Find the angle between direct systems of equations. The simplest problems with a straight line on a plane. Mutual arrangement of lines. Angle between lines. Distance from point to line

Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2 , then sharp corner 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 this 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 their parallelism consists in 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), 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.

With the help of this online calculator find the angle between the lines. A detailed solution with explanations is given. To calculate the angle between the lines, set the dimension (2-if a line is considered on a plane, 3- if a line is considered in space), enter the elements of the equation in the cells and click on the "Solve" button. See the theoretical part below.

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Data entry instruction. Numbers are entered as whole numbers (examples: 487, 5, -7623 etc.), decimal numbers (eg 67., 102.54 etc.) or fractions. The fraction must be typed in the form a/b, where a and b (b>0) are integer or decimal numbers. Examples 45/5, 6.6/76.4, -7/6.7, etc.

1. Angle between lines on a plane

The lines are given by the canonical equations

1.1. Determining the angle between lines

Let the lines in two-dimensional space L 1 and L

Thus, from formula (1.4) one 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 one . If the found angle is greater than 90°, then you can find the minimum angle between the lines L 1 and L 2: φ 1 =180-φ .

From formula (1.4) one can deduce the conditions of parallelism and perpendicularity of two straight lines.

Example 1. Determine the angle between lines

Let's simplify and solve:

1.2. Condition of parallel lines

Let be φ =0. Then cosφ=1. In this case, expression (1.4) will take the following form:

Example 2. Determine if lines are parallel

Equality (1.9) is satisfied, hence the lines (1.10) and (1.11) are parallel.

Answer. The lines (1.10) and (1.11) are parallel.

1.3. The condition of perpendicularity of lines

Let be φ =90°. Then cosφ=0. In this case, expression (1.4) will take the following form:

Example 3. Determine if lines are perpendicular

Condition (1.13) is satisfied, hence the lines (1.14) and (1.15) are perpendicular.

Answer. The lines (1.14) and (1.15) are perpendicular.

The straight lines are given by the general equations

1.4. Determining the angle between lines

Let two 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°. Find the minimum angle between the 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 collinear vectors n 1 and n 2 and can be represented as follows:

Equality (1.24) is satisfied, hence the lines (1.26) and (1.27) are parallel.

Answer. The lines (1.26) and (1.27) are parallel.

1.6. The condition of perpendicularity of lines

The condition of perpendicularity of lines L 1 and L 2 can be extracted from formula (1.20) by substituting cos(φ )=0. Then the scalar product ( n 1 ,n 2)=0. Where

Equality (1.28) is satisfied, hence the lines (1.29) and (1.30) are perpendicular.

Answer. The lines (1.29) and (1.30) are perpendicular.

2. Angle between lines in space

2.1. Determining the angle between lines

Let the lines in space L 1 and L 2 are given by the 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 get:

Let's simplify and solve:

Let's find the corner φ

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 goes through given points$A\left(2,-4,-1\right)$ and $B\left(-3,5,6\right)$, the second line is 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 $\overline(AM)$ vector:

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

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

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 °.

ANGLE BETWEEN PLANES

Let's consider two planes α 1 and α 2 given respectively by the equations:

Under angle between two planes we mean one of the dihedral angles formed by these planes. Obviously, the angle between the normal vectors and the planes α 1 and α 2 is equal to one of the indicated adjacent dihedral angles or . That's why . Because And , then

Example. Determine the angle between planes x+2y-3z+4=0 and 2 x+3y+z+8=0.

Condition of parallelism of two planes.

Two planes α 1 and α 2 are parallel if and only if their normal vectors and are parallel, and hence .

So, two planes are parallel to each other if and only if the coefficients at the corresponding coordinates are proportional:

Condition of perpendicularity of planes.

It is clear that two planes are perpendicular if and only if their normal vectors are perpendicular, and hence, or .

In this way, .

Examples.

DIRECT IN SPACE.

VECTOR EQUATION DIRECT.

PARAMETRIC EQUATIONS DIRECT

The position of a straight line in space is completely determined by specifying any of its fixed points M 1 and a vector parallel to this line.

A vector parallel to a straight line is called guiding the vector of this line.

So let the straight l passes through a point M 1 (x 1 , y 1 , z 1) lying on a straight line parallel to the vector .

Consider an arbitrary point M(x,y,z) on a straight line. It can be seen from the figure that .

The vectors and are collinear, so there is such a number t, what , where is the multiplier t can accept any numerical value depending on the position of the point M on a straight line. Factor t is called a parameter. Denoting the radius vectors of points M 1 and M respectively, through and , we obtain . This equation is called vector straight line equation. It shows that each parameter value t corresponds to the radius vector of some point M lying on a straight line.

We write this equation in coordinate form. Notice, that , and from here

The resulting equations are called parametric straight line equations.

When changing the parameter t coordinates change x, y And z and dot M moves in a straight line.

CANONICAL EQUATIONS DIRECT

Let be M 1 (x 1 , y 1 , z 1) - a point lying on a straight line l, And is its direction vector. Again, take an arbitrary point on a straight line M(x,y,z) and consider the vector .

It is clear that the vectors and are collinear, so their respective coordinates must be proportional, hence

– canonical straight line equations.

Remark 1. Note that the canonical equations of the line could be obtained from the parametric equations by eliminating the parameter t. Indeed, from the parametric equations we obtain or .

Example. Write the equation of a straight line in a parametric way.

Denote , hence x = 2 + 3t, y = –1 + 2t, z = 1 –t.

Remark 2. Let the line be perpendicular to one of the coordinate axes, for example, the axis Ox. Then the direction vector of the line is perpendicular Ox, Consequently, m=0. Consequently, the parametric equations of the straight line take the form

Eliminating the parameter from the equations t, we obtain the equations of the straight line in the form

However, in this case, too, we agree to formally write the canonical equations of the straight line in the form . Thus, if the denominator of one of the fractions is zero, then this means that the line is perpendicular to the corresponding coordinate axis.

Similarly, the canonical equations corresponds to a straight line perpendicular to the axes Ox And Oy or parallel axis Oz.

Examples.

GENERAL EQUATIONS A DIRECT LINE AS A LINE OF INTERCEPTION OF TWO PLANES

Through each straight line in space passes an infinite number of planes. Any two of them, intersecting, define it in space. Therefore, the equations of any two such planes, considered together, are the equations of this line.

In general, any two non-parallel planes given by the general equations

determine their line of intersection. These equations are called general equations straight.

Examples.

Construct a straight line given by equations

To construct a line, it is enough to find any two of its points. The easiest way is to choose the points of intersection of the line with coordinate planes. For example, the point of intersection with the plane xOy we obtain from the equations of a straight line, assuming z= 0:

Solving this system, we find the point M 1 (1;2;0).

Similarly, assuming y= 0, we get the point of intersection of the line with the plane xOz:

From the general equations of a straight line, one can proceed to its canonical or parametric equations. To do this, you need to find some point M 1 on the line and the direction vector of the line.

Point coordinates M 1 we obtain from this system of equations, giving one of the coordinates an arbitrary value. To find the direction vector, note that this vector must be perpendicular to both normal vectors And . Therefore, for the direction vector of the straight line l you can take the cross product of normal vectors:

Example. Give the general equations of the straight line to the canonical form.

Find a point on a straight line. To do this, we choose arbitrarily one of the coordinates, for example, y= 0 and solve the system of equations:

The normal vectors of the planes defining the line have coordinates Therefore, the direction vector will be straight

. Consequently, l: .

ANGLE BETWEEN RIGHTS

corner between straight lines in space we will call any of the adjacent angles formed by two straight lines drawn through an arbitrary point parallel to the data.

Let two straight lines be given in space:

Obviously, the angle φ between the lines can be taken as the angle between their direction vectors and . Since , then according to the formula for the cosine of the angle between the vectors we get