The distance between parallel planes, the intercept form of a plane and mixed problems
For parallel planes, first make sure that both equations have the same normal vector (multiply one equation through if needed). Then pick any point on one plane and find its distance to the other. With ax + by + cz = k₁ and ax + by + cz = k₂ the distance is |k₁ − k₂|/√(a² + b² + c²). The same idea gives the distance from a line to a plane parallel to it: take any point of the line. Intercept form. A plane that does not pass through the origin and is not parallel to an axis meets the three axes at (p, 0, 0), (0, q, 0) and (0, 0, r). Then x/p + y/q + z/r = 1, because each of these three points satisfies it. This is the intercept form (in Arabic الأجزاء المقطوعة من المحاور). To find the intercepts from ax + by + cz = k, put two coordinates equal to 0 and solve for the third: the table shows it. Conversely, divide the equation by k to get the intercept form. Intercepts may be negative. The three intercept points also form a triangle, whose area comes from the vector product, and with the origin a tetrahedron (a triangular pyramid) of volume pqr/6.
| axis | put the other two coordinates = 0 | intercept for 2x + 3y + 6z = 12 |
|---|---|---|
| x-axis | 2x = 12 | 6 |
| y-axis | 3y = 12 | 4 |
| z-axis | 6z = 12 | 2 |
A tent corner. Imagine the corner of a room as the origin and the floor and two walls as the coordinate planes. A slanted board touches the three edges at 6, 4 and 2 units from the corner. Write its equation (x/6 + y/4 + z/2 = 1) and find how far the corner is from the board (12/7 units).