operate on (1) with S.a and S.a,. We thus obtain two equations, which determine x and x,, as y is already known.

A somewhat different mode of treating this problem will be discussed presently.

204. The equation

Sap = 0

imposes on the sole condition of being perpendicular to a; and therefore, being satisfied by the vector drawn from the origin to any point in a plane through the origin and perpendicular to a, is the equation of that plane.

To find this equation by a direct process similar to that usually employed in cöordinate geometry, we may remark that, by § 29, we may write

p = xß+Yy,

where ẞ and y are any two vectors perpendicular to a. In this form the equation contains two indeterminates, and is often useful; but it is more usual to eliminate them, which may be done at once by operating by S.a, when we obtain the equation first written.

It may also be written, by eliminating one of the indeterminates only, as

VBp = ya,

where the form of the equation shows that Saß = 0.

represents a plane drawn through the extremity of ẞ and perpendicular to a. This, of course, may, like the last, be put into various equivalent forms.

206. The line of intersection of the two planes

and

S.a (p-B) = 0,7

S.a,(p-B,) = 0, S

(1)

contains all points whose value of p satisfies both conditions. But we may write (§ 92), since a, a,, and Vaa1 are not coplanar,

pS.aa, Vaa,

Vaa,S.aa,p+ V.a, Vaa, Sap + V.V(aa,) aSa1p,

or, by the given equations,

-pT2 Vaa1 = V.a, Vaa, Saẞ+V.V(aa1)a§ã‚ß‚ +x Vaa1,' (2) where x, a scalar indeterminate, is put for S.aa1p which may have any value. In practice, however, the two definite given scalar equations are generally more useful than the partially indeterminate vector-form which we have derived from them. When both planes pass through the origin we have ẞ=ẞ, = 0, and obtain at once

p = xVaa1

as the equation of the line of intersection.

207. The plane passing through the origin, and through the line of intersection of the two planes (1), is easily seen to have the equation Sa B, Sap-Saß Sa1p = 0,

For this is evidently the equation of a plane passing through the origin. And, if p be such that

is perpendicular to the vector-line of intersection (2) of the two planes (1), and to every vector joining the origin with a point in that line.

The student may verify these statements as an exercise.

208. To find the vector-perpendicular from the extremity of ẞ on the plane Sap=0,

we must note that it is necessarily parallel to a, and hence that the corresponding value of p is

Similarly the vector-perpendicular from the extremity of ẞ on the plane

Sa (p-y) = 0

may easily be shown to be

-a-'Sa (B-7).

209. The equation of the plane which passes through the extremities of a, ß, y may be thus found. If p be the vector of any point in it, p-a, a-ẞ, and B-y lie in the plane, and therefore (§ 101)

or

Hence, if

S.(p-a) (a-B) (B-y) = 0,

Sp(Vaẞ+VBy+ Vya) — S.aßy = 0.

d = x(Vaß + Vẞy + Vya)

be the vector-perpendicular from the origin on the plane containing the extremities of a, ß, y, we have

d = (Vaß + Vẞy + Vya)-1S.aßy.

From this formula, whose interpretation is easy, many curious properties of a tetrahedron may be deduced by the reader.

210. Taking any two lines whose equations are

we see that

p = B + xa,

p = ẞ1+ x1α1,

S.aa, (p-8)= 0

is the equation of a plane parallel to both. Which plane, of course, depends on the value of d.

Now if 86, the plane contains the first line; if d =

the second.

19

Hence, if yVaa, be the shortest vector distance between the lines, we have

S.aa, (B-B,-yVaa,) = 0,

or Ty Vaa,) =TS.(ẞ—ß1) UVaa1,

the result of § 203.

211. Find the equation of the plane, passing through the origin, which makes equal angles with three given lines. Also find the angles in question.

Let a, ẞ, y be unit-vectors in the directions of the lines, and let the equation of the plane be

is the required equation; and the required sine is

S.aßy

T(Vaẞ+Vßy+Vya) *

212. Find the locus of the middle points of a series of straight lines, each parallel to a given plane and having its extremities in two fixed lines.

the fixed lines. Also let x and r, correspond to the extremities

of one of the variable lines, being the vector of its middle point. Then, obviously,

This gives a linear relation between a and a1, so that, if we substitute for x, in the preceding equation, we obtain a result of the form

@=d+x€,

where 8 and are known vectors. The required locus is, therefore, a straight line.

213. Three planes meet in a point, and through the line of intersection of each pair a plane is drawn perpendicular to the third; prove that, in general, these planes pass through the same line.

Let the point be taken as origin, and let the equations of the planes be

Sap = 0,

SBp = 0,
βρ

Syp = 0.

The line of intersection of the first two is || Vaß, and therefore the normal to the first of the new planes is

V.yVaẞ.

Hence the equation of this plane is

or

S.pV.yVaẞ = 0,

Spp Say-Sap Spy = 0,

and those of the other two planes may be easily formed from this by cyclical permutation of a, ß, y.

We see at once that any two of these equations give the third by addition or subtraction, which is the proof of the theorem.

214. Given any number of points A, B, C, &c., whose vectors (from the origin) are a1, a,, a,, &c., find the plane through the origin for which the sum of the squares of the perpendiculars let fall upon it from these points is a maximum or minimum.

U