An inclined support is 5 m long, with its lower end 3 m from the foot of the vertical. The structure’s height is not 5 m. How are these lengths related, and which ground line is perpendicular to the support?

A line perpendicular to a plane makes a right angle with every line in the plane through its foot. Two distinct lines intersecting at the foot suffice to test it. Height, slant and orthogonal projection form a right triangle. We use Pythagoras and lines and planes.

Two directions determine perpendicularity

Let l meet plane α at H. Writing l⊥α means l is perpendicular to every line of α through H. We need not check infinitely many lines. If l⊥a₁ and l⊥a₂, where distinct a₁,a₂ lie in α and intersect at H, then l⊥α.

Lines a₁,a₂ need not be perpendicular to each other. What matters is having two distinct directions in one plane, with l through their common point. Euclid XI.4, the criterion.

One right angle is insufficient: we can draw l⊥a₁ within the plane itself without making l perpendicular to the plane containing it. Two parallel test lines also fail to provide two independent directions. Here perpendicular lines intersect; angles between skew lines come later.

Perpendicular, slant and its projection

Let P be outside α, with PH⊥α and H∈α. Segment PH is the perpendicular from P; H is its foot. Length PH is the distance from P to the plane. Such a perpendicular exists; Euclid XI.11 constructs it.

The foot is unique. If another H′ were also a foot, Pythagoras would give PH′²=PH²+HH′² and PH²=PH′²+HH′². Together these imply HH′=0, so H′=H. Thus any PK with K∈α,K≠H is a slant, with foot K.

Under orthogonal projection onto α, P maps to H while K stays at K. Segment PK therefore projects to HK. Do not confuse HK with PH: PH is the height, while HK lies in the plane.

Two distinct directions in α

la₁a₂PHα

Vertical section: 3–4–5

PHK435

From projection to slant

K453aPHα
Panels 1 and 3 are schematic spatial views; do not measure lengths or angles on them. In panel 1, a₁ and a₂ intersect at H and are both perpendicular to l=PH, but not to each other. Panel 2 is the actual plane of triangle PHK at uniform scale: PH=4 m, HK=3 m, PK=5 m. In panel 3, HK is the orthogonal projection of PK onto α; a lies in α through K, a⊥HK and a⊥PK. Angle markers are calculated projections of actual spatial right angles; they need not appear square on screen. Dashed HK distinguishes projection from slant.

Since PH is perpendicular to all plane lines through H, PH⊥HK. Thus:

(PK)^(2) = (PH)^(2) + (HK)^(2).

In our own example HK=3 m,PK=5 m, so PH=√(25−9)=4 m. With height 4 m and another slant’s projection HL=6 m, its length is PL=√(16+36)=√52≈7.211 m. Each number belongs to its segment, rather than to a screen distance.

Why the perpendicular is shortest

For any K≠H in the plane, HK>0, so PK²=PH²+HK²>PH². Hence PK>PH. This proves PH is the shortest connection from P to the plane, rather than just one possible connection.

Slants from the same point to the same plane share height PH. Equal projections therefore correspond to equal slants, and conversely. A longer projection gives a longer slant. This comparison does not apply without a shared height to segments from different starting points or to different planes.

The theorem of three perpendiculars

Let a be a line in α through K, with HK the nonzero projection of slant PK. Then a⊥HK implies a⊥PK. The converse also holds: a⊥PK implies a⊥HK.

In the third diagram a passes through the support’s lower end K across its projection HK. It thus makes a right angle with the slant itself, although that need not look like 90° in a flat view. Omitting a⊂α or using the wrong projection leaves the conclusion unjustified.

In our model, P is 4 m from the plane but 5 m from line a: PK is the perpendicular to a. For any other X on a, PX²=PK²+KX², so no shorter connection to a exists. Distance to a plane and distance to a line in it are different problems.

A distance-based proofOptional · safe to skip

Why both directions hold

Choose U,V on a on opposite sides of K with KU=KV. In α, HK is perpendicular to a if and only if HU=HV: the perpendicular-bisector property. Height PH gives PU²=PH²+HU² and PV²=PH²+HV². Hence HU=HV if and only if PU=PV.

In triangle PUV, K is the midpoint of UV. Thus PU=PV is equivalent to median PK also being an altitude, or PK⊥a. These equivalences prove both the theorem and its converse. HK≠0 matters: with K=H we already have the perpendicular, while the zero projection has no defined direction.

In a spatial problem, locate foot H first, then projection HK, then slant PK and the required line a. This turns a difficult view into justified right triangles. Checking the drawing checks the model, rather than measuring perspective.