Sinusitis

Definition of slant. Mathematics

TRIANGLES.

§ 31.PERPENDICULAR AND INCLINED TO A STRAIGHT.

1. Projection of a segment onto a straight line.

If through some point taken outside a line we draw a line perpendicular to it, then for brevity the segment from this point to the line is called in one word perpendicular.

Segment CO is perpendicular to line AB. Point O is called base of the perpendicular CO (drawing 168).

If a line drawn through a given point intersects another line, but is not perpendicular to it, then its segment from a given point to the point of intersection with another line is called inclined to this line.

Segment BC - inclined to straight line AO. Point C is called basis inclined (Fig. 169).

If we drop perpendiculars from the ends of some segment onto an arbitrary line, then the line segment enclosed between the bases of the perpendiculars is called projection of the segment to this straight line.

Segment A "B" is the projection of segment AB onto EC. Segment OM" is also called the projection of segment OM onto EC.

The projection of the segment KR perpendicular to the EU will be point K" (Fig. 170).

2. Properties of perpendicular and oblique.

Theorem 1. A perpendicular drawn from a point to a straight line is less than any oblique drawn from the same point to this straight line.

Segment AC (Fig. 171) is perpendicular to straight line OB, and AM is one of the inclined lines drawn from point A to straight line OB. It is required to prove that AM > AC.

IN /\ The MAC segment AM is the hypotenuse, and the hypotenuse is greater than each of the legs of this triangle (§ 30). Therefore, AM > AC. Since we took the oblique AM arbitrarily, we can assert that any oblique to a straight line is greater than a perpendicular to this line (and a perpendicular is shorter than any oblique) if they are drawn to it from the same point.

The converse statement is also true, namely: if the segment AC (Fig. 171) is less than any other segment connecting the point AC with any point on the straight line OB, then it is perpendicular to OB. In fact, segment AC cannot be inclined to OB, since then it would not be the shortest of the segments connecting point A with points of straight line OB. This means that it can only be perpendicular to OB.

The length of the perpendicular dropped from a given point to a straight line is taken as the distance from a given point to this straight line.

Theorem 2. If two oblique lines drawn to a line from the same point are equal, then their projections are equal.

Let BA and BC be inclined lines drawn from point B to straight line AC (Fig. 172), and AB = BC. It is necessary to prove that their projections are also equal.

To prove this, let us lower the perpendicular BO from point B to AC. Then AO and OS will be projections of inclined AB and BC onto straight line AC. Triangle ABC is isosceles according to the theorem. VO is the height of this triangle. But the height is isosceles triangle, drawn to the base, is at the same time the median of this triangle (§ 18).

Therefore AO = OS.

Theorem 3(reverse). If two oblique lines drawn to a straight line from the same point have equal projections, then they are equal to each other.

Let AC and CB be inclined to straight line AB (Fig. 173). CO_|_ AB and AO = OB.

It is required to prove that AC = BC.

In right triangles AOC and BOC, the legs AO and OB are equal. CO is the common leg of these triangles. Hence, /\ AOC = /\ SUN. From the equality of triangles it follows that AC = BC.

Theorem 4. If two inclined lines are drawn from the same point to a straight line, then the one that has a larger projection onto this straight line is larger.

Let AB and BC be inclined to straight line AO; VO_|_AO and AO>SO. It is required to prove that AB > BC.

1) Inclined ones are located on one side of the perpendicular.

Angle ACE external to right triangle OWL (drawing 174), and therefore / DIA > / OWL, i.e. he is stupid. It follows that AB > CB.

2) Inclined ones are located on both sides of the perpendicular. To prove this, let us plot the segment OK = OS on AO from point O and connect point K with point B (Fig. 175). Then, by Theorem 3, we have: VC = BC, but AB > VC, therefore, AB > BC, i.e. the theorem is valid in this case as well.

Theorem 5(reverse). If two inclined lines are drawn from the same point to a straight line, then the larger inclined line also has a larger projection onto this straight line.

Let KS and BC be inclined to the straight line KB (Fig. 176), SO_|_KB and KS > BC. It is required to prove that KO > OB.

Between the segments KO and OB there can be only one of three relationships:

1) KO< ОВ,
2) KO = OV,
3) KO > OV.

KO cannot be less than OB, since then, according to Theorem 4, the inclined KS would be less than the inclined BC, and this contradicts the conditions of the theorem.

In the same way, KO cannot equal OB, since in this case, according to Theorem 3, KS = BC, which also contradicts the conditions of the theorem.

Consequently, only the last relation remains true, namely, that
KO > OV.

A perpendicular dropped from a given point to a given plane is a segment connecting a given point with a point on the plane and lying on a straight line perpendicular to the plane. The end of this segment lying in the plane is called the base of the perpendicular. The distance from a point to a plane is the length of the perpendicular drawn from this point to the plane.

The slope drawn from a given point to a given plane is any segment that connects a given point to a point on the plane and is not perpendicular to this plane. The end of a segment lying in a plane is called the inclined base. A segment connecting the bases of a perpendicular and an oblique drawn from the same point is called an oblique projection.

In Figure 136, from point A, a perpendicular AB and an inclined AC are drawn to the plane. Point B is the base of the perpendicular, point C is the base of the inclined one, BC is the projection of the inclined AC onto plane a.

Since the distances from the points of a line to a plane parallel to it are the same, the distance from a line to a plane parallel to it is the distance from any point of it to this plane.

A straight line drawn on a plane through the base of an inclined plane perpendicular to its projection is also perpendicular to the inclined itself. And vice versa: if a straight line in a plane is perpendicular to an inclined one, then it is also perpendicular to the projection of the inclined one (the theorem of three perpendiculars).

In Figure 137, a perpendicular AB and an inclined AC are drawn to plane a. The straight line o, lying in the plane a, is perpendicular to BC - the projection of the inclined AC onto the plane a. According to T. 2.12, straight line a is perpendicular to inclined AC. If it were known that straight line a is perpendicular to inclined AC, then according to T. 2.12 it would be perpendicular to its projection - BC.

Example. The legs of right triangle ABC are equal to 16 and from the vertex right angle C is drawn perpendicular to the plane of this triangle CD = 35 m (Fig. 138). Find the distance from point D to the hypotenuse AB.

Solution. Let's do it. According to the condition, DC is perpendicular to the plane, i.e. DE is inclined, CE is its projection, therefore, by the theorem about three perpendiculars, it follows from the condition that

From we find To find the height CE in we find

On the other hand, where

From the Pythagorean theorem

46. Perpendicularity of planes.

Two intersecting planes are called perpendicular if any plane perpendicular to the line of intersection of these planes intersects them along perpendicular lines.

Figure 139 shows two planes that intersect along a straight line a. The plane y is perpendicular to the line a and intersects. In this case, the plane y intersects the plane a along the straight line c, and the plane intersects along the straight line d, and i.e., by definition

T. 2.13. If a plane passes through a line perpendicular to another plane, then these planes are perpendicular (a sign of perpendicularity of planes).

In Figure 140, the plane passes through a straight line, i.e., the plane is perpendicular.

Geometry lesson in 10th grade

In one of the previous lessons, you became acquainted with the concept of the projection of a point onto a given plane parallel to a given line.

In this lesson you will continue to study lines and planes; learn how the angle is between a straight line and a plane. You will become familiar with the concept of orthogonal projection onto a plane and consider its properties. The lesson will give definitions of the distance from a point to a plane and from a point to a straight line, the angle between a straight line and a plane. The famous theorem of three perpendiculars will be proven.

Orthographic projection

Orthogonal projection of a point and a figure.

Orthogonal projection of the part.

Orthogonal projection of point A onto a given plane is called the projection of a point onto this plane parallel

a straight line perpendicular to this plane. Orthographic projection

of a figure onto a given plane p consists of orthogonal projections onto the plane p of all points of this figure. Orthographic projection is often used to depict spatial bodies on a plane, especially in technical drawings. It gives a more realistic image than arbitrary parallel projection, especially of round bodies.

Perpendicular and oblique

Let a straight line be drawn through point A, which does not belong to the plane p, perpendicular to this plane and intersecting it at point B. Then

segment AB is called

perpendicular, omitted from the point

And to this plane, and point B itself is the base of this perpendicular. Any segment AC, where C is

an arbitrary point of the plane p, different from B, is called inclined to

this plane.

Note that point B in this definition is orthogonal

projection of point A, and segment AC - Perpendicular and oblique. orthogonal projection of oblique AB.

Orthogonal projections have all the properties of ordinary parallel projections, but also have a number of new properties.

Let a perpendicular and several inclined lines be drawn from one point to the plane. Then the following statements are true.

1. Any inclined plane is longer than both the perpendicular and the orthogonal projection of the inclined plane onto this plane.

2. Equal slopes have equal orthogonal projections, and vice versa, oblique ones having equal projections are also equal.

3. One oblique is longer than the other if and only if the orthogonal projection of the first oblique is longer than the orthogonal projection of the second oblique.

Properties of orthographic projection

Proof.

Let a perpendicular AB and two oblique ones AC and AD be drawn from point A to plane p; then the segments BC and BD are orthogonal projections of these segments onto the plane p.

Let us prove the first statement: any inclined plane is longer than both the perpendicular and the orthogonal projection of the inclined plane onto this plane. Consider, for example, the oblique AC and the triangle ABC formed by the perpendicular AB, this oblique AC, and its orthogonal projection BC. This triangle is right-angled with a right angle at vertex B and hypotenuse AC, which, as we know from planimetry, is longer than each of the legs, i.e. and perpendicular AB, and projection BC.

From point A to the plane pi, a perpendicular AB and two inclined ones AC and AD are drawn.

Properties of orthographic projection

Triangles

ABC and ABD

equal in leg and hypotenuse.

Now we will prove the second statement, namely: equal obliques have equal orthogonal projections, and vice versa, obliques having equal projections are also equal.

Consider right triangles ABC and ABD. They

have a common leg AB. If the oblique AC and AD are equal, then the right triangles ABC and ABD are equal in leg and hypotenuse, and then BC = BD. Conversely, if the projections BC and BD are equal, then these same triangles are equal along two legs, and then their hypotenuses AC and AD are equal. Sun< BD, как мы только что доказали,АС < AD, что опять противоречит условию.

A third possibility remains: BC > BD. The theorem is proven.

If BC is greater than BD,

then AC is larger than the side

AE equal to AD.

GEOMETRY

Section II. STEREOMETRY

§8. PERPENDICULAR AND OBLIQUE. OBLIQUE PROJECTION ON A PLANE.

2. Properties of perpendicular and oblique.

Let's consider the properties of perpendicular and oblique.

1) The perpendicular drawn from a given point to the plane is less than any inclined one drawn from the same point to the plane.

In Figure 411: AN AK.

2) If two inclined slopes drawn from a given point to a plane are equal, then their projections are equal.

K 1 and perpendicular AN and AK = AK 1. Then by property: NK = NK 1.

3) If two inclined slopes drawn from a given point to a given plane have equal projections, then they are equal to each other.

In Figure 412, two inclined planes AK and A are drawn from point A to plane a K 1 and perpendicular to AN, with KH = K 1 N. Then by property: AK = AK 1 .

4) If two inclined planes are drawn from a given point to a plane, then the larger inclined one has a larger projection.

L and perpendicular to AN, A K > AL . Then by property: H K > HL.

5) If two inclined slopes are drawn from a given point to a plane, then the larger of them is the one that has a larger projection onto the given plane.

In Figure 413, two inclined planes AK and A are drawn from point A to plane a L and perpendicular to AN, NK> N L . Then by property: AK> A L .

Example 1. Two inclined slopes are drawn from a point to a plane, the lengths of which are 41 cm and 50 cm. Find the projections of the inclined ones, if they have a ratio of 3:10, and the distance from the point to the plane.

Solutions. 1) A L = 41 cm; AK = 50 cm (Fig. 413). By property we have H L NK. Let us denote H L = 3 x cm, NK = 10 x cm, AN = h see AN - distance from point A to the planeα .

4) Equating, we get 41 2 - 9x 2 = 50 2 - 100 x 2; x 2 = 9; x = 3 (considering x> 0). So, N L = 3 ∙ 3 = 9 (cm), NK = 10 ∙ 3 = 30 (cm).

Example 2. From this point to two inclined planes are drawn, eachin cm. The angle between the inclined ones is 60°, and the angle between their projections is straight. Find the distance from the point to the plane.

If through some point taken outside a line we draw a line perpendicular to it, then for brevity the segment from this point to the line is called in one word perpendicular.

Segment CO is perpendicular to line AB. Point O is called base of the perpendicular CO (rice).

Segment BC - inclined to straight line AO. Point C is called basis inclined (Fig.).

Segment АВ - projection of segment AB onto EC. The segment OM is also called the projection of the segment OM onto the EC.

The projection of the segment KP perpendicular to EC will be the point K (Fig.).

2. Properties of perpendicular and oblique.

Theorem 1. A perpendicular drawn from a point to a straight line is less than any oblique drawn from the same point to this straight line.

The segment AC (Fig.) is perpendicular to straight line OB, and AM is one of the inclined lines drawn from point A to straight line OB. It is required to prove that AM > AC.

In ΔMAC, the segment AM is the hypotenuse, and the hypotenuse is larger than each of the legs of this triangle. Therefore, AM > AC. Since we took the oblique AM arbitrarily, we can assert that any oblique to a straight line is greater than a perpendicular to this line (and a perpendicular is shorter than any oblique) if they are drawn to it from the same point.

The converse statement is also true, namely: if the segment AC (Fig.) is less than any other segment connecting the point AC with any point of the straight line OB, then it is perpendicular to OB. In fact, segment AC cannot be inclined to OB, since then it would not be the shortest of the segments connecting point A with points of straight line OB. This means that it can only be perpendicular to OB.

The length of the perpendicular dropped from a given point to a straight line is taken as the distance from a given point to this straight line.

Theorem 2. If two oblique lines drawn to a line from the same point are equal, then their projections are equal.

Let BA and BC be inclined lines drawn from point B to straight line AC (Fig.), and AB = BC. It is necessary to prove that their projections are also equal.

To prove this, let us lower the perpendicular BO from point B to AC. Then AO and OS will be projections of inclined AB and BC onto straight line AC. Triangle ABC is isosceles according to the theorem. VO is the height of this triangle. But the altitude in an isosceles triangle drawn to the base is at the same time the median of this triangle.

Therefore AO = OS.

Theorem 3 (converse). If two oblique lines drawn to a straight line from the same point have equal projections, then they are equal to each other.

Let AC and CB be inclined to straight line AB (Fig.). CO ⊥ AB and AO = OB.

It is required to prove that AC = BC.

In right triangles AOC and BOC, the legs AO and OB are equal. CO is the common leg of these triangles. Therefore, ΔAOC = ΔBOC. From the equality of triangles it follows that AC = BC.

Theorem 4. If two inclined lines are drawn from the same point to a straight line, then the one that has a larger projection onto this straight line is larger.

Let AB and BC be inclined to straight line AO; VO ⊥ AO and AO>CO. It is required to prove that AB > BC.

1) Inclined ones are located on one side of the perpendicular.

Angle ACE is external with respect to the right triangle COB (Fig.), and therefore ∠ACV > ∠COV, i.e. it is obtuse. It follows that AB > CB.

2) Inclined ones are located on both sides of the perpendicular. To prove this, let’s plot the segment OK = OS on AO from point O and connect point K to point B (Fig.). Then, by Theorem 3, we have: VC = BC, but AB > VC, therefore, AB > BC, i.e. the theorem is valid in this case as well.

Theorem 5 (converse). If two inclined lines are drawn from the same point to a straight line, then the larger inclined line also has a larger projection onto this straight line.

Let KS and BC be inclined to the straight line CV (Fig.), CO ⊥ CV and KS > BC. It is required to prove that KO > OB.

Between the segments KO and OB there can be only one of three relationships:

1) KO< ОВ,

2) KO = OV,

3) KO > OV.

KO cannot be less than OB, since then, according to Theorem 4, the inclined KS would be less than the inclined BC, and this contradicts the conditions of the theorem.

In the same way, KO cannot equal OB, since in this case, according to Theorem 3, KS = BC, which also contradicts the conditions of the theorem.

Consequently, only the last relation remains true, namely, that KO > OB.