Given a straight line in space, to find the angle which it makes with each of the axes of co-ordinates.

But

z = r cos. (rx)

yr cos. (ry)

sr cos. (rz)

z' + y' + z' = 1o {cos. '. (rz) + cos. *. (ry) + cos 1. (rz)}

r2 {cos ' + (rz) + cos. *. (ry) + cos. (rz)}

cos. (ra) + cos. ' (ry) + cos. ' (rz) = 1

A very remarkable result, which shows that the sum of the squares of the cosines of the angles which a straight line in space forms with the three axes is equal to unity.

Given two straight lines in space which intersect each other, to find the angle contained between them in terms of the angles which each of the straight lines makes with the axes of co-ordinates.

Let the angles which r, makes with the axes AX, AY, AZ, be called (áx), (r1y),(r,z). Let the angles which r1⁄2 makes with the axes AX, AY, AZ, be called (r,x), (T2Y), (T22). Now as before

r1r, ens Q = X1Xq + Y1Y2 + %1%2 But

21 = r cos. T1x y1ri cos. (r1y) z=r1 cos. (r,z) .. cos.

cos. (†‚x) cos. (TM‚x) + cos. (r1y) cos. (r2y) + cos. (r,z) cos. (r,z)

To find the equation to a plane.

DEFINITION.-A plane is a surface generated by a straight line of indefinite length, which moves parallel to itself along another straight line, also of indefinite length.

Let BC, BD be the intersections of any plane BCDE, with the co-ordinate planes xz, yz. BC, BD are called the traces of the plane BCDE, on the planes xz, yz.

The plane BCDE may be conceived to be generated by the straight line BC, moving parallel to itself, along the straight line BD.

Since the straight BC lies wholly in the plane of xz, its equations will be

y = 0, x = Ax + C...............(1) where A is the tangent of the angle which BC makes with the axis AX, and where C = AB.

B

B'

B

In like manner, the equation to the straight line BD, which lies wholly in the plane yz, will be

x=0, z=By + C...............

(2)

where B is the tangent of the angle which BD makes with the axis AY.

Now let B'C' be any position of the generating line, then B'C' is parallel to BC, its projection on the plane zy will be parallel to AX, and its projection on the plane az will be parallel to BC, hence the equations to B'C' will be

y = a, % = Ax + ß.

........

(3)

where a, ß are quantities constant for all points of the same position B'C' of the generating line, but variable for any other position, such as B′′ C".

It remains for us to express analytically that the generating line BC inter sects in all its positions the line BD.

In order that this may be the case the equations (2) and (3) must hold good

at the same line; eliminating, therefore, x, y, z, between these four equations, we arrive at the equation of condition :

B = Ba + C.................

.(4) which must hold, in order that the straight lines BC, BD, may always intersect. Now the equations (3) and (4) must hold good together, for each position of the generating line. If, therefore, we eliminate the indeterminate quantities a, ß, between these equations, we conclude that the resulting equa tion

z = Ax + By + C

is that of the plane, since x, y, z, are the co-ordinates of the different points of the generating line in any position whatever.

The meaning of the constants in this equation is manifest from what has been said above. A and B are the tangents of the angles which the traces of the plane on the co-ordinate planes of xz, yz, make with the axes AX, AY; and C, is the distance from the origin of the point in which the plane meets the axis AZ.

If BD be parallel to AY, then the plane is perpendicular to the plane of xz, and its equation becomes

the same as that of its trace.

z = Ax + C

To find the equations of the projection on the co-ordinate planes of the intersections of two given planes.

Let the equations to the two planes be

z = Ax+ By + C, & = A'x + B'y + C'

Since these equations hold good together for the straight line which is their common intersection, if we eliminate z we shall have the equation to the projection on the plane zy, that is

(AA)x+(BB) y + Cσ = 0

In like manner, eliminating x or y, we find

(A- A') z + (AB'

· A'B) y + AC

(B' — B) z + (A'B — AB') x + BC' BC= 0

the equations of the projections on the planes of yz and zɛ.

To find the equation to a plane passing through one, two, or three given points.

The equation will be of the form

2 = Az + By + C ......

(1)

And since it passes through a point whose co-ordinates are x, y, z, it must satisfy the equation

Ax+By+ C

Subtracting (2) from (1)

(2)

z — £′ = A (x — x) + B (y — y') the equation required.

The constants A and B being arbitrary, the problem will be indeterminate, and, in point of fact, we know that any number of planes may be drawn through a given point.

If the plane be required to pass through a second point x", y", z", then it must satisfy the equation

= Ax" + By" + C

(3)

and two of the constants may be eliminated between equations (1), (2), (3), one however will still remain, and the problem will be indeterminate, for any number of planes may be drawn through two given points.

If the plane be required to pass through a third point x", y", z'”, it must satisfy the equation

the three constants may then be eliminated between equations (1), (2), (3), (4), and the result will be the equation of the only plane that can pass through the three given points. For we know that three points suffice to determine the position of a plane.

To find the conditions that must hold good, in order that a straight line and a plane may coincide, or be parallel.

Let the equations to the plane and the straight line be

z = Ax + By + C

x = as + α, y = bz + ß

Substituting as + ∞ and bz + ß for x and y in the equation to the plane, we have

(Aa + BB + C) + (Aa + Bb — 1) = 0

If the straight line and plane have only one point in common, we should thus be able to determine its co-ordinates, but if the straight line be altogether situated in the given plane, the above equation of condition must hold good whatever may be the value of z: hence the two parts of the equation must be independent of each other, and we shall have

Aa + BB + c = 0,

the equations of condition required.

AaBb-1=0

If the straight line be merely parallel to the plane, if we move them in a direction parallel to their original position until they reach the origin, the plane and the straight line will coincide, hence the above equations must be satisfied on the supposition that a and B and C are each = 0, therefore

AaBb1 = 0

will be the equation of condition which must be satisfied, in order that a straight line and a plane may be parallel.

To find the conditions requisite in order that a straight line may be perpendicular to a plane.

If a straight line be perpendicular to a plane, the projection of the straight line and the trace of the plane upon any of the co-ordinate planes will be perpendicular to one another.

Let the equation to the given plane and straight line be

z = Ax + By + C

x = az + α........

y = bx + ß

(1)

(2)

where it must be remembered that (1) is the projection of the given straight line on xz, and (2) its projection on yz.

Hence the straight line (3) is perpendicular to (1), and (4) is perpendicular

to (2).

.: A+ a = 0 and

which are the equations of condition required.

B+6=0

To find the equations to a straight line which passes through a given point, and is perpendicular to a given plane.

Let the equation to the plane be

2 = Ax+ By + C

The equations to the required line, since it passes through a point (x, y, z',) must be of the form

}

y - y = b (x — z)

and, since it is perpendicular to the plane

@= A, b = − - B,

To find the distance (d) of the given point, in the last problem, from the plane.

The equations to the straight line are

· ≈ + A (≈ — z) = 0 .......... (1) y −y + B (z − x') = 0 .......................... (2)

and the equation to the plane is

z = Ax+ By + C

which may be put under the form

or

where

z-% = A (xx) + By — y) + C + A2 + By — 2