The theorem of the three perpendiculars
A line is perpendicular to a plane if it is at 90° to every line of the plane that passes through its foot. Take a plane α, a point P outside it, and let PO be perpendicular to α with foot O.
Now take a line l in α. Let OH be the perpendicular from O to l (H is on l).
Theorem. Then PH is perpendicular to l. We call it "three perpendiculars" because PO ⊥ α, OH ⊥ l and PH ⊥ l.
Why it is true. l is at 90° to OH (we drew it so). l is also at 90° to PO, because PO is perpendicular to every line of the plane. So l is at 90° to two crossing lines, PO and OH, which both lie in the plane POH. Then l is at 90° to the whole plane POH, and PH lies in that plane. So l ⊥ PH.
The converse is also true. If PO ⊥ α and PH ⊥ l, then OH ⊥ l.
Distance from a point to a plane
From P there are many segments to the plane. The shortest one is the perpendicular, PO. Its length is the distance from P to the plane.
Why shortest? Take any other point M of the plane. In the triangle POM the angle at O is 90°, so PM is the longest side (the hypotenuse): PM > PO.
To find this distance in a solid, look for a line that stands at 90° to the plane. In a cube or box, an upright edge is perpendicular to the floor, so its length is the distance from its top corner to the floor.
Distance from a point to a line
The distance from a point P to a line l is the length of the perpendicular from P to l. In space, finding that perpendicular is the job of the theorem.
- Drop the perpendicular PO to the plane that holds l.
- In the plane, draw OH ⊥ l.
- Join PH. By the theorem, PH ⊥ l, so PH is the distance.
- Triangle POH has a right angle at O, so PH² = PO² + OH².
Any other point K on l gives PK² = PO² + OK², and OK > OH, so PK > PH. That is why PH is the shortest.
Example: a corner V is 8 above corner A of a square floor of side 6, straight up. The distance from V to the edge CD: AD ⊥ CD and VA ⊥ floor, so VD ⊥ CD, and VD = √(8² + 6²) = 10.
Distance between parallel planes
Two planes α and β are parallel if they never meet. Take any point A of β and drop the perpendicular AA′ to α. The length AA′ is the distance between the parallel planes.
It is the same for every point A of β. All the perpendiculars between two parallel planes are equal (they are opposite sides of rectangles). So the distance does not depend on where you measure.
If a slanted segment joins the planes, its length c, its projection p on the lower plane and the distance h are linked by c² = h² + p².
Examples: the floor and ceiling of a room, the top and bottom faces of a box, two shelves in a cupboard.
Try it: pencil, ruler and string
Use a book as the plane, an upright pencil as PO, a ruler on the book as the line l, and a string from the pencil tip to the ruler. Check the right angles with a set square. Then measure PO, OH and the string. Does the string match the square root of PO² plus OH²?
Key formulas and definitions
- PO ⊥ plane, OH ⊥ l in the plane ⇒ PH ⊥ l
- Distance from P to the plane = PO (the perpendicular)
- Distance from P to the line l = PH, with PH² = PO² + OH²
- Distance between parallel planes = any common perpendicular
- Slanted segment between parallel planes: c² = h² + p²
Worked examples
1. PO = 4 cm is perpendicular to a plane. O is 3 cm from the line l in the plane (OH ⊥ l). Find the distance from P to l.
By the theorem PH ⊥ l, so PH is the distance. PH² = 4² + 3² = 25, so PH = 5 cm.
2. A point P is 12 cm above a plane. Its foot O is 5 cm from the line l. Find the distance from P to l.
PH² = 12² + 5² = 144 + 25 = 169. PH = 13 cm.
3. ABCD is a square of side 6 cm. VA = 8 cm is perpendicular to the plane of the square. Find the distance from V to the line CD.
AD ⊥ CD (square) and VA ⊥ plane. By the theorem VD ⊥ CD. So the distance is VD = √(VA² + AD²) = √(64 + 36) = 10 cm.
4. In the same figure, find the distance from V to the line BC.
AB ⊥ BC and VA ⊥ plane, so VB ⊥ BC. VB = √(VA² + AB²) = √(64 + 36) = 10 cm.
5. A segment AB of length 13 cm joins two parallel planes. Its projection on the lower plane is 5 cm. Find the distance between the planes.
The projection, the distance and AB make a right triangle. h² = 13² − 5² = 169 − 25 = 144, so h = 12 cm.
6. Triangle ABC is right-angled at A with AB = 3 cm and AC = 4 cm. MA = 12 cm is perpendicular to the plane ABC. Find the distance from M to BC.
BC = 5. Draw AD ⊥ BC. AD = (AB × AC) ÷ BC = 12 ÷ 5 = 2.4 cm (area two ways). By the theorem MD ⊥ BC. MD² = 12² + 2.4² = 144 + 5.76 = 149.76, so MD ≈ 12.24 cm.
Common mistakes
- Using the slanted segment PO′ as the distance to the plane. The distance is only the perpendicular.
- Forgetting the first perpendicular: OH must be drawn from the foot O, not from P.
- Writing PH² = PO² + PH² or mixing up the hypotenuse. PH is the longest side of triangle POH.
- Thinking the distance between parallel planes changes with the place you measure. It is the same everywhere.