Proposition 4. Perpendicular through Internal Point

41. Theorem. Through a given internal point there can pass one and only one line perpendicular to a plane.

Given the plane m and the internal point P.

Prove that through P there can pass one and only one line which is to m.

Proof. If a and b are any two lines in the plane m passing through P, then through P there is a plane x which is

a and a plane y which is to b.

to

$ 39

Since a and b are not identical but meet at P, x and y are not identical and must intersect in a st. line PQ.

$ 32

$ 33

$ 37

Now if another line PQ' could pass through P and be to m, it would be to a and to b.

$33

Hence

and

PQ' would lie in both x and y,

$ 38

PQ' would coincide with PQ.

.. PQ is the one and only 1 to m through P.

Proposition 5. Perpendicular through External Point 42. Theorem. Through a given external point there can pass one and only one line perpendicular to a plane.

Given the plane m and the external point P.

Prove that through P there can pass one and only one line which is to m.

Proof. Let be any line in m, and let PA be to l.

In m construct AR to l, and in the plane of AR and P construct POL to AR.

Produce PO to P', making OP' = OP.

§§ 13, 1; 13, 2

Post. 2

Let OB be any other line from 0 to 1, and draw PB, P'A, and P'B.

Then

l is to plane AP'P.

$ 37

Now prove that rt. ▲APB is congruent to rt. ▲AP'B, hence that AOPB is congruent to AOP'B, and hence that PO is 1 to OB.

Then

PO is to m.

$ 33

Further, if PQ is any other line from P to m and QO is drawn, then ≤QOP is a rt. Z.

$ 33

Hence PQ is not to m, and PO is the only 1.

Proposition 6. A Perpendicular and Obliques

43. Theorem. If from an external point a perpendicular and obliques are drawn to a plane,

1. The perpendicular is shorter than any oblique; 2. Obliques meeting the plane at equal distances from the foot of the perpendicular are equal;

3. Of two obliques meeting the plane at unequal distances from the foot of the perpendicular, the more remote is the longer.

Given the plane m, the external point P, PO tom, and the obliques PA, PB, PC drawn to m so that OA>OB=OC.

Prove that PO<PC, PB=PC, and PA> PC.

Proof. Produce PO to P', making OP' = OP.

Post. 2

Draw

P'C.

Post, 1

Now prove that PP'<PC+CP', and hence that PO<PC.

Then prove that

and hence that

Finally, prove that

and hence that

AOBP is congruent to AOCP,

PB PC.

PA>PB,

PA>PC.

44. Distance. The length of the perpendicular from a point to a plane is called the distance from the point to the plane.

The following corollaries (§§ 45, 46) extend the idea of a locus with which the student is familiar from plane geometry.

45. Corollary. The locus of points equidistant from the vertices of a triangle is the line through the center of the circumscribed circle, and perpendicular to the plane of the triangle.

We have first to prove (§ 19, 1) that any point P on the LOY satisfies the conditions; that is, that AP BP CP. But this follows

=

from §§ 16, 4 and 43, 2.

We have then to prove that any point P which satisfies the conditions

=

B

Y

m

that AP BPCP is on the line OY. Now AO=BO= CO (§ 16, 4). Then since OP is a common side, A AOP, BOP, and COP are congruent (§ 7, 5). Then the 4 made by OP with any lines in m are equal, and hence they are rt. 4. Hence OP is to m at O (§ 33), and since OP is part of OY (§ 41), P is on OY.

46. Corollary. The lotus of points equidistant from two given points is the plane perpendicular at the midpoint to the line segment joining the points.

We have first to prove (§ 19, 1) that any point P in m satisfies the condition that PA PB. We have then to prove that any point P' such that P'A = P'B lies in m, which is best done by an indirect proof.

The proof of each of these steps is left for the student.

We here meet a case in which the locus

A

m

B

is a plane instead of a line. In plane geometry, as the student has already found, a locus is usually a line; in solid geometry a locus may be a line or it may be a surface.

Exercises. Lines and Planes

1. Equal oblique lines drawn from a point to a plane meet the plane at equal distances from the foot of the perpendicular from the point to the plane; and of two unequal oblique lines the longer meets the plane at the greater distance from the foot of the perpendicular.

2. The locus of points equidistant from all points on a circle is a line through the center, perpendicular to the plane of the circle.

3. Find the locus of points at a given distance from each of two given points.

4. Explain how a carpenter might proceed to set a joist so that it shall be perpendicular to a horizontal floor. Draw a figure to illustrate any method which seems practical to you for the carpenter to use.

5. What geometric principle is involved in the statement that if the spoke of a wheel is perpendicular to the axle, the spoke determines a plane as the wheel revolves?

6. A steel smokestack 80 ft. high is braced by four steel wires each 100 ft. long and reaching from the top of the stack to the ground. If the wires are straight, at what distance from the foot of the stack does each reach the ground? Would three wires serve as well? Would two serve as well? Would one serve as well? State the geometric principle involved in each case.

7. Through a point P there pass four lines such that no three are in the same plane. Find the number of planes determined by the four lines.

8. In a plane P there lie four lines such that no three pass through the same point. Find the greatest number of points that can be determined by the four lines.