Commentary on Proposition XI.19¶
Back to Proposition XI.19.
Legendre, followed by other writers already quoted, uses a preliminary proposition equivalent to Euclid’s definition of planes at right angles to one another.
If two planes are perpendicular to one another, a straight line drawn in one of them perpendicular to their common section will be perpendicular to the other.
Let the perpendicular planes CE, MN (figure of last note) intersect in CD, and let AB be drawn in CE perpendicular to CD.
In the plane MN draw BF at right angles to CD.
Then, since the planes are perpendicular, the angle ABF (their inclination) is a right angle.
Therefore AB is perpendicular to both CD and BF, and therefore to the plane MN.
We are now in a position to prove XI. 19, viz. If two planes be perpendicular to a third, their intersection is also perpendicular to that third plane.
Let each of the two planes AC, AD intersecting in AB be perpendicular to the plane MN. Let AC, AD intersect MN in BC, BD respectively. In the plane MN draw BE at right angles to BC and BF at right angles to BD.
Now, since the planes AC, MN are at right angles, and BE is drawn in the latter perpendicular to BC, BE is perpendicular to the plane AC. Hence AB is perpendicular to BE. [XI. 4] Similarly AB is perpendicular to BF. Therefore AB is perpendicular to the plane through BE, BF, i.e. to the plane MN.
An useful problem is that of drawing a common perpendicular to two straight lines not in one plane, and in connexion with this the following proposition may be given.
Given a plane and a straight line not perpendicular to it, one plane, and only one, can be drawn through the straight line perpendicular to the plane.
Let AB be the given straight line, MN the given plane. From any point C in AB draw CD perpendicular to the plane MN. Through AB and CD draw a plane AE. Then the plane AE is perpendicular to the plane MN. [XI. 18]
If any other plane could be drawn through AB perpendicular to MN, the intersection AB of the two planes perpendicular to MN would itself be perpendicular to MN: [XI. 19] which contradicts the hypothesis.
To draw a common perpendicular to two straight lines not in the same plane.
Let AB, CD be the given straight lines. Through CD draw the plane MN parallel to AB (Prop. 2 in note to XI. 15). Through AB draw the plane AF perpendicular to the plane MN (see the last preceding proposition).
Let the planes AF, MN intersect in EF, and let EF meet CD in G. From G, in the plane AF, draw GH at right angles to EF, meeting AB in H. GH is then the required perpendicular. For AB is parallel to EF (Prop. 4 in note to XI. 15); therefore GH, being perpendicular to EF, is also perpendicular to AB. But, the plane AF being perpendicular to the plane MN, and GH being perpendicular to EF, their intersection, GH is perpendicular to the plane MN, and therefore to CD. Therefore GH is perpendicular to both AB and CD.
Only one common perpendicular can be drawn to two straight lines not in one plane. For, if possible, let KL also be perpendicular to both AB and CD. Let the plane through KL, AB meet the plane MN in LQ. Then AB is parallel to LQ (Prop. 4 in note to XI. 15), so that KL, being perpendicular to AB, is also perpendicular to LQ. Therefore KL is perpendicular to both CL and LQ, and consequently to the plane MN. But, if KP be drawn in the plane AF perpendicular to EF, KP is also perpendicular to the plane MN.
Thus there are two perpendiculars from the point K to the plane MN: which is impossible.
Rausenberger’s construction for the same problem is more elegant. Draw, he says, through each straight line a plane parallel to the other. Then draw through each straight line a plane perpendicular to the plane through the other. The two planes last drawn will intersect in a straight line, and this straight line is the common perpendicular required.
The form of the construction best suited for examination purposes, because the most self-contained, is doubtless that given by Smith and Bryant.
Let AB, CD be the two given straight lines.
Through any point E in CD draw EF parallel to AB.
From any point G in AB draw GH perpendicular to the plane CDF, meeting the plane in H.
Through H in the plane CDF draw HK parallel to FE or AB, to cut CD in K.
Then, since AB, HK are parallel, AGHK is a plane.
Complete the parallelogram GHKL.
Now, since LK, GH are parallel, and GH is perpendicular to the plane CDF, LK is perpendicular to the plane CDF.
Therefore LK is perpendicular to CD and KH, and therefore to AB which is parallel to KH.