Instruction

1

If you set the General equation of a plane AX+BY+CZ+D=0 or form A(x-x0)+B(y-y0)+C(z-z0)=0, we can immediately write the answer is n(A, b, C). The fact that this equation was derived, as the problem of determining the equation of a plane normal and a point.

2

For General answer, you will need a vector product due to the fact that the latter is always perpendicular to the original vectors. So, vector product of vectors, is a certain vector, the module of which is equal to the product of the first module (a) on the second module (b) and the sine of the angle between them. In this case the vector (denote it by n) are orthogonal to a and b is important. Three of these vectors are right, that is, from the end of the n shortest rotation from a to b is performed counterclockwise.

[a,b] is one of the common symbols of the cross product. To calculate the vector product in coordinate form, a vector is used determinant (see Fig.1)

[a,b] is one of the common symbols of the cross product. To calculate the vector product in coordinate form, a vector is used determinant (see Fig.1)

3

In order to avoid confusion with minus sign, rewrite the result in the form: n={nx, ny, nz}=i(aybz-azby)+j(azbx-axbz)+k(axby-aybx), and the coordinates {nx, ny, nz}={(aybz-azby), (azbx-axbz), (axby-aybx)}.

Moreover, in order to avoid confusion with numerical examples will write all the values obtained separately: nx=aybz-azby, ny=azbx-axbz, nz=axby-aybx.

Moreover, in order to avoid confusion with numerical examples will write all the values obtained separately: nx=aybz-azby, ny=azbx-axbz, nz=axby-aybx.

4

Return to the solution of the problem. The plane can be set in various ways. Let the normal to the plane defined by two non-collinear vectors, and numerical.

Let the vectors a(2, 4, 5) and b(3, 2, 6). The normal to the plane coincides with their vector artwork and as soon as it was found to be equal to n(nx, ny, nz),

nx=aybz-azby, ny=azbx-axbz, nz=axby-aybx. In this case ax=2, ay=4, az=5, bx=3, by=2, bz=6. Thus,

nx=24-10=14, ny=12-15=-3, nz=4-8=-4. Normal found: n(14, -3, -4). While it is normal to a family of planes.

Let the vectors a(2, 4, 5) and b(3, 2, 6). The normal to the plane coincides with their vector artwork and as soon as it was found to be equal to n(nx, ny, nz),

nx=aybz-azby, ny=azbx-axbz, nz=axby-aybx. In this case ax=2, ay=4, az=5, bx=3, by=2, bz=6. Thus,

nx=24-10=14, ny=12-15=-3, nz=4-8=-4. Normal found: n(14, -3, -4). While it is normal to a family of planes.

# Advice 2: How to find the normal vector

The task of finding

**the vector****of the normal**straight line on the plane and the plane in space is too easy. In fact, it ends with the entry of General equations of straight line or plane. Because the curve on the plane just a special case of a surface in space, it is about the normals to the surface will be discussed.Instruction

1

The first method This method is the easiest, but its understanding requires knowledge of the concept of the scalar field. However, inexperienced in this matter, the reader will be able to use the resulting formulas of this issue.

2

It is known that the scalar field f is defined as f=f(x, y, z) and any surface is the level surface f(x, y, z)=C (C=const). In addition, the normal surface level coincides with the gradient of the scalar field at a given point.

3

The gradient of the scalar field (function of three variables) is called a vector g=gradf=iдf/DH+jдf/DN+kдf/дz={дf/HH, дf/du, дf/дz}. Since the length

**of the normals**is irrelevant, it remains only to record the answer. Normal to поверхностиf(x, y, z)-C=0 in точкеМ0(x0, y0, z0) n=gradf=iдf/DH+jдf/DN+kдf/дz={дf/HH, дf/du, дf/дz}.4

*The second method Let surface given by the equation F(x, y, z)=0. To be able to continue to draw analogies with the first method, note that the derivative of a constant is zero and F is given as f(x, y, z)-C=0 (C=const). If you hold a section of this surface by an arbitrary plane, the resulting spatial curve can be considered as the locus of any vector function r(t)= ix(t)x+jy(t)+kz(t). Then the derivative*

**vector**r’(t)= ix’(t)+jy’(t)+kz’(t) is tangential at a point M0(x0, y0, z0) of the surface (see Fig.1).

5

In order to avoid confusion, the current coordinates of the tangent line should be marked, e.g. in italics (x, y, z). The canonical equation of the tangent line, given that r’(t0) is the direction vector, is written as (x-x(t0))/(dx(t0)/dt)= (y-y(t0))/(dy(t0)/dt)= (z-z(t0))/(dz(t0)/dt).

6

Substituting the coordinates of the vector-function in the equation of the surface f(x, y, z)-C=0 and differentiated with respect to t you will receive (дf/DH)(DH/дt)+(дf/du) (du/дt)+(дf/дz)(дz/дt)=0. The equality is a scalar product of some

**vector**n(дf/HH, дf/du, дf/дz) and r’(x’(t), y’(t), z’(t)). Since it is zero, then n(дf/HH, дf/du, дf/дz) is the desired vector**normals**. It is obvious that the results of both methods are identical.7

Example (has a theoretical value). To find a vector

**normal**to the surface given by the classical equation of two variables function z=z(x, y). Solution. Rewrite this equation in the form z-z(x, y)=F(x, y, z)=0. Following any of the suggested methods, it turns out that n(-дz/DX, -дz/dy, 1) is the sought vector**of the normal**.