The problem of search of a vector of a normal of a straight line on the plane and the plane in space is too simple. Actually it comes to the end with record of the general equations of a straight line or plane. As the curve on the planes only a special case of a surface in space, about normals to a surface will also go the speech.
Instruction
1. The first way This way the simplest, but for his understanding is required knowledge of a concept of the scalar field. However, and the reader, inexperienced in this question, will be able to use the resulting matter formulas.
2. It is known that scalar field f is set as f=f (x, y, z), and any surface at the same time is a surface of level f (x, y, z) = C (C=const). Besides, the normal of a surface of level coincides with a gradient of the scalar field in the set point.
3. A gradient skalyarno of the field (functions of three variables) is called g=gradf=idf/dx+jdf/dy+kdf/dz= vector {df/dx, df/dy, df/dz }. As length of a normal does not matter, it is necessary only to write down the answer. A normal to poverkhnostif (x, y, z)-C=0 in tochkem0 (x0, y0, z0) n=gradf=idf/dx+jdf/dy+kdf/dz= {df/dx, df/dy, df/dz }.
4. The second way Let the surface is set by F equation (x, y, z) =0. That it was possible to draw further analogies to the first way, it is necessary to consider that the derivative of a constant is equal to zero, and F is set as f (x, y, z)-C=0 (C=const). If to carry out the section of this surface by any plane, then the arisen spatial curve can be considered a godograf of any vector function of r(t) = ix(t)x+jy(t) +kz(t). Then the derivative of a vector of r’ (t) = ix of’ (t) of +jy’ (t) of +kz’ is sent (t) on a tangent in some point of M0 (x0, y0, z0) to a surface (see fig. 1).
5. That there was no confusion, the current coordinates of a tangent straight line should be designated, for example, in the italics (x, y, z). Initial the equation of a tangent straight line, taking into account that r’ (t0) is the directing vector, registers as (x-x (t0)) / (dx(t0)/dt) = (y-y(t0))/(dy(t0)/dt) = (z-z(t0))/(dz(t0)/dt).
6. Having substituted vector function coordinates in f surface equation (x, y, z)-C=0 and having differentiated on t you receive (дf/дx)(дx/дt)+ (df/dy) (du / дt)+ (df/dz) (dz/dt) =0. Equality represents a scalar product of some vector of n (df/dx, df/dy, df/dz) and r’ (x’ (t), y’ (t), z’ (t)). As it is equal to zero, n (df/dx, df/dy, df/dz) and there is a required vector of a normal. It is obvious that results of both ways are identical.
7. An example (has theoretical value). To find a vector of a normal to a surface of the function of two z=z variables set by the classical equation (x, y). Decision. Rewrite this equation in the form of z-z (x, y) = F (x, y, z) =0. Following any of prepositional ways, it turns out that n (-dz/dx, - dz/dy, 1) - a required vector of a normal.