At the solution of geometrical and practical tasks sometimes it is required to find distance between the parallel planes. So, for example, height of the room is, actually, a distance between a ceiling and a floor which represent the parallel planes. An example of the parallel planes are also opposite walls, covers of the book, a wall of boxes and many other things.

## It is required to you

- - ruler;
- - drawing triangle with a right angle;
- - calculator;
- - compasses.

## Instruction

1. To find distance between two parallel planes: • draw the straight line perpendicular to one of the plane; • define points of intersection of this straight line from each of the planes; • measure distances between these points.

2. To draw the straight line perpendicular to the plane, use the following method borrowed from descriptive geometry: • choose any point on the plane; • draw through this point two the crossed straight lines; • construct a straight line perpendicular to at the same time both crossed straight lines.

3. If the parallel planes are located horizontally, for example, a floor and a ceiling of the house, then for measurement of distance use a plumb. For this purpose: • take a thread, length obviously bigger the measured distance; • tie a small small weight to one of its ends; • throw thread through the tack or a wire located near a ceiling, or hold thread with a finger; • lower a small weight until it does not concern a floor; • record a thread point when the small weight falls to a floor (for example, tie a small knot); • measure distance between a mark and the end of thread with cargo.

4. If the planes are set by the analytical equations, then find distance between them as follows: • let A1*kh + V1*u + C1*z + D1 = 0 and A2*kh + V2*u + C2*z + D2 = 0 – the equations of the planes in space; • as for the parallel planes multipliers at coordinates are equal, rewrite these equations in the following look: A*kh + V*u + C*z + D1 = 0 and A*kh + V*u + C*z + D2 = 0; • use the following formula for finding of distance between these parallel planes: s = |D2-D1| / √ (A²+B²+C²), where: | | - standard designation of the module (absolute value) of expression.

5. Example: Define distance between the parallel planes set by the equations: the 6th +6u-3z+10=0 and the 6th +6u-3z+28=0. Decision: Substitute parameters from the equations for the planes in the above-stated formula. It will turn out: s = |28-10| / √ (6²+6²+ (-3)²) = 18 / √ 81 = 18/9 = 2. Answer: Distance between the parallel planes – 2 (units of measure).