Find the cosine of the angle between two lines online. Angle between two lines. How to find the distance between two parallel lines

Task 1

Find the cosine of the angle between the lines $\frac(x+3)(5) =\frac(y-2)(-3) =\frac(z-1)(4) $ and $\left\(\begin(array )(c) (x=2\cdot t-3) \\ (y=-t+1) \\ (z=3\cdot t+5) \end(array)\right.$.

Let two lines be given in space: $\frac(x-x_(1) )(m_(1) ) =\frac(y-y_(1) )(n_(1) ) =\frac(z-z_(1 ) )(p_(1) ) $ and $\frac(x-x_(2) )(m_(2) ) =\frac(y-y_(2) )(n_(2) ) =\frac(z- z_(2) )(p_(2) ) $. We choose an arbitrary point in space and draw two auxiliary lines through it, parallel to the data. The angle between the given lines is any of the two adjacent angles formed by the auxiliary lines. The cosine of one of the angles between the lines can be found using the well-known formula $\cos \phi =\frac(m_(1) \cdot m_(2) +n_(1) \cdot n_(2) +p_(1) \cdot p_( 2) )(\sqrt(m_(1)^(2) +n_(1)^(2) +p_(1)^(2) ) \cdot \sqrt(m_(2)^(2) +n_( 2)^(2) +p_(2)^(2) ) ) $. If the value $\cos \phi >0$, then an acute angle between the lines is obtained, if $\cos \phi

Canonical equations of the first line: $\frac(x+3)(5) =\frac(y-2)(-3) =\frac(z-1)(4) $.

The canonical equations of the second straight line can be obtained from the parametric ones:

\ \ \

Thus, the canonical equations of this line are: $\frac(x+3)(2) =\frac(y-1)(-1) =\frac(z-5)(3) $.

We calculate:

\[\cos \phi =\frac(5\cdot 2+\left(-3\right)\cdot \left(-1\right)+4\cdot 3)(\sqrt(5^(2) +\ left(-3\right)^(2) +4^(2) ) \cdot \sqrt(2^(2) +\left(-1\right)^(2) +3^(2) ) ) = \frac(25)(\sqrt(50) \cdot \sqrt(14) ) \approx 0.9449.\]

Task 2

The first line passes through the given points $A\left(2,-4,-1\right)$ and $B\left(-3,5,6\right)$, the second line passes through the given points $C\left (1,-2,8\right)$ and $D\left(6,7,-2\right)$. Find the distance between these lines.

Let some line be perpendicular to lines $AB$ and $CD$ and intersect them at points $M$ and $N$, respectively. Under these conditions, the length of the segment $MN$ is equal to the distance between the lines $AB$ and $CD$.

We build the vector $\overline(AB)$:

\[\overline(AB)=\left(-3-2\right)\cdot \bar(i)+\left(5-\left(-4\right)\right)\cdot \bar(j)+ \left(6-\left(-1\right)\right)\cdot \bar(k)=-5\cdot \bar(i)+9\cdot \bar(j)+7\cdot \bar(k ).\]

Let the segment representing the distance between the lines pass through the point $M\left(x_(M) ,y_(M) ,z_(M) \right)$ on the line $AB$.

We build the vector $\overline(AM)$:

\[\overline(AM)=\left(x_(M) -2\right)\cdot \bar(i)+\left(y_(M) -\left(-4\right)\right)\cdot \ bar(j)+\left(z_(M) -\left(-1\right)\right)\cdot \bar(k)=\] \[=\left(x_(M) -2\right)\ cdot \bar(i)+\left(y_(M) +4\right)\cdot \bar(j)+\left(z_(M) +1\right)\cdot \bar(k).\]

The vectors $\overline(AB)$ and $\overline(AM)$ are the same, hence they are collinear.

It is known that if the vectors $\overline(a)=x_(1) \cdot \overline(i)+y_(1) \cdot \overline(j)+z_(1) \cdot \overline(k)$ and $ \overline(b)=x_(2) \cdot \overline(i)+y_(2) \cdot \overline(j)+z_(2) \cdot \overline(k)$ are collinear, then their coordinates are proportional, then is $\frac(x_((\it 2)) )((\it x)_((\it 1)) ) =\frac(y_((\it 2)) )((\it y)_( (\it 1)) ) =\frac(z_((\it 2)) )((\it z)_((\it 1)) ) $.

$\frac(x_(M) -2)(-5) =\frac(y_(M) +4)(9) =\frac(z_(M) +1)(7) =m$, where $m $ is the result of the division.

From here we get: $x_(M) -2=-5\cdot m$; $y_(M) +4=9\cdot m$; $z_(M) +1=7\cdot m$.

Finally, we obtain expressions for the coordinates of the point $M$:

We build the $\overline(CD)$ vector:

\[\overline(CD)=\left(6-1\right)\cdot \bar(i)+\left(7-\left(-2\right)\right)\cdot \bar(j)+\ left(-2-8\right)\cdot \bar(k)=5\cdot \bar(i)+9\cdot \bar(j)-10\cdot \bar(k).\]

Let the segment representing the distance between the lines pass through the point $N\left(x_(N) ,y_(N) ,z_(N) \right)$ on the line $CD$.

We construct the vector $\overline(CN)$:

\[\overline(CN)=\left(x_(N) -1\right)\cdot \bar(i)+\left(y_(N) -\left(-2\right)\right)\cdot \ bar(j)+\left(z_(N) -8\right)\cdot \bar(k)=\] \[=\left(x_(N) -1\right)\cdot \bar(i)+ \left(y_(N) +2\right)\cdot \bar(j)+\left(z_(N) -8\right)\cdot \bar(k).\]

The vectors $\overline(CD)$ and $\overline(CN)$ are the same, hence they are collinear. We apply the condition of collinear vectors:

$\frac(x_(N) -1)(5) =\frac(y_(N) +2)(9) =\frac(z_(N) -8)(-10) =n$ where $n $ is the result of the division.

From here we get: $x_(N) -1=5\cdot n$; $y_(N) +2=9\cdot n$; $z_(N) -8=-10\cdot n$.

Finally, we obtain expressions for the coordinates of point $N$:

We build the $\overline(MN)$ vector:

\[\overline(MN)=\left(x_(N) -x_(M) \right)\cdot \bar(i)+\left(y_(N) -y_(M) \right)\cdot \bar (j)+\left(z_(N) -z_(M) \right)\cdot \bar(k).\]

We substitute the expressions for the coordinates of the points $M$ and $N$:

\[\overline(MN)=\left(1+5\cdot n-\left(2-5\cdot m\right)\right)\cdot \bar(i)+\] \[+\left(- 2+9\cdot n-\left(-4+9\cdot m\right)\right)\cdot \bar(j)+\left(8-10\cdot n-\left(-1+7\cdot m\right)\right)\cdot \bar(k).\]

After completing the steps, we get:

\[\overline(MN)=\left(-1+5\cdot n+5\cdot m\right)\cdot \bar(i)+\left(2+9\cdot n-9\cdot m\right )\cdot \bar(j)+\left(9-10\cdot n-7\cdot m\right)\cdot \bar(k).\]

Since the lines $AB$ and $MN$ are perpendicular, the scalar product of the corresponding vectors is equal to zero, i.e. $\overline(AB)\cdot \overline(MN)=0$:

\[-5\cdot \left(-1+5\cdot n+5\cdot m\right)+9\cdot \left(2+9\cdot n-9\cdot m\right)+7\cdot \ left(9-10\cdot n-7\cdot m\right)=0;\] \

After completing the steps, we get the first equation for determining $m$ and $n$: $155\cdot m+14\cdot n=86$.

Since the lines $CD$ and $MN$ are perpendicular, the scalar product of the corresponding vectors is equal to zero, i.e. $\overline(CD)\cdot \overline(MN)=0$:

\ \[-5+25\cdot n+25\cdot m+18+81\cdot n-81\cdot m-90+100\cdot n+70\cdot m=0.\]

After completing the steps, we obtain the second equation for determining $m$ and $n$: $14\cdot m+206\cdot n=77$.

Find $m$ and $n$ by solving the system of equations $\left\(\begin(array)(c) (155\cdot m+14\cdot n=86) \\ (14\cdot m+206\cdot n =77) \end(array)\right.$.

We apply the Cramer method:

Find the coordinates of points $M$ and $N$:

\ \

Finally:

Finally, we write the vector $\overline(MN)$:

$\overline(MN)=\left(2.691-\left(-0.6215\right)\right)\cdot \bar(i)+\left(1.0438-0.7187\right)\cdot \bar (j)+\left(4,618-2,6701\right)\cdot \bar(k)$ or $\overline(MN)=3,3125\cdot \bar(i)+0,3251\cdot \bar( j)+1.9479\cdot\bar(k)$.

The distance between lines $AB$ and $CD$ is the length of the vector $\overline(MN)$:$d=\sqrt(3.3125^(2) +0.3251^(2) +1.9479^( 2) ) \approx 3.8565$ lin. units

Definition

A geometric figure consisting of all points of a plane enclosed between two rays emanating from one point is called flat corner.

Definition

Angle between two intersecting direct called the value of the smallest plane angle at the intersection of these lines. If two lines are parallel, then the angle between them is assumed to be zero.

The angle between two intersecting lines (if measured in radians) can take values from zero to $\dfrac(\pi)(2)$.

Definition

Angle between two intersecting lines is called the value equal to the angle between two intersecting straight lines parallel to the skew ones. The angle between lines $a$ and $b$ is denoted by $\angle (a, b)$.

The correctness of the introduced definition follows from the following theorem.

Plane angle theorem with parallel sides

The values of two convex plane angles with correspondingly parallel and equally directed sides are equal.

Proof

If the angles are straight, then they are both equal to $\pi$. If they are not developed, then we plot equal segments $ON=O_1ON_1$ and $OM=O_1M_1$ on the corresponding sides of the angles $\angle AOB$ and $\angle A_1O_1B_1$.

The quadrilateral $O_1N_1NO$ is a parallelogram because its opposite sides $ON$ and $O_1N_1$ are equal and parallel. Similarly, the quadrilateral $O_1M_1MO$ is a parallelogram. Hence $NN_1 = OO_1 = MM_1$ and $NN_1 \parallel OO_1 \parallel MM_1$, hence $NN_1=MM_1$ and $NN_1 \parallel MM_1$ by transitivity. The quadrilateral $N_1M_1MN$ is a parallelogram because its opposite sides are equal and parallel. Hence, the segments $NM$ and $N_1M_1$ are also equal. Triangles $ONM$ and $O_1N_1M_1$ are equal according to the third triangle equality criterion, hence the corresponding angles $\angle NOM$ and $\angle N_1O_1M_1$ are also equal.

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 form

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.

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 directing 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 win 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

rewriting these equations differently, we have