Who has more points in a straight line and in space? Prove it. 20

This involves basic limit thinking. The answer is just as much.

The key to solving this kind of problem is to establish a one-to-one correspondence between the points in the space and the points in the line. If this one-to-one correspondence can be established, then there are as many points in space as there are in a line. On the contrary, it is not the same.

Because all the points in the line are the same, and all the points in the space are the same. So we only need to prove that a line of length one and a cube with one side have the same number of points to prove the whole proposition.

The specific methods of establishing correspondence are as follows:1Establish an origin on a straight line and a Cartesian coordinate system in space.

Any point on a straight line can be represented by a number x. Any point in space can be represented by a set of numbers (a, b, c).

2.For any point in space, such as (,, we take a nested approach to convert it into a point on a straight line (. In the same way, for any point in a straight line, we take the opposite approach to transform it into a point in space.

For example, (, i.e., (, so that any point in the line and any point in space can find their corresponding points, and this transformation has one and only one result. This is a strict one-to-one correspondence.

From this we show that there are as many points in a straight line as there are points in space.

As for the fact that all the points on the line are the same and that all the points in the space are the same, the landlord can prove it in a similar way. The key is to establish a one-to-one correspondence.

The number of points in space is the higher order infinity relative to a straight line.

There are many points in space!

There are infinite points in a straight line, and the straight line can be infinitely extended, but there is only one, but there can be countless straight lines in space, so there are many points in space!

It's all infinite points, and it's not proven

The equation of a straight line determined by 2 points in space is solved as follows:

Preparation of materials: coordinate system, direction vector.

1. In a planar Cartesian coordinate system.

1. Draw a planar Cartesian coordinate system and mark the two known points.

2. Connect two points, and make a perpendicular line perpendicular to the horizontal axis at each point, and make a parallel line parallel to the x-axis with the point closest to the x-axis.

3. Among the resulting triangles, <>

4. The corresponding linear equation can be obtained by using the slope of the straight line equal to the tangent value.

Second, in a three-dimensional Cartesian coordinate system.

1. Draw two points in the three-dimensional Cartesian coordinate system and connect the two points.

2. Subtract the coordinates of the two points to obtain a vector width that is the direction vector of a straight line in space.

3. Using the symmetry of the linear equation, that is, each coordinate of the direction vector, as the corresponding denominator, the unknown number subtracts the corresponding known number of the cautious school, and as the numerator, the spatial linear equation can be obtained.

Whether it's a plane or a space, any two pin points can determine a straight line.

in spatial coordinates.

For example, the spatial coordinates of point A are (x1, y1, z1) and (x2, y2, z2) all have three unknown quantities, so the straight line they form has three unknown quantities.

You can also compare it with the plane, there are only two coordinate quantities in the plane, a(x1,y1) and b(x2,y2) in the plane, so of course their straight lines only have two unknown quantities.

Equation of a two-point straight line in space a, b.

Calculation formula: (x-x1) (x2-x1)=(y-y1) (y2-y1)=(z-z1) (z2-z1).

How did this formula come about, you can go up and down the ppt of the next space straight line, and it's OK.

Proof of Proof: Use the method of counter-proof.

Suppose there is a straight line ab in the space and another point c, suppose there is a straight line cd when the ridge bridge passes through c, and ce is parallel to ab

According to the nature of the parallel lines, CE is parallel to CD, which contradicts that they both pass the point C, i.e., the assumption is not true, and the proof is required by the title.

Use the method of counter-evidence. Proof: Suppose that there are 2 straight lines A and B parallel to the straight line C at a point outside the straight line, then A C, B C, and there are parallel transmissions to know a B, and the parallel lines do not intersect, which contradicts both A and B a little. So the assumption is wrong.

Therefore, there is only one and only one straight line parallel to the straight line.

Isn't that a mathematical theorem? Do you still have to prove the theorem? How can your teacher come up with such a substandard question?

If it must be proven, use the method of counter-proof.

Proof: If a point is passed, there are 2 (or n) straight lines parallel to the known straight lines.

According to the axiom of parallelism, these 2 (or n) lines should also be parallel to each other.

But they are all in the same place.

So it is impossible to be parallel.

So the assumption is not valid.

That is, there is only one and only one straight line parallel to the straight line.

Establish. Yes, there are three ways in which two straight lines are related in space:1Parallel 2Intersect 3Opposite.

Parallel to intersect then they must be on the same plane.

Otherwise they are other.

You can prove it with the anti-way method;

Hehe, I went to college, and I studied literature, and now I look at this a little nostalgic.

Happy studying.

It's not nature, it's an axiom of parallelism, it's a conclusion that is generally accepted as correct.

Use the common method of counterproof in mathematics:

Suppose there is another straight line parallel to the known straight line at this point, that is, there are two straight lines parallel to the straight line at a point outside the straight line.

Then, the two lines passing through the point are parallel to the known lines at the same time.

Because two straight lines that are parallel to one straight line at the same time are parallel to each other.

If the two straight lines pass the same point, the two straight lines intersect.

Contradiction. So the assumption is not valid.

So there is only one straight line parallel to the straight line.

And that's true, because this point and the known line determine a plane, and the line that is parallel to the known line through this point is also in this plane

In Euclidean geometry, it is an axiom that Euclid himself thinks is a bit problematic, so he uses this axiom very little. This is not the case in non-planar geometries such as Riemannian geometry.

This is the axiom of parallel lines.

A little bit p beyond the straight line l can make a straight line pq straight line l, and it can only be a straight line like a straight line.

a.The projection of the point is on the same surface projection of the straight line.

b.The ratio of the spatial point to the line segment is equal to the ratio of the projection point to the projection line segment.

c.Target. d.Target.

Correct backup key answer: The projection of the point is on the same surface projection of the straight line; The ratio of the space point to the projection segment is equal to the ratio of the projection point to the projection segment.

Substituting x-2=(z-4) 2 y-3=(z-4) 2 into 2x=y=z-6=0 together to get z=2 and z=2 back to x=1 y=2, so the intersection point is (1,2,2).

Existence: There may be zero, one, or infinite points at the intersection of a line and a plane. Feasibility: If the two points do not coincide with the known straight line, a straight line can be determined, and only the known straight line and the plane can be used to obtain the relationship between the two.

Vector method: When the general equation of the plane is known (ax+by+cz+d=0), n =a,b,c) is the normal vector of the plane, and it is easy to find the distance from the point to the plane and the projection of a vector to the normal vector.

The plane equation in the space Cartesian coordinate system is the general equation of the space straight line ax+by+cz+d=0: the two plane equations are combined, indicating a straight line (intersecting line) The plane equation ax+by+cz+d=0 in the space Cartesian coordinate system is: a1x+b1y+c1z+d1=0, a2x+b2y+c2z+d2=0, the standard formula of the parallel (the result of the concurrent can be expressed as a determinant) space straight line:

Similar to the point oblique in a planar coordinate system) (x-x0) a (y-y0) b (z-z0) c

The analysis is as follows:

1. The two-point formula of the spatial straight line: (similar to the two-point formula in the plane coordinate system) (x-x1) (x2-x1) (y-y1) (y2-y1) (z-z1) (z2-z1) (z2-z1) can be substituted to obtain, the plane equation in the space Cartesian coordinate system is ax+by+cz+d=0 The general equation of the space straight line: the two plane equations are combined, indicating that the plane equation in a straight line (intersection line) space Cartesian coordinate system is ax+by+cz+d=0 The linear equation is:

a1x+b1y+c1z+d1=0, a2x+b2y+c2z+d2=0, the standard formula for parallel (the result of the syndicate can be expressed as a determinant) spatial straight line: (similar to the point oblique in a plane coordinate system) (x-x0) a (y-y0) b (z-z0) c

where (a, b, c) is the two-point formula of the straight line in the direction vector space: (similar to the two-point formula in the plane coordinate system) (x-x1) (x-x2) (y-y1) (y-y2) (z-z1) (z-z2).

2. The cylindrical coordinates ( , z) are. An expression for a point in a cylindrical coordinate system. Let p(x,y,z) be a point in space, then the point p can also be determined by three ordered numbers, z, where is the distance between the projection m of the point p in the xoy plane and the origin point, and is the angle between the projection mo of the directed line segment po in the xoy plane and the positive direction of the x-axis.

The correspondence between the coordinates of the points of the cylindrical coordinate system and the three-dimensional Cartesian coordinate system is, x= cos, y= sin, z=z.

First of all, I think there is no only answer, if it is parallel, the answer is unique.

In fact, it is required that a line of a straight line in vertical space only needs to find a vector whose space vector is perpendicular to the known straight line, and the direction vector of the known straight line in the problem is (3,2, 1) Let the vector (x,y,z) be perpendicular to (3,2, 1), then 3x 2y z=0, and arbitrarily find a set of values that satisfy the secondary equation ( , then the equation is: (x 2) y 1) z 3) For example, (1, 1,1) then the result is: (x 2) 1=(y 1) 1=(z 3) 1

There should be a problem with the answer x item given by the landlord, and the straight line given by the answer is just the point sought. I guess it's a typing error.