Definition. If two lines are given y = k 1 x + b 1, y = k 2 x + b 2, then acute angle between these straight 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 lines Ax + Bу + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A 1 = λA, B 1 = λB are proportional. If also C 1 = λC, 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 a given line

Definition. A straight line passing through the point M 1 (x 1, y 1) and perpendicular to the straight line y = 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 + Bу + C = 0 is determined as

.

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

(1)

The coordinates x 1 and y 1 can be found by solving the system of equations:

The second equation of the system is the equation of a line passing through a given point M 0 perpendicular to a given 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 = -3; k 2 = 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 = 3/5, k 2 = -5/3, k 1* k 2 = -1, therefore, the lines are perpendicular.

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

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

2 x – 3 y + 3 = 0;

The required height equation has the form: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: from where b = 17. Total: .

Answer: 3 x + 2 y – 34 = 0.

The equation of a line passing through a given point in a given direction. Equation of a line passing through two given points. The angle between two straight lines. The condition of parallelism and perpendicularity of two straight 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 beam center.

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

The angular coefficient 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 straight lines are given by equations with a slope

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 line is subtracted from the slope of the second line.

If the equations of a 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 an angular coefficient, then the necessary and sufficient condition for their parallelism is the equality of their angular coefficients:

k 1 = k 2 . (8)

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

5. Conditions for perpendicularity of two straight lines:

a) In the case when the lines are given by equations (4) with an angular coefficient, a necessary and sufficient condition for their perpendicularity is that their angular coefficients are inverse 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 lines are given in general form (6), then the condition for their perpendicularity (necessary and sufficient) is to satisfy 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 lines passing through the point M, one of which is parallel and the other perpendicular to the given line l.

This material is devoted to such a concept as the angle between two intersecting lines. In the first paragraph we will explain what it is and show it in illustrations. Then we will look at 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 used in practice.

Yandex.RTB R-A-339285-1

In order to understand what the angle formed when two lines intersect is, we need to remember the very definition of angle, perpendicularity and point of intersection.

Definition 1

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

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

Let's say we know that one of the angles is equal to α. In this case, the angle that is vertical with respect 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 angles. 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's move on to formulating 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 chosen from several options.

To begin with, we can take geometric methods. If we know something about supplementary angles, then we can relate them to the angle we need using the properties of equal or similar figures. For example, if we know the sides of a triangle and 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 the condition right triangle, then for calculations we will also need knowledge of sine, cosine and tangent of an angle.

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

We have a rectangular (Cartesian) coordinate system O x y, in which two straight lines are given. Let's denote them by letters a and b. The straight lines can be described using some equations. The original lines have an intersection point M. How to determine the required angle (let's denote it α) between these straight lines?

Let's start by formulating the basic principle of finding an angle under given conditions.

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

The angle subtended 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. Let us assume that we have a line a with a direction vector a → = (a x, a y) and a line b with a direction vector b → (b x, b y). Now let's plot two vectors a → and b → from the intersection point. After this we will see that they will each be located on their own straight line. Then we have four options for their relative arrangement. 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 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. Thus,

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 straight 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 straight 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's give an example of solving the problem.

Example 1

In a rectangular coordinate system on a plane, two intersecting lines a and b are given. They can be described by the 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 our condition, which means that for this line we can immediately write down the coordinates of its direction vector. To do this, we need to take the values ​​of the coefficients for the parameter, i.e. the straight line x = 1 + 4 · λ y = 2 + λ λ ∈ R will have a direction vector a → = (4, 1).

The second 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 move directly to finding the angle. To do this, simply substitute the existing 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 straight 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 n a → = (n a x , n a y) and a line b with a normal vector n b → = (n b x , n b y), then the angle between them will be equal to the angle between n a → and n b → or the angle that will be adjacent to n a →, n b → ^. This method is shown in the picture:

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 α = a r c cos 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

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

Solution

The original lines are specified using normal line equations of the form A x + B y + C = 0. We denote the normal vector as n → = (A, B). Let's find the coordinates of the first normal vector for one line and write them: n a → = (3, 5) . For the second line x + 4 y - 17 = 0, the normal vector will have coordinates n b → = (1, 4). Now let’s 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 = 1 - cos 2 α = 1 - 23 2 34 2 = 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 straight lines if we know the coordinates of the direction vector of one straight line and the normal vector of the other.

Let us assume that straight line a has a direction vector a → = (a x , a y) , and straight line b has a normal vector n b → = (n b x , n b y) . We need to set these vectors aside from the intersection point and consider all options for their relative positions. See in the 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 for a → , n b → ^ > 90 ° .

Thus,

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

Let us formulate a conclusion.

Definition 4

To find the sine of the angle between two lines intersecting on 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 angle 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 guide and normal vector from the given equations. It turns out a → = (- 5, 3) and n → b = (1, 4). We take the formula 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 and calculate:

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

Please note that we took the equations from the previous problem and obtained exactly the same result, but in a different way.

Answer:α = a r c sin 7 2 34

Let us present another way to find the desired angle using the angular coefficients of given straight 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 slopes. To find the angle of intersection, we 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 lines intersecting in a plane, given by the equations y = - 3 5 x + 6 and y = - 1 4 x + 17 4. Calculate the value of the intersection angle.

Solution

The angular coefficients 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 enough to know the coordinates of the guides and/or normal vectors of given lines and be able to determine them by different types equations. But it’s better to remember or write down the formulas for calculating the cosine of an angle.

How to calculate the angle between intersecting lines in space

The calculation of such an angle can be reduced to calculating the coordinates of the direction vectors and determining the magnitude of the angle formed by these vectors. For such examples, the same reasoning that we gave before is used.

Let's assume that we have a rectangular coordinate system located at three-dimensional space. It contains two straight lines a and b with an intersection point M. To calculate the coordinates of the direction vectors, we need to know the equations of these lines. Let us 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 line defined in three-dimensional space using the equation x 1 = y - 3 = z + 3 - 2. It is known that it intersects with the O z axis. Calculate the intercept angle and the cosine of that angle.

Solution

Let us denote the angle that needs 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 found that the angle we need will be equal to a r c cos 1 2 = 45 °.

Answer: cos α = 1 2 , α = 45 ° .

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If on a straight line in space we mark two arbitrary points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), then the coordinates of these points must satisfy the straight line equation obtained above:

In addition, for point M 1 we can write:

.

Solving these equations together, we get:

.

This is the equation of a line passing through two points in space.

General equations of a straight line in space.

The equation of a straight line can be considered as the equation of the line of intersection of two planes.

General equations of a straight line in coordinate form:

The practical task often consists of reducing equations of lines in general form to canonical form.

To do this, you need to find an arbitrary point on the line and the numbers m, n, p.

In this case, the directing vector of the straight line can be found as the vector product of the normal vectors to the given planes.

Example. Find the canonical equation if the line is given in the form:

To find an arbitrary point on a line, we take its coordinate x = 0, and then substitute this value into the given system of equations.

Those. A(0, 2, 1).

Find the components of the directing vector of the straight line.

Then the canonical equations of the line:

Example. Bring to canonical form the equation of a line given in the form:

To find an arbitrary point on a straight line, which is the line of intersection of the above planes, we take z = 0. Then:

;

2x – 9x – 7 = 0;

We get: A(-1; 3; 0).

Direct vector: .

Angle between planes.

The angle between two planes in space  is related to the angle between the normals to these planes  1 by the relation:  =  1 or  = 180 0 -  1, i.e.

cos = cos 1 .

Let's determine the angle  1. It is known that planes can be specified by the relations:

, Where

(A 1, B 1, C 1), (A 2, B 2, C 2). We find the angle between the normal vectors from their scalar product:

.

Thus, the angle between the planes is found by the formula:

The choice of the sign of the cosine depends on which angle between the planes should be found - acute or adjacent to it obtuse.

Conditions for parallelism and perpendicularity of planes.

Based on the formula obtained above for finding the angle between planes, one can find the conditions for parallelism and perpendicularity of planes.

In order for the planes to be perpendicular, it is necessary and sufficient that the cosine of the angle between the planes be equal to zero. This condition is met if:

The planes are parallel, the normal vectors are collinear:  .This condition is satisfied if: .

The angle between straight lines in space.

Let two lines be given in space. Their parametric equations are:

The angle between straight lines  and the angle between direction vectors  of these straight lines are related by the relation:  =  1 or  = 180 0 -  1. The angle between the direction vectors is found from the scalar product. Thus:

.

Conditions for parallelism and perpendicularity of lines in space.

For two lines to be parallel it is necessary and sufficient that the direction vectors of these lines be collinear, i.e. their corresponding coordinates were proportional.

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

Note that point A can be chosen arbitrarily; in particular, it can lie on one of these 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 adjacent angles formed by intersecting lines l 1 And m 1 (l 1 || l, m 1 || m). The angle between parallel lines is considered equal to zero.

Angle between straight 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 .

Task. Given a cube ABCDA 1 B 1 C 1 D 1 (Fig. 139).

Find the angle between straight lines AB and DC 1.

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

Therefore, \(\widehat((AB;DC_1))\) = 45°.

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

Calculation of the angle between straight lines.

The problem of calculating the angle between two straight lines in space is solved in the same way as in a plane. Let us denote by φ the magnitude of the angle between the lines l 1 And l 2, and through ψ - the magnitude of the angle between the direction vectors A 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. Obviously, 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

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

hence,

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

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 the 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 = (-√2 ; √2 ; -2), b = (√3 ; √3 ; √6 ).

Using 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 the 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 A On the first line we take the vector product of normal vectors n 1 = (3; 0; -12) and n 2 = (1; 1; -3) planes defining this line. Using 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 using formula (1) we calculate 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 MABC, the edges MA, MB and MC are mutually perpendicular (Fig. 207);

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

Let CA and DB be the direction vectors of straight lines CA and DB.

Let's take point M as the origin of coordinates. By the condition of the equation 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). Let's use formula (1):

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

Using the cosine table, we find that the angle between straight lines CA and DB is approximately 72°.