There are several ways of a task of the plane: the general equation, the directing normal vector cosines, the equation in pieces and so forth. Using elements of concrete record, it is possible to find distance between the planes.
Instruction
1. The plane in geometry can be defined differently. For example, it is a surface which any two points are connected by a straight line which also consists of plane points. By other definition, this set of the points which are on equal removal from any two set, not belonging to it.
2. The plane – the simplest concept of stereometry meaning the flat figure beyond all bounds directed extensively. The sign of parallelism of two planes consists in lack of crossings, i.e. two set spatial figures have no general points. The second sign: if one plane is parallel to the crossed straight lines belonging another, then these planes are parallel.
3. To find distance between two parallel planes, it is necessary to determine length of a piece perpendicular of the Ends of this piece the points belonging to each plane are. Besides, normal a vector are also parallel, so if the planes are set by the general equation, then equality of the relations of coordinates of normals will be necessary and sufficient sign of their parallelism.
4. So, let A1 planes are set • x + B1 • at + C1 • z + D1 = 0 and A2 • x + B2 • at + C2 • z + D2 = 0 where Ai, Bi, Ci are coordinates of normals, and D1 and D2 – distance from a point of intersection of coordinate axes. The planes are parallel if: A1/A2 = B1/B2 = C1/C2, and distance between them it is possible to find on a formula: d = | D2 – D1 | / √ (|A1•A2| + B1•B2 + C1•C2).
5. Example: two planes x + 4 are given • at - 2•z + 14 = 0 and-2 • x - 8 • at + 4•z + 21 = 0. To define whether they are parallel. If yes, that to find distance between them.
6. Decision. A1/A2 = to B1/B2 = =-1/2 – the planes are parallel to C1/C2. Pay attention to presence of coefficient-2. If D1 and D2 correspond with each other to the same coefficient, then the planes coincide. In our case it not so, as 21 • (-2) ≠ 14, therefore, it is possible to find distance between the planes.
7. Divide for convenience the second equation into size of coefficient-2: x + 4 • at - 2•z + 14 = 0; x + 4 • at - 2•z – 21/2 = 0. Then the formula will take a form: d = | D2 – D1 | / √ (A² + B² + C²) = | 14 + 21/2 | / √ (1 + 16 + 4) ≈ 5.35.