GEOMETRY OF THREE DIMENSIONS.

BOOK VII.

OF PLANES AND LINES.

553. Already, in 50, a plane has been defined as the surface generated by the motion of a line always passing through a fixed point while it slides along a fixed line.

554. Already, in 97, the theorem has been assumed, that, if two points of a line are in a plane, the whole line lies in that plane.

Two other assumptions will now be made:

555. Any number of planes may be passed through any line.

556. A plane may be revolved on any line lying in it.

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THEOREM I.

557. Through two intersecting lines, one plane, and only one,

passes.

A

HYPOTHESIS. Two lines AB, BC, meeting in B.

CONCLUSION. One plane, and only one, passes through them. PROOF. By 555, let any plane FG be passed through AB, and, by 556, be revolved around on AB as an axis until it meets any point C of the line BC.

The line BC then has two points in the plane FG; and therefore, by 554, the whole line BC is in this plane.

Also, any plane containing AB and BC must coincide with FG.

For let be any point in a plane containing AB and BC.

Draw QMN in this plane to cut AB, BC, in M and N. Then, since M and N are points in the plane EF, therefore, by 554, Q is a point in the plane EF.

Similarly, any point in a plane containing AB, BC, must lie in EF; therefore any plane containing AB, BC, must coincide with EF.

558. COROLLARY I. Two lines which intersect lie in one plane, and a plane is completely determined by the condition that it passes through two intersecting lines.

559. COROLLARY II. Any number of lines, each of which intersects all the others at different points, lie in the same plane; but a line may pass through the intersection of two others without being in their plane.

1

560. COROLLARY III. A line, and a point without that line, determine a plane.

A

B

PROOF. Suppose AB the line, and C the point without AB. Draw the line CD to any point D in AB. Then one plane contains AB and CD, therefore one plane contains AB and C.

Again, any plane containing AB must contain D; therefore, any

plane containing AB and C must contain CD also.

But there is only one plane that can contain AB and CD.
Therefore there is only one plane that can contain AB and C.
Hence the plane is completely determined.

561. COROLLARY IV. Three points not in the same line determine a plane.

C

For let A, B, C be three such points. Draw the line AB. Then a plane which contains A, B, and C must contain AB and C; and a plane which contains AB and C must contain A, B, and C. Now, AB and C are contained by one plane, and one only; therefore A, B, and C are contained by one plane, and one only. Hence the plane is completely determined.

562. Two parallel lines determine a plane.

For, by the definition of parallel lines, the two lines are in the same plane; and, as only one plane can be drawn to contain one of the lines and any point in the other line, it follows that only one plane can be drawn to contain both lines.

THEOREM II.

563. If two planes cut one another, their common section must be a straight line.

HYPOTHESIS. Let AB and CD be two intersecting planes.
CONCLUSION. Their common section is a straight line.

PROOF. Let M and N be two points common to both planes. Draw the straight line MN. Therefore, by 554, since M and N are in both planes, the straight line MN lies in both planes.

And no point out of this line can be in both planes; because then two planes would each contain the same line and the same point without it, which, by 560, is impossible.

Hence every point in the common section of the planes lies in the straight line MN.

564. CONTRANOMINAL OF 554. A line which does not lie altogether in a plane may have no point, and cannot have more than one point, in common with the plane,

Therefore three planes which do not pass through the same line cannot have more than one point in common; for, by 563, the points common to two planes lie on a line, and this line can have only one point in common with the third plane.

565. All planes are congruent; hence properties proved for one plane hold for all. A plane will slide upon its trace.

PRINCIPLE OF DUALITY.

566. When any figure is given, we may construct a reciprocal figure by taking planes instead of points, and points instead of planes, but lines where we had lines.

The figure reciprocal to four points which do not lie in a plane will consist of four planes which do not meet in a point. From any theorem we may infer a reciprocal theorem.

Two points determine a line.
Three points which are not in

a line determine a plane.

A line and a point without it. determine a plane.

Two lines in a plane determine a point.

Two planes determine a line.

Three planes which do not pass through a line determine a point.

A line and a plane not through it determine a point.

Two lines through a point determine a plane.

There is also a more special principle of duality, which, in the plane, takes points and lines as reciprocal elements; for they have this fundamental property in common, that two elements of one kind determine one of the other. Thus, from a proposition relating to lines or angles in axial symmetry, we get a proposition relating to points or sects in central symmetry.

The angle between two corresponding lines is bisected by the axis.

The sect between two corresponding points is bisected by the

center.