Find the acute angle between the lines. Finding the angle between lines

Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2 , then the acute angle between these lines will be defined as

Two lines are parallel if k 1 = k 2 . Two lines are perpendicular if k 1 = -1/ k 2 .

Theorem. The straight lines Ax + Vy + C \u003d 0 and A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients A 1 \u003d λA, B 1 \u003d λB are proportional. If also С 1 = λС, then the lines coincide. The coordinates of the point of intersection of two lines are found as a solution to the system of equations of these lines.

Equation of a line passing through a given point

Perpendicular to this line

Definition. The line passing through the point M 1 (x 1, y 1) and perpendicular to the line y \u003d kx + b is represented by the equation:

Distance from point to line

Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Vy + C \u003d 0 is defined as

Proof. Let the point M 1 (x 1, y 1) be the base of the perpendicular dropped from the point M to the given line. Then the distance between points M and M 1:

(1)

The x 1 and y 1 coordinates can be found as a solution to the system of equations:

The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicular to a given straight line. If we transform the first equation of the system to the form:

A(x - x 0) + B(y - y 0) + Ax 0 + By 0 + C = 0,

then, solving, we get:

Substituting these expressions into equation (1), we find:

The theorem has been proven.

Example. Determine the angle between the lines: y = -3 x + 7; y = 2 x + 1.

k 1 \u003d -3; k2 = 2; tgφ = ; φ= p /4.

Example. Show that the lines 3x - 5y + 7 = 0 and 10x + 6y - 3 = 0 are perpendicular.

Solution. We find: k 1 \u003d 3/5, k 2 \u003d -5/3, k 1 * k 2 \u003d -1, therefore, the lines are perpendicular.

Example. The vertices of the triangle A(0; 1), B (6; 5), C (12; -1) are given. Find the equation for the height drawn from vertex C.

Solution. We find the equation of the side AB: ; 4 x = 6 y - 6;

2x – 3y + 3 = 0;

The desired height equation is: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: whence b = 17. Total: .

Answer: 3x + 2y - 34 = 0.

Equation of a line passing through a given point in a given direction. Equation of a straight line passing through two given points. Angle between two lines. Condition of parallelism and perpendicularity of two lines. Determining the point of intersection of two lines

1. Equation of a line passing through a given point A(x 1 , y 1) in a given direction, determined by the slope k,

y - y 1 = k(x - x 1). (1)

This equation defines a pencil of lines passing through a point A(x 1 , y 1), which is called the center of the beam.

2. Equation of a straight line passing through two points: A(x 1 , y 1) and B(x 2 , y 2) is written like this:

The slope of a straight line passing through two given points is determined by the formula

3. Angle between straight lines A And B is the angle by which the first straight line must be rotated A around the point of intersection of these lines counterclockwise until it coincides with the second line B. If two lines are given by slope equations

y = k 1 x + B 1 ,

y = k 2 x + B 2 , (4)

then the angle between them is determined by the formula

It should be noted that in the numerator of the fraction, the slope of the first straight line is subtracted from the slope of the second straight line.

If the equations of a straight line are given in general view

A 1 x + B 1 y + C 1 = 0,

A 2 x + B 2 y + C 2 = 0, (6)

the angle between them is determined by the formula

4. Conditions for parallelism of two lines:

a) If the lines are given by equations (4) with a slope, then the necessary and sufficient condition for their parallelism is the equality of their slopes:

k 1 = k 2 . (8)

b) For the case when the lines are given by equations in general form (6), the necessary and sufficient condition for their parallelism is that the coefficients at the corresponding current coordinates in their equations are proportional, i.e.

5. Conditions for perpendicularity of two lines:

a) In the case when the lines are given by equations (4) with a slope, the necessary and sufficient condition for their perpendicularity is that their slopes are reciprocal in magnitude and opposite in sign, i.e.

This condition can also be written in the form

k 1 k 2 = -1. (11)

b) If the equations of straight lines are given in general form (6), then the condition for their perpendicularity (necessary and sufficient) is to fulfill the equality

A 1 A 2 + B 1 B 2 = 0. (12)

6. The coordinates of the point of intersection of two lines are found by solving the system of equations (6). Lines (6) intersect if and only if

1. Write the equations of the lines passing through the point M, one of which is parallel and the other is perpendicular to the given line l.

This material is devoted to such a concept as the angle between two intersecting straight lines. In the first paragraph, we will explain what it is and show it in illustrations. Then we will analyze how you can find the sine, cosine of this angle and the angle itself (we will separately consider cases with a plane and three-dimensional space), we will give the necessary formulas and show with examples how exactly they are applied in practice.

Yandex.RTB R-A-339285-1

In order to understand what an angle formed at the intersection of two lines is, we need to recall the very definition of an angle, perpendicularity, and an intersection point.

Definition 1

We call two lines intersecting if they have one common point. This point is called the point of intersection of the two lines.

Each line is divided by the point of intersection into rays. In this case, both lines form 4 angles, of which two are vertical and two are adjacent. If we know the measure of one of them, then we can determine the other remaining ones.

Let's say we know that one of the angles is equal to α. In such a case, the angle that is vertical to it will also be equal to α. To find the remaining angles, we need to calculate the difference 180 ° - α . If α is equal to 90 degrees, then all angles will be right. Lines intersecting at right angles are called perpendicular (a separate article is devoted to the concept of perpendicularity).

Take a look at the picture:

Let us proceed to the formulation of the main definition.

Definition 2

The angle formed by two intersecting lines is the measure of the smaller of the 4 angles that form these two lines.

An important conclusion must be drawn from the definition: the size of the angle in this case will be expressed by any real number in the interval (0, 90] . If the lines are perpendicular, then the angle between them will in any case be equal to 90 degrees.

The ability to find the measure of the angle between two intersecting lines is useful for solving many practical problems. The solution method can be selected from several options.

For starters, we can take geometric methods. If we know something about additional angles, then we can connect them to the angle we need using the properties of equal or similar shapes. For example, if we know the sides of a triangle and we need to calculate the angle between the lines on which these sides are located, then the cosine theorem is suitable for solving. If we have a right triangle in the condition, then for calculations we will also need to know the sine, cosine and tangent of the angle.

The coordinate method is also very convenient for solving problems of this type. Let's explain how to use it correctly.

We have a rectangular (cartesian) coordinate system O x y with two straight lines. Let's denote them by letters a and b. In this case, straight lines can be described using any equations. The original lines have an intersection point M . How to determine the desired angle (let's denote it α) between these lines?

Let's start with the formulation of the basic principle of finding an angle under given conditions.

We know that such concepts as directing and normal vector are closely related to the concept of a straight line. If we have the equation of some straight line, we can take the coordinates of these vectors from it. We can do this for two intersecting lines at once.

The angle formed by two intersecting lines can be found using:

angle between direction vectors;
angle between normal vectors;
the angle between the normal vector of one line and the direction vector of the other.

Now let's look at each method separately.

1. Suppose we have a line a with direction vector a → = (a x , a y) and a line b with direction vector b → (b x , b y) . Now let's set aside two vectors a → and b → from the intersection point. After that, we will see that they will each be located on their own line. Then we have four options for their relative position. See illustration:

If the angle between two vectors is not obtuse, then it will be the angle we need between the intersecting lines a and b. If it is obtuse, then the desired angle will be equal to the angle adjacent to the angle a → , b → ^ . Thus, α = a → , b → ^ if a → , b → ^ ≤ 90 ° , and α = 180 ° - a → , b → ^ if a → , b → ^ > 90 ° .

Based on the fact that the cosines of equal angles are equal, we can rewrite the resulting equalities as follows: cos α = cos a → , b → ^ if a → , b → ^ ≤ 90 ° ; cos α = cos 180 ° - a → , b → ^ = - cos a → , b → ^ if a → , b → ^ > 90 ° .

In the second case, reduction formulas were used. In this way,

cos α cos a → , b → ^ , cos a → , b → ^ ≥ 0 - cos a → , b → ^ , cos a → , b → ^< 0 ⇔ cos α = cos a → , b → ^

Let's write the last formula in words:

Definition 3

The cosine of the angle formed by two intersecting lines will be equal to the modulus of the cosine of the angle between its direction vectors.

The general form of the formula for the cosine of the angle between two vectors a → = (a x, a y) and b → = (b x, b y) looks like this:

cos a → , b → ^ = a → , b → ^ a → b → = a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2

From it we can derive the formula for the cosine of the angle between two given lines:

cos α = a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2 = a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2

Then the angle itself can be found using the following formula:

α = a r c cos a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2

Here a → = (a x , a y) and b → = (b x , b y) are the direction vectors of the given lines.

Let us give an example of solving the problem.

Example 1

In a rectangular coordinate system, two intersecting lines a and b are given on the plane. They can be described by parametric equations x = 1 + 4 · λ y = 2 + λ λ ∈ R and x 5 = y - 6 - 3 . Calculate the angle between these lines.

Solution

We have a parametric equation in the condition, which means that for this straight line we can immediately write down the coordinates of its direction vector. To do this, we need to take the values of the coefficients at the parameter, i.e. the line x = 1 + 4 λ y = 2 + λ λ ∈ R will have a direction vector a → = (4 , 1) .

The second straight line is described using the canonical equation x 5 = y - 6 - 3 . Here we can take the coordinates from the denominators. Thus, this line has a direction vector b → = (5 , - 3) .

Next, we proceed directly to finding the angle. To do this, simply substitute the available coordinates of the two vectors into the above formula α = a r c cos a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2 . We get the following:

α = a r c cos 4 5 + 1 (- 3) 4 2 + 1 2 5 2 + (- 3) 2 = a r c cos 17 17 34 = a r c cos 1 2 = 45°

Answer: These lines form an angle of 45 degrees.

We can solve a similar problem by finding the angle between normal vectors. If we have a line a with a normal vector na → = (nax , nay) and a line b with a normal vector nb → = (nbx , nby) , then the angle between them will be equal to the angle between na → and nb → or the angle that will be adjacent to na → , nb → ^ . This method is shown in the picture:

The formulas for calculating the cosine of the angle between intersecting lines and this angle itself using the coordinates of normal vectors look like this:

cos α = cos n a → , n b → ^ = n a x n b x + n a y + n b y n a x 2 + n a y 2 n b x 2 + n b y 2

Here n a → and n b → denote the normal vectors of two given lines.

Example 2

Two straight lines are given in a rectangular coordinate system using the equations 3 x + 5 y - 30 = 0 and x + 4 y - 17 = 0 . Find the sine, cosine of the angle between them, and the magnitude of that angle itself.

Solution

The original straight lines are given using normal straight line equations of the form A x + B y + C = 0 . Denote the normal vector n → = (A , B) . Let's find the coordinates of the first normal vector for one straight line and write them down: n a → = (3 , 5) . For the second line x + 4 y - 17 = 0 the normal vector will have coordinates n b → = (1 , 4) . Now add the obtained values to the formula and calculate the total:

cos α = cos n a → , n b → ^ = 3 1 + 5 4 3 2 + 5 2 1 2 + 4 2 = 23 34 17 = 23 2 34

If we know the cosine of an angle, then we can calculate its sine using the basic trigonometric identity. Since the angle α formed by straight lines is not obtuse, then sin α \u003d 1 - cos 2 α \u003d 1 - 23 2 34 2 \u003d 7 2 34.

In this case, α = a r c cos 23 2 34 = a r c sin 7 2 34 .

Answer: cos α = 23 2 34 , sin α = 7 2 34 , α = a r c cos 23 2 34 = a r c sin 7 2 34

Let's analyze the last case - finding the angle between the lines, if we know the coordinates of the direction vector of one line and the normal vector of the other.

Assume that line a has a direction vector a → = (a x , a y) , and line b has a normal vector n b → = (n b x , n b y) . We need to postpone these vectors from the intersection point and consider all options for their relative position. See picture:

If the angle between the given vectors is no more than 90 degrees, it turns out that it will complement the angle between a and b to a right angle.

a → , n b → ^ = 90 ° - α if a → , n b → ^ ≤ 90 ° .

If it is less than 90 degrees, then we get the following:

a → , n b → ^ > 90 ° , then a → , n b → ^ = 90 ° + α

Using the rule of equality of cosines of equal angles, we write:

cos a → , n b → ^ = cos (90 ° - α) = sin α for a → , n b → ^ ≤ 90 ° .

cos a → , n b → ^ = cos 90 ° + α = - sin α at a → , n b → ^ > 90 ° .

In this way,

sin α = cos a → , nb → ^ , a → , nb → ^ ≤ 90 ° - cos a → , nb → ^ , a → , nb → ^ > 90 ° ⇔ sin α = cos a → , nb → ^ , a → , nb → ^ > 0 - cos a → , nb → ^ , a → , nb → ^< 0 ⇔ ⇔ sin α = cos a → , n b → ^

Let's formulate a conclusion.

Definition 4

To find the sine of the angle between two lines intersecting in a plane, you need to calculate the modulus of the cosine of the angle between the direction vector of the first line and the normal vector of the second.

Let's write down the necessary formulas. Finding the sine of an angle:

sin α = cos a → , n b → ^ = a x n b x + a y n b y a x 2 + a y 2 n b x 2 + n b y 2

Finding the corner itself:

α = a r c sin = a x n b x + a y n b y a x 2 + a y 2 n b x 2 + n b y 2

Here a → is the direction vector of the first line, and n b → is the normal vector of the second.

Example 3

Two intersecting lines are given by the equations x - 5 = y - 6 3 and x + 4 y - 17 = 0 . Find the angle of intersection.

Solution

We take the coordinates of the directing and normal vector from the given equations. It turns out a → = (- 5 , 3) and n → b = (1 , 4) . We take the formula α \u003d a r c sin \u003d a x n b x + a y n b y a x 2 + a y 2 n b x 2 + n b y 2 and consider:

α = a r c sin = - 5 1 + 3 4 (- 5) 2 + 3 2 1 2 + 4 2 = a r c sin 7 2 34

Note that we took the equations from the previous problem and got exactly the same result, but in a different way.

Answer:α = a r c sin 7 2 34

Here is another way to find the desired angle using the slope coefficients of given lines.

We have a line a , which is defined in a rectangular coordinate system using the equation y = k 1 · x + b 1 , and a line b , defined as y = k 2 · x + b 2 . These are equations of lines with a slope. To find the angle of intersection, use the formula:

α = a r c cos k 1 k 2 + 1 k 1 2 + 1 k 2 2 + 1 , where k 1 and k 2 are the slopes of the given lines. To obtain this record, formulas for determining the angle through the coordinates of normal vectors were used.

Example 4

There are two straight lines intersecting in the plane, given by the equations y = - 3 5 x + 6 and y = - 1 4 x + 17 4 . Calculate the angle of intersection.

Solution

The slopes of our lines are equal to k 1 = - 3 5 and k 2 = - 1 4 . Let's add them to the formula α = a r c cos k 1 k 2 + 1 k 1 2 + 1 k 2 2 + 1 and calculate:

α = a r c cos - 3 5 - 1 4 + 1 - 3 5 2 + 1 - 1 4 2 + 1 = a r c cos 23 20 34 24 17 16 = a r c cos 23 2 34

Answer:α = a r c cos 23 2 34

In the conclusions of this paragraph, it should be noted that the formulas for finding the angle given here do not have to be learned by heart. To do this, it is sufficient to know the coordinates of the guides and/or normal vectors of the given lines and be able to determine them from different types equations. But the formulas for calculating the cosine of an angle are better to remember or write down.

How to calculate the angle between intersecting lines in space

The calculation of such an angle can be reduced to the calculation of the coordinates of the direction vectors and the determination of the magnitude of the angle formed by these vectors. For such examples, we use the same reasoning that we have given before.

Let's say we have a rectangular coordinate system located in 3D space. It contains two lines a and b with the intersection point M . To calculate the coordinates of the direction vectors, we need to know the equations of these lines. Denote the direction vectors a → = (a x , a y , a z) and b → = (b x , b y , b z) . To calculate the cosine of the angle between them, we use the formula:

cos α = cos a → , b → ^ = a → , b → a → b → = a x b x + a y b y + a z b z a x 2 + a y 2 + a z 2 b x 2 + b y 2 + b z 2

To find the angle itself, we need this formula:

α = a r c cos a x b x + a y b y + a z b z a x 2 + a y 2 + a z 2 b x 2 + b y 2 + b z 2

Example 5

We have a straight line defined in 3D space using the equation x 1 = y - 3 = z + 3 - 2 . It is known that it intersects with the O z axis. Calculate the angle of intersection and the cosine of that angle.

Solution

Let's denote the angle to be calculated by the letter α. Let's write down the coordinates of the direction vector for the first straight line - a → = (1 , - 3 , - 2) . For the applicate axis, we can take the coordinate vector k → = (0 , 0 , 1) as a guide. We have received the necessary data and can add it to the desired formula:

cos α = cos a → , k → ^ = a → , k → a → k → = 1 0 - 3 0 - 2 1 1 2 + (- 3) 2 + (- 2) 2 0 2 + 0 2 + 1 2 = 2 8 = 1 2

As a result, we got that the angle we need will be equal to a r c cos 1 2 = 45 °.

Answer: cos α = 1 2 , α = 45 ° .

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

Let lines be given in space l And m. Through some point A of the space we draw straight lines l 1 || l And m 1 || m(Fig. 138).

Note that the point A can be chosen arbitrarily, in particular, it can lie on one of the given lines. If straight l And m intersect, then A can be taken as the point of intersection of these lines ( l 1 = l And m 1 = m).

Angle between non-parallel lines l And m is the value of the smallest of the adjacent angles formed by intersecting straight lines l 1 And m 1 (l 1 || l, m 1 || m). The angle between parallel lines is assumed to be zero.

Angle between lines l And m denoted by $\widehat((l;m)) $. From the definition it follows that if it is measured in degrees, then 0 ° < $\widehat((l;m)) $ < 90°, and if in radians, then 0 < $\widehat((l;m)) $ < π / 2 .

A task. The cube ABCDA 1 B 1 C 1 D 1 is given (Fig. 139).

Find the angle between straight lines AB and DC 1 .

Straight AB and DC 1 crossing. Since the line DC is parallel to the line AB, the angle between the lines AB and DC 1, according to the definition, is equal to $\widehat(C_(1)DC)$.

Hence $\widehat((AB;DC_1))$ = 45°.

Direct l And m called perpendicular, if $\widehat((l;m)) $ = π / 2. For example, in a cube

Calculation of the angle between lines.

The problem of calculating the angle between two straight lines in space is solved in the same way as in the plane. Denote by φ the angle between the lines l 1 And l 2 , and through ψ - the angle between the direction vectors but And b these straight lines.

Then if

ψ <90° (рис. 206, а), то φ = ψ; если же ψ >90° (Fig. 206.6), then φ = 180° - ψ. It is obvious that in both cases the equality cos φ = |cos ψ| is true. According to the formula (the cosine of the angle between non-zero vectors a and b is equal to the scalar product of these vectors divided by the product of their lengths) we have

$$ cos\psi = cos\widehat((a; b)) = \frac(a\cdot b)(|a|\cdot |b|) $$

Consequently,

$$ cos\phi = \frac(|a\cdot b|)(|a|\cdot |b|) $$

Let the lines be given by their canonical equations

$$ \frac(x-x_1)(a_1)=\frac(y-y_1)(a_2)=\frac(z-z_1)(a_3) \;\; And \;\; \frac(x-x_2)(b_1)=\frac(y-y_2)(b_2)=\frac(z-z_2)(b_3) $$

Then the angle φ between the lines is determined using the formula

$$ cos\phi = \frac(|a_(1)b_1+a_(2)b_2+a_(3)b_3|)(\sqrt((a_1)^2+(a_2)^2+(a_3)^2 )\sqrt((b_1)^2+(b_2)^2+(b_3)^2)) (1)$$

If one of the lines (or both) is given by non-canonical equations, then to calculate the angle, you need to find the coordinates of the direction vectors of these lines, and then use formula (1).

Task 1. Calculate angle between lines

$$ \frac(x+3)(-\sqrt2)=\frac(y)(\sqrt2)=\frac(z-7)(-2) \;\;and\;\; \frac(x)(\sqrt3)=\frac(y+1)(\sqrt3)=\frac(z-1)(\sqrt6) $$

Direction vectors of straight lines have coordinates:

a \u003d (-√2; √2; -2), b = (√3 ; √3 ; √6 ).

By formula (1) we find

$$ cos\phi = \frac(|-\sqrt6+\sqrt6-2\sqrt6|)(\sqrt(2+2+4)\sqrt(3+3+6))=\frac(2\sqrt6)( 2\sqrt2\cdot 2\sqrt3)=\frac(1)(2) $$

Therefore, the angle between these lines is 60°.

Task 2. Calculate angle between lines

$$ \begin(cases)3x-12z+7=0\\x+y-3z-1=0\end(cases) and \begin(cases)4x-y+z=0\\y+z+1 =0\end(cases) $$

Behind the guide vector but the first straight line we take the vector product of normal vectors n 1 = (3; 0; -12) and n 2 = (1; 1; -3) planes defining this line. By the formula $=\begin(vmatrix) i & j & k \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end(vmatrix) $ we get

$$ a==\begin(vmatrix) i & j & k \\ 3 & 0 & -12 \\ 1 & 1 & -3 \end(vmatrix)=12i-3i+3k $$

Similarly, we find the direction vector of the second straight line:

$$ b=\begin(vmatrix) i & j & k \\ 4 & -1 & 1 \\ 0 & 1 & 1 \end(vmatrix)=-2i-4i+4k $$

But formula (1) calculates the cosine of the desired angle:

$$ cos\phi = \frac(|12\cdot (-2)-3(-4)+3\cdot 4|)(\sqrt(12^2+3^2+3^2)\sqrt(2 ^2+4^2+4^2))=0 $$

Therefore, the angle between these lines is 90°.

Task 3. In the triangular pyramid MAVS, the edges MA, MB and MC are mutually perpendicular, (Fig. 207);

their lengths are respectively equal to 4, 3, 6. Point D is the middle [MA]. Find the angle φ between lines CA and DB.

Let SA and DB be the direction vectors of the lines SA and DB.

Let's take the point M as the origin of coordinates. By the task condition, we have A (4; 0; 0), B(0; 0; 3), C(0; 6; 0), D (2; 0; 0). Therefore $\overrightarrow(CA)$ = (4; - 6;0), $\overrightarrow(DB)$= (-2; 0; 3). We use formula (1):

$$ cos\phi=\frac(|4\cdot (-2)+(-6)\cdot 0+0\cdot 3|)(\sqrt(16+36+0)\sqrt(4+0+9 )) $$

According to the table of cosines, we find that the angle between the straight lines CA and DB is approximately 72 °.

but. Let two lines be given. These lines, as it was indicated in Chapter 1, form various positive and negative angles, which, in this case, can be both acute and obtuse. Knowing one of these angles, we can easily find any other.

By the way, for all these angles, the numerical value of the tangent is the same, the difference can only be in the sign

Equations of lines. The numbers are the projections of the directing vectors of the first and second lines. The angle between these vectors is equal to one of the angles formed by straight lines. Therefore, the problem is reduced to determining the angle between the vectors, We get

For simplicity, we can agree on an angle between two straight lines to understand an acute positive angle (as, for example, in Fig. 53).

Then the tangent of this angle will always be positive. Thus, if a minus sign is obtained on the right side of formula (1), then we must discard it, i.e., keep only the absolute value.

Example. Determine the angle between lines

By formula (1) we have

from. If it is indicated which of the sides of the angle is its beginning and which is its end, then, counting always the direction of the angle counterclockwise, we can extract something more from formulas (1). As is easy to see from Fig. 53 the sign obtained on the right side of the formula (1) will indicate which one - acute or obtuse - the angle forms the second line with the first.

(Indeed, from Fig. 53 we see that the angle between the first and second direction vectors is either equal to the desired angle between the lines, or differs from it by ±180°.)

d. If the lines are parallel, then their direction vectors are also parallel. Applying the condition of parallelism of two vectors, we get!

This is a necessary and sufficient condition for two lines to be parallel.

Example. Direct

are parallel because

e. If the lines are perpendicular, then their direction vectors are also perpendicular. Applying the condition of perpendicularity of two vectors, we obtain the condition of perpendicularity of two lines, namely

Example. Direct

perpendicular because

In connection with the conditions of parallelism and perpendicularity, we will solve the following two problems.

f. Draw a line parallel to a given line through a point

The decision is made like this. Since the desired line is parallel to the given one, then for its directing vector we can take the same one as that of the given line, i.e., a vector with projections A and B. And then the equation of the desired line will be written in the form (§ 1)

Example. Equation of a straight line passing through a point (1; 3) parallel to a straight line

will be next!

g. Draw a line through a point perpendicular to the given line

Here, it is no longer suitable to take a vector with projections A and as a directing vector, but it is necessary to winnow a vector perpendicular to it. The projections of this vector must therefore be chosen according to the condition that both vectors are perpendicular, i.e., according to the condition

This condition can be fulfilled in an infinite number of ways, since here there is one equation with two unknowns. But the easiest way is to take it. Then the equation of the desired straight line will be written in the form

Example. Equation of a line passing through a point (-7; 2) in a perpendicular line

will be the following (according to the second formula)!

h. In the case when the lines are given by equations of the form

Instruction

note

The period of the trigonometric function tangent is 180 degrees, which means that the angles of inclination of the straight lines cannot, in absolute value, exceed this value.

Useful advice

If the slope coefficients are equal to each other, then the angle between such lines is 0, since such lines either coincide or are parallel.

To determine the angle between intersecting lines, it is necessary to transfer both lines (or one of them) to a new position by the method of parallel transfer to the intersection. After that, you should find the angle between the resulting intersecting lines.

You will need

Ruler, right triangle, pencil, protractor.

Instruction

So, let the vector V = (a, b, c) and the plane A x + B y + C z = 0 be given, where A, B and C are the coordinates of the normal N. Then the cosine of the angle α between the vectors V and N is: cos α \u003d (a A + b B + c C) / (√ (a² + b² + c²) √ (A² + B² + C²)).

To calculate the value of the angle in degrees or radians, you need to calculate the function inverse to the cosine from the resulting expression, i.e. arccosine: α \u003d arscos ((a A + b B + c C) / (√ (a² + b² + c²) √ (A² + B² + C²))).

Example: find injection between vector(5, -3, 8) and plane, given by the general equation 2 x - 5 y + 3 z = 0. Solution: write down the coordinates of the normal vector of the plane N = (2, -5, 3). Substitute all known values into the above formula: cos α = (10 + 15 + 24)/√3724 ≈ 0.8 → α = 36.87°.