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 straight lines be given by their own 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 take the first straight line vector product 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 triangular pyramid MABC edges MA, MB and MS 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°.
With the help of this online calculator you can find the angle between straight lines. Given detailed solution with explanations. To calculate the angle between straight lines, set the dimension (2 if a straight line on a plane is considered, 3 if a straight line in space is considered), enter the elements of the equation into the cells and click on the “Solve” button. See the theoretical part below.
×
Warning
Clear all cells?
Close Clear
Data entry instructions. Numbers are entered as integers (examples: 487, 5, -7623, etc.), decimals (ex. 67., 102.54, etc.) or fractions. The fraction must be entered in the form a/b, where a and b (b>0) are integers or decimals. Examples 45/5, 6.6/76.4, -7/6.7, etc.
1. Angle between straight lines on a plane
Lines are defined by canonical equations
1.1. Determining the angle between straight lines
Let the lines in two-dimensional space L 1 and L
Thus, from formula (1.4) we can find the angle between the lines L 1 and L 2. As can be seen from Fig. 1, intersecting lines form adjacent angles φ And φ 1 . If the angle found is greater than 90°, then you can find the minimum angle between straight lines L 1 and L 2: φ 1 =180-φ .
From formula (1.4) we can derive the conditions for parallelism and perpendicularity of two straight lines.
Example 1. Determine the angle between lines
Let's simplify and solve:
1.2. Condition for parallel lines
Let φ =0. Then cosφ=1. In this case, expression (1.4) will take the following form:
| , |
| , |
Example 2. Determine whether the lines are parallel
Equality (1.9) is satisfied, therefore lines (1.10) and (1.11) are parallel.
Answer. Lines (1.10) and (1.11) are parallel.
1.3. Condition for perpendicularity of lines
Let φ =90°. Then cosφ=0. In this case, expression (1.4) will take the following form:
Example 3. Determine whether the lines are perpendicular
Condition (1.13) is satisfied, therefore lines (1.14) and (1.15) are perpendicular.
Answer. Lines (1.14) and (1.15) are perpendicular.
Lines are defined by general equations
1.4. Determining the angle between straight lines
Let two straight lines L 1 and L 2 are given by general equations
From the definition of the scalar product of two vectors, we have:
Example 4. Find the angle between lines
Substituting values A 1 , B 1 , A 2 , B 2 in (1.23), we get:
This angle is greater than 90°. Let's find the minimum angle between straight lines. To do this, subtract this angle from 180:
On the other hand, the condition of parallel lines L 1 and L 2 is equivalent to the condition of collinearity of vectors n 1 and n 2 and can be represented like this:
Equality (1.24) is satisfied, therefore lines (1.26) and (1.27) are parallel.
Answer. Lines (1.26) and (1.27) are parallel.
1.6. Condition for perpendicularity of lines
Condition for perpendicularity of lines L 1 and L 2 can be extracted from formula (1.20) by substituting cos(φ )=0. Then scalar product (n 1 ,n 2)=0. Where
Equality (1.28) is satisfied, therefore lines (1.29) and (1.30) are perpendicular.
Answer. Lines (1.29) and (1.30) are perpendicular.
2. Angle between straight lines in space
2.1. Determining the angle between straight lines
Let there be straight lines in space L 1 and L 2 are given by canonical equations
where | q 1 | and | q 2 | direction vector modules q 1 and q 2 respectively, φ -angle between vectors q 1 and q 2 .
From expression (2.3) we obtain:
 .
|
Let's simplify and solve:
 .
|
Let's find the angle φ
A. Let two straight lines be given. These straight lines, as indicated in Chapter 1, form various positive and negative angles, which can be either acute or 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 direction vectors of the first and second straight lines. The angle between these vectors is equal to one of the angles formed by straight lines. Therefore, the problem comes down to determining the angle between the vectors. We get
For simplicity, we can agree that the angle between two straight lines is understood as an acute positive angle (as, for example, in Fig. 53).
Then the tangent of this angle will always be positive. Thus, if there is a minus sign on the right side of formula (1), then we must discard it, i.e., save only the absolute value.
Example. Determine the angle between straight lines
According to formula (1) we have
With. If it is indicated which of the sides of the angle is its beginning and which is its end, then, always counting the direction of the angle counterclockwise, we can extract something more from formula (1). As can be easily seen from Fig. 53, the sign obtained on the right side of formula (1) will indicate what kind of angle - acute or obtuse - the second straight line forms 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 straight lines, or differs from it by ±180°.)
d. If the lines are parallel, then their direction vectors are parallel. Applying the condition of parallelism of two vectors, we get!
This is a necessary and sufficient condition for the parallelism of two lines.
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 straight lines, namely
Example. Direct
are perpendicular due to the fact that
In connection with the conditions of parallelism and perpendicularity, we will solve the following two problems.
f. Draw a line through a point parallel to the given line
The solution is carried out like this. Since the desired line is parallel to this one, then for its direction 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 line passing through the point (1; 3) parallel to the line
there will be next!
g. Draw a line through a point perpendicular to the given line
Here, it is no longer suitable to take the vector with projections A and as the directing vector, but it is necessary to take the vector perpendicular to it. The projections of this vector must therefore be chosen according to the condition of perpendicularity of both vectors, i.e. according to the condition
This condition can be fulfilled in countless ways, since here is one equation with two unknowns. But the easiest way is to take or Then the equation of the desired line will be written in the form
Example. Equation of a line passing through the point (-7; 2) in a perpendicular line
there will be the following (according to the second formula)!
h. In the case when the lines are given by equations of the form
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 the ways in which 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 exactly how 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 it. 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 them 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 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 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 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 sort it out 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 given vectors 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 straight lines with an angular coefficient. To find the intersection angle, 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 a plane, given by 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 in 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's 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 axis applicate we can take 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 ° .
If you notice an error in the text, please highlight it and press Ctrl+Enter
Instructions
note
Period trigonometric function The tangent is equal to 180 degrees, which means that the slope angles of straight lines cannot, in absolute value, exceed this value.
If slopes are equal to each other, then the angle between such lines is equal to 0, since such lines either coincide or are parallel.
To determine the value of the angle between intersecting lines, it is necessary to move both lines (or one of them) to a new position using the parallel translation method until they intersect. After this, you should find the angle between the resulting intersecting lines.
You will need
- Ruler, right triangle, pencil, protractor.
Instructions
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 equal to: cos α = (a A + b B + c C)/(√(a² + b² + c²) √(A² + B² + C²)).
To calculate the angle in degrees or radians, you need to calculate the inverse to cosine function from the resulting expression, i.e. arccosine:α = аrsсos ((a A + b B + c C)/(√(a² + b² + c²) √(A² + B² + C²))).
Example: find corner between vector(5, -3, 8) and plane, given 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 everything known values into the given formula: cos α = (10 + 15 + 24)/√3724 ≈ 0.8 → α = 36.87°.
Video on the topic
A straight line that has one common point with a circle is tangent to the circle. Another feature of the tangent is that it is always perpendicular to the radius drawn to the point of contact, that is, the tangent and radius form a straight line corner. If two tangents to a circle AB and AC are drawn from one point A, then they are always equal to each other. Determining the angle between tangents ( corner ABC) is made using the Pythagorean theorem.
Instructions
To determine the angle, you need to know the radius of the circle OB and OS and the distance of the starting point of the tangent from the center of the circle - O. So, angles ABO and ASO are equal, the radius OB is, for example, 10 cm, and the distance to the center of the circle AO is 15 cm. Determine the length of the tangent using formula in accordance with the Pythagorean theorem: AB = Square root from AO2 – OB2 or 152 - 102 = 225 – 100 = 125;