations in two unknowns Are two intersecting (most often) straight lines on a plane.

Example 4

Find the point of intersection of lines

Decision: There are two ways of solving - graphical and analytical.

The graphical way is to simply draw the data lines and find out the intersection point directly from the drawing:

Here's our point:. To check it, you should substitute its coordinates in each equation of the straight line, they should fit both there and there. In other words, the coordinates of a point are the solution of the system. Basically, we looked at a graphical way to solve systems of linear equations with two equations, two unknowns.

The graphical method, of course, is not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide so, the point is that it will take time to get a correct and EXACT drawing. In addition, some straight lines are not so easy to construct, and the point of intersection itself may be located somewhere in the thirty kingdom outside the notebook sheet.

Therefore, it is more expedient to search for the intersection point using the analytical method. Let's solve the system:

To solve the system, the method of term-by-term addition of equations was used. Visit the lesson to build relevant skills. How to solve a system of equations?

Answer:

The check is trivial - the coordinates of the intersection point must satisfy every equation in the system.

Example 5

Find the intersection point of the lines if they intersect.

This is an example for a do-it-yourself solution. It is convenient to divide the task into several stages. Analysis of the condition suggests what is needed:
1) Make the equation of the straight line.
2) Make the equation of the straight line.
3) Find out the relative position of the straight lines.
4) If the lines intersect, then find the intersection point.

The development of an algorithm of actions is typical for many geometric problems, and I will repeatedly focus on this.

Complete solution and the answer at the end of the tutorial:

A pair of shoes has not yet been worn down, as we got to the second section of the lesson:

Perpendicular straight lines. Distance from point to line.
Angle between straight lines

Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to this one, and now the hut on chicken legs will turn 90 degrees:

How to build a line perpendicular to a given one?

Example 6

The straight line is given by the equation. Equate a perpendicular line through a point.

Decision: By condition it is known that. It would be nice to find the direction vector of the straight line. Since the lines are perpendicular, the trick is simple:

From the equation "remove" the normal vector:, which will be the direction vector of the straight line.

Let us compose the equation of a straight line by a point and a direction vector:

Answer:

Let's expand the geometric sketch:

Hmmm ... Orange sky, orange sea, orange camel.

Analytical verification of the solution:

1) Take out the direction vectors from the equations and with the help dot product of vectors we come to the conclusion that the straight lines are indeed perpendicular:.

By the way, you can use normal vectors, it's even easier.

2) Check if the point satisfies the obtained equation .

The check, again, is easy to do orally.

Example 7

Find the point of intersection of perpendicular lines if the equation is known and point.

This is an example for a do-it-yourself solution. There are several actions in the task, so it is convenient to draw up the solution point by point.

Our exciting journey continues:

Distance from point to line

Before us is a straight strip of the river and our task is to reach it by the shortest way. There are no obstacles, and the most optimal route will be movement along the perpendicular. That is, the distance from a point to a straight line is the length of a perpendicular line.

Distance in geometry is traditionally denoted by the Greek letter "ro", for example: - the distance from the point "em" to the straight line "de".

Distance from point to line expressed by the formula

Example 8

Find the distance from point to line

Decision: all you need is to carefully substitute the numbers into the formula and perform the calculations:

Answer:

Let's execute the drawing:

The distance from the point to the line found is exactly the length of the red line. If you draw up a drawing on checkered paper on a scale of 1 unit. \u003d 1 cm (2 cells), then the distance can be measured with an ordinary ruler.

Consider another task for the same blueprint:

The task is to find the coordinates of a point that is symmetrical to a point with respect to a straight line ... I propose to perform the actions yourself, but I will outline the solution algorithm with intermediate results:

1) Find a line that is perpendicular to the line.

2) Find the point of intersection of the lines: .

Both steps are detailed in this tutorial.

3) The point is the midpoint of the line segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the midpoint of the segment we find.

It will not be superfluous to check that the distance is also 2.2 units.

Difficulties here may arise in calculations, but in the tower, a micro calculator helps out great, allowing you to count ordinary fractions. Repeatedly advised, will advise and again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for an independent solution. Let me give you a little hint: there are infinitely many ways to solve it. Debriefing at the end of the lesson, but better try to guess for yourself, I think your ingenuity was dispersed quite well.

Angle between two straight lines

Every angle is a jamb:


In geometry, the angle between two straight lines is taken as the SMALLEST angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered the angle between intersecting straight lines. And his "green" neighbor is considered as such, or oppositely oriented "Crimson" corner.

If the straight lines are perpendicular, then any of the 4 angles can be taken as the angle between them.

How do angles differ? Orientation. First, the direction of the corner "scrolling" is of fundamental importance. Second, a negatively oriented angle is written with a minus sign, for example, if.

Why did I tell this? It seems that you can get by with the usual concept of an angle. The fact is that in the formulas by which we will find the angles, you can easily get a negative result, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing, for a negative angle, be sure to indicate its orientation with an arrow (clockwise).

How to find the angle between two straight lines? There are two working formulas:

Example 10

Find the angle between straight lines

Decision and Method one

Consider two straight lines given by equations in general form:

If straight not perpendicularthen oriented the angle between them can be calculated using the formula:

Let's pay close attention to the denominator - this is exactly scalar product direction vectors of straight lines:

If, then the denominator of the formula vanishes, and the vectors will be orthogonal and the straight lines are perpendicular. That is why a reservation was made about the non-perpendicularity of the straight lines in the formulation.

Based on the above, it is convenient to arrange a solution in two steps:

1) Calculate the scalar product of the direction vectors of straight lines:
, therefore, the straight lines are not perpendicular.

2) The angle between the straight lines is found by the formula:

Through inverse function the corner itself is easy to find. In this case, we use the oddness of the arctangent (see. Graphs and properties of elementary functions):

Answer:

In the answer, we indicate the exact value, as well as the approximate value (preferably both in degrees and in radians), calculated using a calculator.

Well, minus, so minus, that's okay. Here's a geometric illustration:

It is not surprising that the angle turned out to have a negative orientation, because in the problem statement, the first number is a straight line and the "twisting" of the angle began with it.

If you really want to get a positive angle, you need to swap the straight lines, that is, take the coefficients from the second equation , and the coefficients are taken from the first equation. In short, you must start with a straight line .

The distance from a point to a straight line is the length of the perpendicular dropped from a point to a straight line. In descriptive geometry, it is determined graphically using the algorithm below.

Algorithm

1. The straight line is transferred to a position in which it will be parallel to any projection plane. For this, methods of transforming orthogonal projections are used.
2. From a point, a perpendicular is drawn to a straight line. This construction is based on the projection theorem right angle.
3. The length of a perpendicular is determined by transforming its projections or using the right triangle method.

The following figure shows a complex drawing of point M and line b defined by segment CD. It is required to find the distance between them.

According to our algorithm, the first thing to do is to move the line to a position parallel to the projection plane. It is important to understand that after the transformations, the actual distance between the point and the line should not change. That is why it is convenient to use the method of replacing planes here, which does not involve moving figures in space.

The results of the first stage of construction are shown below. The figure shows how an additional frontal plane P 4 is introduced parallel to b. In the new system (P 1, P 4) points C "" 1, D "" 1, M "" 1 are at the same distance from the X axis 1 as C "", D "", M "" from the axis X.

Performing the second part of the algorithm, from M "" 1 we lower the perpendicular M "" 1 N "" 1 to the straight line b "" 1, since the right angle MND between b and MN is projected onto the plane P 4 in full size. On the communication line, we determine the position of point N "and carry out the projection M" N "of the segment MN.

At the final stage, you need to determine the value of the segment MN by its projections M "N" and M "" 1 N "" 1. For this we build right triangle M "" 1 N "" 1 N 0, in which the leg N "" 1 N 0 is equal to the difference (Y M 1 - Y N 1) of the removal of points M "and N" from the X 1 axis. The length of the hypotenuse M "" 1 N 0 of the triangle M "" 1 N "" 1 N 0 corresponds to the desired distance from M to b.

Second solution

• Parallel to CD, we introduce a new frontal plane P 4. It intersects П 1 along the X 1 axis, and X 1 ∥C "D". In accordance with the method of replacing planes, we determine the projections of points C "" 1, D "" 1 and M "" 1, as shown in the figure.
• Perpendicular to C "" 1 D "" 1 we build an additional horizontal plane P 5, onto which the straight line b is projected to the point C "2 \u003d b" 2.
• The distance between point M and line b is determined by the length of the segment M "2 C" 2, marked in red.

Similar tasks:

155 *. Determine the actual size of the line segment AB 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 transforming the drawing it is necessary to achieve the parallelism of this segment of the square. V or pl. H or supplement the V, H system with another plane perpendicular to pl. V or to pl. H and at the same time parallel to this segment.

In fig. 153, in shows the introduction of an additional plane S, perpendicular to pl. H and parallel to a given segment AB.

The projection a s b s is equal to the natural value of the segment AB.

In fig. 153, d shows another technique: segment AB is rotated around a straight line passing through point B and perpendicular to pl. H, to a position parallel

pl. V. In this case, point B remains in place, and point A takes a new position A 1. The horizon is in the new position. projection а 1 b || x-axis. The projection a "1 b" is equal to the natural value of the segment AB.

156. Given a pyramid SABCD (fig. 154). Determine the actual size of the edges of the pyramid AS and CS, using the method of changing projection planes, and the edges BS and DS, using the method of rotation, and take the axis of rotation perpendicular to the square. H.

157 *. Determine the distance from point A to straight line BC (Fig. 155, a).

Decision. The distance from a point to a straight line is measured by a perpendicular segment drawn from a point to a straight line.

If the straight line is perpendicular to any plane (Fig. 155.6), then the distance from the point to the straight line is measured by the distance between the projection of the point and point-projection straight 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, it is necessary to introduce two additional planes into the V, H system.

First (Fig. 155, c) we enter pl. S parallel to the BC segment (the new S / H axis is parallel to the bc projection), and construct the b s c s and a s projections. Then (Fig. 155, d) we introduce another pl. T perpendicular to line BC (new T / S axis perpendicular to b s c s). We build projections of a line and a point - with t (b t) and a t. The distance between points a t and with t (b t) is equal to the distance l from point A to line BC.

In fig. 155e, the same task is accomplished using the rotation method in its form, which is called the parallel movement method. First, the straight line BC and point A, keeping their mutual position unchanged, turn around some (not indicated in the drawing) straight line perpendicular to pl. H, so that the line BC is parallel to the square. V. This is tantamount to moving points A, B, C in planes parallel to square. H. In this case, 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. We position the horizon. the projection of the straight line BC parallel to the x axis (position b 1 c 1) and define the projection a 1, putting 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. projection b "1, a" 1, c "1. Next, we move 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 will be located perpendicular to x, b axes 2 c "2 \u003d b" 1 c "1, and to construct the projection a" 2, take b "2 2" 2 \u003d 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 from 1 to 2 and a 1 to 2 || x 1 we get projections b 2 with 2 and a 2 and the required distance l from point A to line BC. You can determine the distance from A to BC by rotating the plane defined by point A and line BC around the horizontal of this plane to the position T || pl. H (Fig. 155, f).

In the plane set by point A and straight line BC, draw a horizontal line A-1 (Fig. 155, g) and turn point B around it. Point B moves to square. R (given in the drawing by the trace of R h), perpendicular to A-1; at point O is the center of rotation of point B. We now determine the actual value of the radius of rotation of the VO, (Fig. 155, c). In the required position, i.e. when pl. T, defined by point A and line BC, will become || pl. 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 line BC took position T.

Having drawn (Fig. 155, i) straight line b 1 1, we get the horizon. the projection of the straight line BC, already located || pl. H in the same plane with 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 pl. H (Fig. 155, j), turning pl. The horizon around it. trace. Moving from specifying the plane by point A and straight line BC to specifying straight lines BC and A-1 (Fig. 155, l), we find traces of these straight lines and draw traces P ϑ and P h through them. We build (Fig. 155, m) combined with pl. H position front. trace - P ϑ0.

Draw the horizon through point a. frontal projection; the aligned frontal passes through point 2 on the track Р h parallel to Р ϑ0. Point A 0 - combined with pl. H is the position of point A. Similarly, we find point B 0. Direct sun combined with pl. H position passes through point B 0 and point m (horizontal line trace).

The distance from point A 0 to line B 0 C 0 is equal to the required distance l.

You can perform the indicated construction, finding only one trace P h (Fig. 155, n and o). The whole construction is similar to a turn around a horizontal (see Fig. 155, g, c, i): the trace Р h is one of the contour lines of the square. R.

Of the methods for transforming a drawing given for solving this problem, the method of rotation around a horizontal or frontal is preferred.

158. Given pyramid SABC (fig. 156). Determine distances:

a) from the top B of the base to its side AC by parallel movement;

b) from the top of the S pyramid to the sides BC and AB of the base by rotating around the horizontal;

c) from the top S to the side AC of the base by changing the projection planes.

159. A prism is given (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 frontal;

c) between the edges AD and BE by parallel movement.

160. Determine the actual size of the quadrangle ABCD (Fig. 158) by aligning it with pl. H. Use only horizontal plane trace.

161 *. Determine the distance between the crossing lines AB and CD (Fig. 159, a) and construct projections of the perpendicular common to them.

Decision. The distance between crossing lines is measured by the segment (MN) of the perpendicular to both lines (Fig. 159, b). Obviously, if one of the straight lines is placed perpendicular to any square. T then

the segment MN of the perpendicular to both lines will be parallel to square. T projection on this plane will display the desired distance. The projection of the right angle menad MN n AB on the square. T is also 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 pl. T.

In 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 pl. H and parallel to the straight line CD (Fig. 159, c). Then we introduce another additional square. T, perpendicular to pl. S and perpendicular to the same straight line CD (Fig. 159, d). Now you can construct 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 points of intersection of this perpendicular with straight lines AB and CD. At 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, by m s and n s we find m and n on ab and cd, and on them m "and n" on a "b" and c "d".

In fig. 159, c shows the solution to this problem by the method of parallel movements. First, we put a straight CD parallel to the square. V: projection c 1 d 1 || x. Next, we move straight 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 with "2 d" 2 ⊥ x. The segment of the required perpendicular is located || pl. H, and therefore m 2 n 2 expresses the desired distance l between AB and CD. 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, and finally the projections m "and n ", m and n.

162. Given pyramid SABC (fig. 160). Determine the distance between the edge SB and the side AC of the base of the pyramid and build projections of the common perpendicular to SB and AC, applying the method of changing the projection planes.

163. Given a pyramid SABC (fig. 161). Determine the distance between the edge SH and the BC side of the base of the pyramid and construct the projection of the common perpendicular to SX and BC, applying the method of parallel movement.

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 value of a perpendicular drawn from a point to a plane. This distance is projected onto any square. life-size projections, if this plane is perpendicular to the square. projections (Fig. 162, c). This situation can be achieved by transforming the drawing, for example, by changing the square. projections. We introduce pl. S (Fig. 16c, d), perpendicular to pl. triangle BCD. To do this, we spend in pl. triangle horizontal B-1 and place the projection axis S perpendicular to the projection b-1 of the 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 required distance l of the point to the plane.

On rio. 162, e the method of parallel movement is applied. We move the whole system until the horizontal of the B-1 plane is perpendicular to the V plane: the projection b 1 1 1 should be perpendicular to the x axis. In this position, the plane of the triangle will become frontally projection, and the distance l from point A to it will turn out to be square. V without distortion.

In fig. 162, b the plane is defined by traces. We introduce (Fig. 162, e) an additional square. S, perpendicular to pl. P: S / H axis perpendicular to P h. The rest is clear from the drawing. In fig. 162, the problem was solved with one movement: pl. P goes into position P 1, that is, it becomes front-projection. Track. Р 1h is perpendicular to the x-axis. We build a front in this position of the plane. horizontal trace - point n "1, n 1. 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. Given a pyramid SABC (see fig. 160). Determine the distance from point A to the SBC face of the pyramid using the parallel movement method.

166. Given a pyramid SABC (see fig. 161). Determine the height of the pyramid using the parallel movement method.

167 *. Determine the distance between crossing lines AB and CD (see Fig. 159, a) as the distance between parallel planes drawn through these lines.

Decision. In fig. 163, and shows parallel planes P and Q, of which pl. Q is carried out through CD parallel to AB, and pl. R - through AB parallel to pl. Q. The distance between such planes is the distance between the crossing lines AB and CD. However, you can limit yourself to building only one plane, for example, Q, parallel to AB, and then determine the distance from at least point A to this plane.

In fig. 163c shows the Q plane drawn through CD parallel to AB; in projections drawn with "e" || a "b" and ce || ab. Applying the method of changing the square. projections (Fig. 163, c), we introduce an additional square. S, perpendicular to pl. V and at the same time

perpendicular to pl. Q. To draw the S / V axis, 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 pl. S as a straight line with s d s. The rest is clear from the drawing.

168. Given the pyramid SABC (see fig. 160). Determine the distance between the ribs SC and AB. Apply: 1) the method of changing the square. projections, 2) a method of parallel movement.

169 *. Determine the distance between the 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 carry out the 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. In fig. 164, g introduced an additional pl. S perpendicular to pl. H and to both given planes. The S.H axis is perpendicular to the horizon. horizontal projection drawn in one of the planes. We build a projection of this plane and point In another plane on the square. 5. The distance of the point d s to the straight line l s a s is equal to the required distance between the parallel planes.

In fig. 164, d another construction is given (according to the method of parallel movement). In order for the plane, expressed by intersecting straight lines AB and AC, to be perpendicular to pl. V, horizon. we put the projection of the horizontal of this plane perpendicular to the x-axis: 1 1 2 1 ⊥ x. Distance between the front. projection d "1 point D and straight line a" 1 2 "1 (front. projection of the plane) is equal to the required distance between the planes.

In fig. 164, e shows the introduction of an additional pl. S, perpendicular to the area H and to the given planes P and Q (the S / H axis is perpendicular to the tracks P h, and Q h). We build traces P s, and Q s. The distance between them (see Fig. 164, c) is equal to the required distance l between the planes P and Q.

In 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 tracks turn out to be perpendicular to the x-axis. Distance between new front. by 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 the faces ABFE and DCGH - l 2; c) between the edges ADHE and BCGF-l 3.