CHAP. II.

INCLINATIONS OF LINES AND PLANES TO ONE ANOTHER.

PROPOSITION I.-LEMMA.

159. IF a line, in space, be perpendicular to any plane, the orthogonal projection of the line on either of three rectangular co-ordinate planes will be perpendicular to the intersection of the plane with that co-ordinate plane.

P

M

.p

Y

N

Let MP, MN be the lines in which the plane intersects the co-ordinate planes zox, ZoY, and let pq be the projection of the perpendicular line upon zoY. Then, since the projecting plane passing through the original line is perpendicular to ZoY, and by passing through that original line it is perpendicular to the plane MNP; it follows that the two planes ZOY and MNP are perpendicular to the projecting plane: hence MN, the common section of the first planes is (Geom. Planes, Prop. XIX.) perpendicular to the last plane, and therefore perpendicular to pq, which it meets in that plane. In like manner it may be proved that the projections of the perpendicular line on zOX, XOY, are respectively perpendicular to MP and PN, the intersections of MNP with those planes.

PROPOSITION II.

X

160. To find the equations for a line passing through a given point and perpendicular to a given plane.

Let the co-ordinates of the given point be x', y', z', and those of the point in which the line intersects the plane be X, Y, Z. Now, the equation for the plane being (Art. 152.)

Ax+By+Cz=D;

if lines on the plane, parallel to the intersections of the latter

with the co-ordinate planes zox, zoy, and passing through the point whose co-ordinates are x, y, z, be projected on those planes; the lines so projected, being (Art. 154. c.) parallel to the intersections, their equations will be,

But the projections of the perpendicular line on zox, ZOY are, by the preceding lemma, perpendicular to the intersections of the given plane with the co-ordinate planes, and to the lines represented by the equations (e); therefore the equations for the projections of the perpendicular line (x, y, z being the co-ordinates of its intersection with the given plane) will be (Art. 16. a.)

B
x=2z+P, and y==z+Q

(f)

(P and Q being put for the constant terms). And the equations for the projections of the same line, at points which are the projections of that point whose co-ordinates are x', y', z', will be

These are the equations for the perpendicular to the given plane when that perpendicular is projected on the planes

ZOX, ZOY.

The values of x and y determined from the equations (g) being substituted in the equation of the plane, from the latter the value of z may be determined; and thus the point is found at which a line perpendicular to the plane, and passing through a given point, will intersect the plane.

*

*Since in the equation Ax+By+cz D, for a plane (AO, or D, Fig. to Art. 150., being a line let fall from o perpendicularly on the plane)

A=COS. AOX, B=COS. AOY, C=COS. AOZ;

161. COR. The length of the perpendicular between the points whose co-ordinates are x, y, z and x'y' z' is evidently equal to

√ {(x' — x)2+(y' − y)2 + (z' —z)2 }

Now, if r represent the length of a perpendicular let fall from o on the plane, and p be the length of the perpendicular last mentioned, we shall have [(d) Art. 152.]

and

whence

ax+by+cz=r2,

ax'+by+cz' = (r+p)2;

a(x' —x)+b (y'—y)+c (z' —z)=2rp+p2.

Substituting in this equation the values of x-x and y'—y in (g) we shall have,

from which, by substitution in (g), we shall obtain corresponding values of x-x and y'-y. Lastly, substituting all these values in the radical expression above, we obtain

162. To find the inclination of a given line to a plane.

x

in which is manifestly the tangent of the angle which the projection on

zox of a perpendicular to the plane makes with oz, and is the tangent

of the angle which the projection on zor of a perpendicular to the plane makes with the same line oz; or, are the cotangents of the angles

which the projections of the perpendicular to the plane make with ox and or respectively. The coefficients of z in the equations (e) are evidently tangents of the angles which the intersections of the given plane with the co-ordinate planes make with oz; or they are respectively the cotangents of the angles which those intersections make with xo and yo produced.

Imagine a line, as OA (Fig. to Art. 150.), to be drawn from the origin of the rectangular co-ordinates perpendicularly on the given plane, and let op be parallel to the given line; then the angle AOP will be equal to the complement of the required inclination of the line to the plane. Now, if the equation for the given plane be represented by

Ax+By+Cz=D,

[(d') Art. 152.]

the equations for a line, as OA, perpendicularly to it, if the perpendicular be supposed to pass through the origin of the co-ordinates, will be [(f) Art. 158.)]

also the equations of the given line, supposed to pass through O, will be, on zox, x'=a'z', and on zOY, y' =b′ z'.

A

B
C

In these equations and are the quantities designated,

C

respectively, a and b in the equations for x and y (Art. 151.), while a' and b' have the same signification as have a' and b' in the values of x and y' in that Article. Therefore, on sub

B

stituting for a and for b in the formula (c') in the same

C

Article, that formula will become an expression for the cosine of the angle between the given line and a perpendicular to the plane; and consequently for the sine of the required angle between the given line and the plane.

If the line were parallel to the plane; since the sine of the angle between the line and the plane would then be zero, the numerator of the expression (c) would be zero, or we should

have

aa'+bb'+1=0;

whence, by the substitution, just mentioned,

Aa' + Bb'+C=0.

PROPOSITION IV.

163. To find the inclination of a given plane to each of three rectangular co-ordinate planes.

Since the angle which two planes make with one another is equal to the angle contained between two lines which are perpendicular to the planes, and intersect one another; it

follows that, if a line pass through the origin of the coordinates perpendicularly to the given plane, the angles which it makes with ox, OY, oz will, respectively, be equal to the angles which the given plane makes with the planes ZOY, ZOX, XOY.

Now the equation for the plane being [(d) Art. 152.] represented by

a

ax+by+cz=r2 (=a2 + b2+c2)

in which r is the length of a perpendicular let fall on the plane from the origin of the co-ordinates; it follows that +dénotes the cosine of the angle which the perpendicular makes with Ox; it is therefore equal to the cosine of the angle which the given plane makes with zoY. In like manner, the cosine of the angle which the plane makes with

b

zox is denoted by ± and the cosine of that which it

makes with xox by ±

PROPOSITION V.

164. The inclinations of two planes to each of three rectangular co-ordinate planes being given; to find the inclination of the planes to one another.

Imagine a line to be drawn through the origin of the coordinates perpendicular to each of the given planes; then the angle contained between these perpendiculars will be equal to the required angle of inclination. Now let x, y, z be the co-ordinates of any point in one of the perpendiculars, and x', y', z' the co-ordinates of any point in the other; then representing the required inclination, we have [(c) Art. 150.] xx' +yy' + zz' xx′+yy' +zz' rr' √ {(x2+y2+z2) (x'2+y'2+z'2)} which, in terms of the tangents of the angles made by the projections of the perpendiculars on the co-ordinate planes, is [(c) Art. 151.]

COS.

COS. Ө

or = >

aa' + bb'+1

When one of the planes is at right angles to the other, cos. = 0;