The equation of two straight lines is known to find the equation of the plane parallel to it

Two straight-line equations are known.

Find the plane equation parallel to it.

Solution: Let the equation for the line l be (x-x) a =(y-y) b =(z-z) c, l crossing point (x, y, z), direction vector n =;

Then the vector n=n n perpendicular to l and l can be used as the normal vector of the plane sought, i.e. ij

kn=n₁×n₂=∣a₁

b₁c₁∣=(b₁c₂-c₁b₂)i-(a₁c₂-c₁a₂)j+(a₁b₂-b₁a₂)ka₂b₂

c Let the plane crossing point (xo, yo, zo) be found, then the plane equation is:

b₁c₂-c₁b₂)(x-xo)-(a₁c₂-c₁a₂)(y-yo)+(a₁b₂-b₁a₂)(z-zo)=0

1. The general form of the equation of the plane is ax+by+cz+d=0, and its normal vector is (a, b, c).

Then find the vectors of the two known linear equations.

Then they are perpendicular to (a, b, c) respectively, multiplied by 0

Here we get 2 equations, because the straight line belongs to the plane, and the points on the straight line also belong to the plane, so we find the two points from these two straight lines respectively and substitute them into the plane equation, and we also get the 2 equations.

From these 4 equations, ABCD can be found.

2: The same algorithm for parallelism. Because the equations of the two parallel lines are different, only the coefficients are proportional.

(x+14)/13=(y-7)/(-8)=(z-1)/(-1)。

The concatenation of two equations is an expression of a straight line.

To get out of the point-oriented equation, you can first use the normal products of two planes to obtain the direction vector of the straight line, take a random z in the system of simultaneous equations, and solve the corresponding x, y to get a point on the straight line.

Such as two planes:

x+2y-3z+3=0。

2x+3y+2z+5=0。

The direction vector of a straight line is (1,2,-3) (2,3,2)=(13,-8,-1).

Let z=1 give x=-14,y=7, i.e., one point on a straight line is (-14,7,1).

So the point-oriented linear equation is:

x+14)/13=(y-7)/(-8)=(z-1)/(-1)。

If d is not equal to 0, take a=-d a, b=-d b, c=-d b, c=-d c, then the intercept equation for the plane is obtained: x a+y b+z c=1.

Its intersection points with the three coordinate axes are p(a,0,0), q(0,b,0), r(0,0,c), where a, b, c are called the intercepts of the plane on the x, y, and z axes, respectively.

Use two plane equations to eliminate one independent variable to obtain a binary equation, assign any value (0, 1, etc., easy to calculate) to one of the variables, solve the other two variables, and obtain the coordinates of a point on the line. Then calculate the normal vectors of the two planes and do the cross product, that is, the direction vector of the straight line. With the coordinates and direction vectors of a point on a line, we can substitute the equations for intersecting lines in the point direction.

If the equations for two straight lines are known, such as (x-x1) u1=(y-y1) u2=(z-z1) u3 and (x-x2) v1=(y-y2) v2=(z-z2) v3, then their square code slow-down vectors.

a=(u1,u2,u3) and b=(v1,v2,v3) respectively, therefore, the normal vector of the plane is n=a b=(u2v3-u3v2 ,-u1v3-u3v1) ,u1v2-u2v1)=(a,b,c), and then the plane equation can be obtained by combining the plane with the fixed point (x1,y1,z1) on the straight line.

is a(x-x1)+b(y-y1)+c(z-z1)=0

This can be done in two ways:

1. Find three different points on two straight lines (two on one and one on the other), and use the three-point equation formula to find the equation.

2. If the linear equation is given by the "point formula" (i.e., "symmetry"), then the given conditions have "two points + one direction", which can be substituted into the "general" equation of the plane to obtain three equations and solve the plane equation.

3. The general form of the equation of the plane is ax+by+cz+d=0, and its normal vector is (a, b, c), and then find the vectors of the two known linear equations, and then multiply them by (a, b, c) respectively, equal to 0

Here we get 2 equations.

4. Because the straight line belongs to the plane, and the points on the straight line also belong to the plane, two points are found from these two straight lines, and the plane equation is substituted to obtain two equations, and the ABCD can be found through these four equations.

Extended Materials. 1. "Plane equation" refers to the equation corresponding to all points in the same plane in space, and its general formula is like ax+by+cz+d=0.

2. In the spatial coordinate system, the equations of the plane can be expressed by the ternary linear equation ax+by+cz+d=0.

The basic idea: find the normal vector of this plane.

Let one line be a and the other be b

Steps: The first step is to take a point A on the straight line A and a point B on B to get the vector AB The second step is to obtain the direction vector of a through the equation of the line A (or B), and the third step of the vector t is to calculate the cross product of the vector t and the vector AB to obtain the plane normal vector N.

The point French expression of the plane is obtained from the coordinates of point A (or point B) and the normal vector n.

Point French example:

Set a coordinates. x0, y0, z0), normal vectors.

n=(r,s,t), then the point French is .

r(x-x0)+s(y-y0)+t(z-z0)=0

Two straight-line equations are known.

Find the plane equation parallel to it.

Solution: Let the equation for the line l be (x-x) a =(y-y) b =(z-z) c, l crossing point (x, y, z), direction vector n =;

Then the vector n=n n perpendicular to l and l can be used as the normal vector of the plane sought, i.e., ij

k∣n=n₁×n₂=∣a₁

b₁c₁∣=(b₁c₂-c₁b₂)i-(a₁c₂-c₁a₂)j+(a₁b₂-b₁a₂)k

a b c Let the plane crossing point (xo, yo, zo) be found, then the plane equation is:

b₁c₂-c₁b₂)(x-xo)-(a₁c₂-c₁a₂)(y-yo)+(a₁b₂-b₁a₂)(z-zo)=0

There are several ways to do this.

1) Find three different points on two straight lines (two on one and one on the other), and use the three-point formula.

Equation to find the equation;

2. If the linear equation is given by the "point formula" (i.e., "symmetry"), then the given conditions have "two points + one direction", which can be substituted into the "general" equation of the plane to obtain three equations and solve the plane equation.

Knowing the equation of a two-plane straight line, the plane equation that requires the determination of the two straight lines can be found by mathematical formulas.

Take two careful counts on the plane a(-1,2,-3) b(-2,1,2) to get a vector in this plane ab=(1,1,-5) combined with another source slide vector in the plane s=(2,-1,3) to obtain the normal vector n = sxab = (2,13,3) the plane is obtained: 2 (x-1) + 13 (y-2) + 3 (z+3) = 0 whole hail wax to obtain.

2x+13y+3z=19

What about the topic? For example.

l1；x-3\/2=y/1=z-1/2 l2:x+1/2=y-1/1=z/2

a(3,0,1) and b(-1,1,0 ) are the points on l1,l2, respectively, and the directional derivatives of the two lines l= (2,1,2).

Let and rotten p(x,y,z) be a point on the plane sought, then the vectors pa,ab,l are coplanar, i.e.

pa x ab)*l ＝ 0

This is a determinant, from which the equation is obtained.

Varies depending on the question.

Summary. If the slope of the equation for the perpendicular lines is equal, i.e., the two lines are parallel, the two planes are also parallel.

In this case, you can find the equation of the perpendicular line of two planes.

If the slope of the equation for the perpendicular lines is equal, i.e., the two lines are parallel, the two planes are also parallel.

If one of the perpendicular lines is perpendicular to the straight line given in the title, then it means that all three are parallel.

Second question. Is it still there?

Writing process is underway.

Have you learned how to find the vertical line of a face?

No. What about normals?

Learned. The normal vectors of the two planes can be seen directly through the two equations, right?

Right. You take the equation of the straight line and put it in the format of the first question.

Their denominator is the direction vector of the straight line.

This is the characteristic of parametric equations.

For the next line, you can set a vector (x,y,z) and assume that one of the unknowns is 1

Find the other two unknowns.

Get this vector.

Then use this vector and the direction vector of the previous line to prove that it is perpendicular or parallel.

I assume x=1 and get this vector as (4,1,-2) multiplying not 0 so not perpendicular.

Students, have you learned how to find the direction vector of a plane intersection line?

These three conditions are calculated using the equation of the volcanic trace.

1) The plane sought passes through a point on the straight or only line;

2) The product of the "normal vector" of the plane and the "direction vector" of the straight line is zero;

3) The point product of the plane "Normal Vector" and the parallel line "Direction Vector" is zero.

Symmetry is determined by a point on a straight line and the direction vector of the straight line.

1) Find an intersection point first, and solve x and y by taking the random value of z

Let z=0

by x+2y=7

2x+y=7

The solution is x=-7 5, y=21 and Zhaoqing 5

So (-7 5,21 5,0) is a point in a straight line.

2) Find the direction vector.

Because the normal vectors of the two known planes are the direction vectors of the straight line before the balance (1,2,-1),(2,1,1) perpendicular to the two normals.

It can be found from the outer product.

Direction vector = (1,2,-1) (2,1,1)i j k3i + j + 5k

So the straight direction vector guess key is (3,1,5).

Therefore, the linear symmetry is (x+7 5) 3=(y-21 5) 1=z 5