Excuse me, in three dimensional space, you already know two equations, how to find their intersectio

This is the equation for two planes, and their intersection point is a straight line. In three-dimensional space, the equation for a straight line is.

x-x0) u1=(y-y0) u2=(z-z0) u3, where (x0,y0,z0) is any point on the line and u1,u2,u3) is the direction vector of the line (i.e., the vector parallel to the line).

For this problem, you can first find a special point, such as taking x0=1 4 , y0=0, z0=5 6 , then equation (1) can be reduced to 4(x-1 4)=8(y-0) , and 2) formula can be reduced to 8(y-0)=-6(z-5 6), therefore, the intersection equation is 4(x-1 4)=8(y-0)=-6(z-5 6), and it can also be written as (x-1 4) 6=(y-0) 3=(z-5 6) (-4).

The common solution of two equations is their intersection.

There are several intersections with several solutions, and there are no intersections without solutions.

Simultaneous equations can be obtained.

4x=1-8y=6z-4

Written in the form of a straight line equation, yes.

x (1 4) = (y-1 8) (-8) = (z-2 3) 6 is a straight line with a directional vector of (1 4, -8, 6) at the crossing point (0, 1 8, 2 3).

Also, the owner of the building should pay attention to the fact that the plane intersection in the space is a straight line rather than a point I hope it will help the landlord, hope!

This is a problem of an infinite number of solutions, considering the relationship between the rank and the unknowns of the coefficient matrix of the system of equations, to understand it simply, if the coefficients of two equations are proportional, then two unknowns are used to represent the third, and if the column is not proportional, then one unmass is used to represent the other two.

It is to solve a system of ternary equations.

We know the plane equations.

can be expressed as:

ax+b y+cz+d=0

For example, know Lachun.

The three planes are.

3x+2y-z-4=0

x+y+z-6=0

2x+y=z-7=0

Simultaneous equations are solved.

x=1,y=2,z=3

That is. 1,2,3) is the intersection of the three flat wheel ridges.

The coordinates of the three points are known to be p1(x1,y1,z1), p2(x2,y2,z2), p3(x3,y3,z3).

So you can set the equation a(x - x1) + b(y - y1) + c(z - z1) = 0 (point franc) (you can also set it to pass the other two points).

Core **: Before that, write the ** of the three 3D points, and then deal with the pending coefficients, as follows:

a = (y3 - y1)*(z3 - z1) -z2 -z1)*(y3 - y1);

b = (x3 - x1)*(z2 - z1) -x2 - x1)*(z3 - z1);

c = (x2 - x1)*(y3 - y1) -x3 - x1)*(y2 - y1);

That is, the plane equations of p1, p2, and p3 are obtained.

The equation can also be written as ax + by + cz + d = 0 (general formula) where d = -(a * x1 + b * y1 + c * z1).

The method is calculated according to the mathematical vector cross product, if you are interested, you can look up some APIs or classes on the Internet, I hope to help you solve the problem!

(1) Let any point o(x,y,z), vector ao (x,y 2,z) plane normal vector n, vector ab (2, 2,0), vector ac (0,0,2), so vector n vector ab vector ac (0, 4, 4), there is because, vector ao · vector n 0, you can get y z 2 0 (2) It is simpler to use a vector mixed product, I don't know if you have learned, let the plane arbitrary point p(x,y,z), vector ap (x,y 2,z), vector bp (x 2,y,z), Vector cp (x,y-2,z-2)[vector ap vector bp vector cp] 0, you can also get the relation of x y z, i.e. the equation y z 2 0! (I've been playing on my phone for a long time, I hope to adopt).

Solution: Let the equation for the plane be .

ax+by+cz+d=0

Substituting a, b, and c into the equation respectively, there is.

2b+d=0

2a+d=0

2b+2c+d=0

Easy to solve. a=b=-d/2

c=0 is substituted into the plane equation and sorted out.

x+y-2=0

That is, what is sought.

The coordinates of the three points are known to be p1(x1,y1,z1), p2(x2,y2,z2), p3(x3,y3,z3).

So you can set the equation a(x - x1) + b(y - y1) + c(z - z1) = 0 (point franc) (you can also set it to pass the other two points).

Core **: Before that, write the ** of the three 3D points, and then deal with the pending coefficients, as follows:

a = (y3 - y1)*(z3 - z1) -z2 -z1)*(y3 - y1);

b = (x3 - x1)*(z2 - z1) -x2 - x1)*(z3 - z1);

c = (x2 - x1)*(y3 - y1) -x3 - x1)*(y2 - y1);

That is, the plane equation of P1, P2, P3 can also be written as: ax + by + cz + d = 0 (general formula) where d = -(a * x1 + b * y1 + c * z1).

C++ is the inheritance of the C language, which can not only carry out the process programming of the C language, but also carry out the object-based programming characterized by abstract data types, and can also carry out the object-oriented programming characterized by inheritance and polymorphism.

C++ excels at object-oriented programming as well as process-based programming, so C++ is based on the size of the problem it can adapt to.

Three planes intersect in pairs to obtain three straight lines, and it is verified that these three straight lines intersect at the same point or burn cherry blossoms or two pairs are parallel.

Known: Plane Plane = a, Plane Flat Segment Noisy Plane = B, Plane Plane = c

Verification: a, b, and c intersect at one point in the same cluster, or a b c

Proof: a, b

a, b a, b intersect or a b

1) When a and b intersect, you may wish to let a b = p, that is, p a, p b and a, b, a

p , p , so p is the common point of and .

and c is known by axiom 2 p c

a, b, and c all pass through the point p, that is, a, b, and c are common points.

2) When a b.

c and a , a

a c and a b

a b c so a, b, c are parallel to each other.

From this, it can be seen that a, b, and c intersect at one point or two pairs are parallel.

Note: This conclusion is often used as a theorem and is often used in judging problems.

Three planes intersect in pairs to obtain three straight lines, and it is verified that these three straight lines intersect at the same point or are parallel to each other.

Known: Plane Plane = a, Plane Plane = b, Plane Flat Bush Plane = c

Verification: a, b, c intersect at the same point, or a b c

Proof: a, b

a, b a, b intersect or a b

1) When a and b intersect, you may wish to let a b = p, that is, p a, p b and a, b, a

p , p , so p is the common point of and .

and c is known by axiom 2 p c

a, b, and c all pass through the point p, that is, a, b, and c are common points.

2) When a b.

c and a , a

a c and a b

a b c so a, b, c are parallel to each other.

From this, it can be seen that a, b, and c intersect at one point or two pairs are parallel.

Note: This conclusion is often used as a theorem, and is often used by or Burning Sakura in judgment section noise problems.