THE DISTANCE
Introducing terms used in the lesson
perpendicular
parallel
projector
right triangle
square
distance
I. The distance from a point to a line:
Give the line a and one point O.
We get the distance from O to the line a is the length of OH. The notation is: d(O,a) = OH = d.
For all M on the line (a), we can see that OH OM
(OH is less than or equal to OM). Hence, d is the least length from O to any point on the line.
Let H is the projector of O on the line a.
II. The distance from a point to a plane:
Give the plane (P) and one point O, H is the projector of O on the plane.
We get the distance from O to (P) is the length of OH. The notation is d(O;(P)) = d = OH.
For all M on the plane, we can see that OH OM (OH is less than or equal OM).
Hence, d is the least length from O to any point on the plane.
III. The distance from a line to a plane that is parallel to that line.
Let d is the distance from the line a to the plane (P) that is parallel to a.
We define that d is equal to the distance from any point on the line a to the plane (P).
The notation is: d(a,(P)) = MH
Example
Let S.ABCD be pyramid with ABCD is a square edge a, SA is perpendicular to the plane (ABCD) and the length of the line SA = .
a) Calculate the distance from A to plane (SCD).
b) Calculate the distance between the straight line CD and mp (SAB).
1) Let H is the projector of A on the line SD.
We see: SD  AH (1)
Moreover CD  AD, CD  AD so CD  (SAD) imply CD  AH. (2)
From (1), (2), We get AH  (SCD)
Hence AH = d(A,(SCD))
2) We calculated AH = 2) We see CD parallel to (SAB) so d(CD, (SAB)) = d(D,(SAB)) = AD = a
Exercise 1: Let ABC.ABC be prismatic with AA  (ABC), AA = a, right triangle ABC at A where BC = 2a, AB = a. Calculate:
a) The distance from line AA to the plane (BCCB).
b) The distance from A to (A  BC).
c) Prove that AB  (ACC’A’) and the distance from point A to the plane (ABC’).