How to find the space normal equation?

How to find the normal equation in high mathematics

First, we need to establish a space rectangular coordinate system, and then get two points (a1, b1, c1) (a2) on the plane , b2, c2) The vector is (x, y, z), let z=1. If it is a plane parallel to the z-axis, let x or y be 1.

Then it is the same as the plane on the plane The vector is vertical and the inner product is zero

In fact, two intersecting vectors on the plane can determine the normal of the plane

Now that we know the coordinates of each point on the plane, You can write the vectors on the two planes. The dot multiplied by (x, y, 1) is equal to 0

Solving these two equations you can get the normal vector

Method How to find the equation of a line

(0,1) is on the curve, so it is the tangent point y'=e^x x=0.y=1, so the slope of the tangent line is 1, and it passes (0,1), so it is x -y+1=0 The normal line is perpendicular to the tangent line, the slope is -1, and it also passes through the tangent point, so it is x+y-1=0

Find the normal vector and a point on the normal line to find the normal equation

Normal vector is a concept in space analytical geometry. The vector represented by a straight line perpendicular to a plane is the normal vector of the plane. Since there are countless straight lines in space perpendicular to the known plane, and each straight line can have different normal vector; therefore, there are countless normal vectors on a plane, but these normal vectors are parallel to each other. From the above theory, the space zero vector is the normal vector of any plane, but because the zero vector cannot represent the information of the plane, it is generally not selected The zero vector is the normal vector of the plane.

If the straight line is known to be perpendicular to the plane, the vector composed of two points of the known straight line can be taken as the normal vector; if there is no such straight line, the element method can be used to find it. The normal vector of a plane; the steps are as follows: first set the normal vector of the plane m (x, y, z), and then find any two non-linear vectors AB (x1, y1, z1) and CD (x2) in the plane ,y2,z2). Since the plane normal vector is perpendicular to all vectors in the plane, we get x*x1+y*y1+z*z1=0 and x*x2+y*y2+z*z2=0. Due to the above There are three unknowns and two equations in the solution (which cannot be solved by adding new vectors and equations, because other equations are equivalent to the above two equations), and the unique normal vector cannot be obtained (because the normal vector is not unique). In order to get To determine the normal vector, you can use fixed z=1 (you can also fix x=1 or y=1) or the method of modulus equal to 1 (unit normal vector), but this step is not necessary. Because determining the normal vector and the uncertainty method The functions of vectors are the same.

The main applications of normal vectors are as follows:

1. Find the angle between the oblique line and the plane: find the angle between the normal vector of the plane and the oblique line Angle, this angle is complementary to the angle formed by the oblique line and the plane. This principle can also be used to prove that the lines and planes are parallel;

2. Find the dihedral angle: find the normal vector of the two planes The angle is equal to or complementary to the dihedral angle;

3. The distance from the point to the surface: the projection of any oblique line (the plane is the line connecting a point and the plane) in the direction of the normal vector; For example, the distance d from point B to plane α = |BD·n|/|n| (the right side of the equation is all vectors, D is any point in the plane, and the vector n is the normal vector of plane α). This principle can also be used to find The distance of straight lines on different planes

The normal vector method is one of the methods that can be used in college entrance examination mathematics. Its advantage is that it is simple in idea and easy to operate. As long as a rectangular coordinate system can be established, the final answer can be written .The disadvantage is that compared with the general solid geometry method, the calculation amount is huge, especially when calculating dihedral angles.

For high numbers, normals of space curves, please tell me how to do it, thank you< /p>

Select B

First find the normal vector of any point on the surface

The normal is perpendicular to the plane

This normal is parallel to the plane Normal vector

List the equation and get the coordinates of the point

The process is as follows:

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Use the parametric equation of a space curve to find its normal

The intersection line between a surface represented by a general equation and a surface represented by a parametric equation is generally a space curve, with the root

According to the specific mathematical representation form and difficulty level of the two surface equations, the method of finding the tangent vector of the intersection line must also be flexible. This article points out

three methods of finding tangent vectors

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How to do this question? Normal equation and derivation problem, urgent need, thank you

1. y=√x, derivation gives y'=1/ 2√x

Then when x=1, y'=1/2

That is, the slope of the tangent line is 1/2, so the slope of the normal line is -1/(1/2)= -2

Passing through the point (1,1 ), so the normal equation is y-1= -2(x-1), that is, y=-2x+3

2. y=sinax

Derive to get y' =a *cosax

Then the derivative is y''= -a^2 *sinax