contains all points whose value of p satisfies both conditions. But we may write (§ 92), since a, a,, and Vaa, are not coplanar, pS.aa1Vaa1 = Vaa1S.aa1p+V.a1Vaa1Sap + V.V (aa1) aSa1p,

or, by the given equations,

(2)

—pT2 Vaa1 = V.a1 Vaa,Saß + V.V (aa1) a§а1ß1+xVaa12, where x, a scalar indeterminate, is put for S.aap 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 3 B1 =0, and obtain at once

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 B1Sap-SaẞSa1p = 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 value of p for its foot is ρ p = ß+xa,

where xa is the vector-perpendicular in question.

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

209.] The equation of the plane which passes through the extremities of a, B, y may be thus found. If p be the vector of any point in it, p—a, a—ß, and 6-7 lie in the plane, and therefore (§ 101)

S.(p—a) (a—ß) (B—y) = 0,

or Sp (Vaß+VBy+ Vya)—S.aẞy = 0.

Hence, if

d = x (Vaß+Vßy+Vya)

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

From this formula, whose interpretation is easy, many curious properties of a tetrahedron may be deduced by the reader. Thus, for instance, if we take the tensor of each side, and remember the result of § 100, we see that

T(Vaß+Vßy+Vya)

is twice the area of the base of the tetrahedron. simply proved thus. The vector area of base is

This may be more

faces of a tetrahedron, This is the hydrostatic

V (a —ẞ) (y—ẞ) = −1 (Vaß+ Vẞy+Vya). Hence the sum of the vector areas of the and therefore of any solid whatever, is zero. proposition for solids immersed in a fluid subject to no external forces.

210.] Taking any two lines whose equations are

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

d =

=

Now if ẞ, the plane contains the first line; if dẞ1, the second.

Hence, if y Vaa, be the shortest vector distance between the lines, we have S.aa (B-B1-y Vaa1) = 0,

or TyVaa) TS. (B — ß1) UVaα1,

=

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

Sop = 0.

Hence

S.p (Vaẞ+ By + Tya) = 0

is the required equation; and the required sine is

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 a and a correspond to the extremities of one of the variable lines, being the vector of its middle point. Then, obviously,

Also

@

This gives a linear relation between x and x, so that, if we substitute for a1 in the preceding equation, we obtain a result of the X1 form

☎ = d +x €,

where 8 and e 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.

be

Let the point be taken as origin, and let the equations of the planes Sap = 0, SBp = 0, Syp = 0.

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

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, α2, az, &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.

be the required equation, with the condition (evidently allowable) To = 1.

The perpendiculars are (§ 208) - 1Swa1, &c.

and the condition that is a unit-vector gives

Sada = 0.

Hence, as da may have any of an infinite number of values, these equations cannot be consistent unless

The values of a are known, so that if we put

Σ.αδασ = φω,

is a given self-conjugate linear and vector function, and therefore a has three values (91, 92, 93, § 164) which correspond to three mutually perpendicular values of w. For one of these there is a maximum, for another a minimum, for the third a maximumminimum, in the most general case when 91, 92, 93 are all different.

215.] The following beautiful problem is due to Maccullagh. Of a system of three rectangular vectors, passing through the origin, two lie on given planes, find the locus of the third.

Let the rectangular vectors be a, p, σ. Then by the conditions of the problem

and

Sop

=

Spo

Saw = 0,

= Sow =

SBp = 0.

@

The solution depends on the elimination of p and among these five equations. [This would, in general, be impossible, as p and between them involve six unknown scalars; but, as the tensors are (by the very form of the equations) not involved, the five given equations are necessary and sufficient to eliminate the four unknown scalars which are really involved. Formally to complete the requisite number of equations we might write

To α, Tp = b, but a and may have any values whatever.]

b

the required equation. As will be seen in next Chapter, this is a cone of the second order whose circular sections are perpendicular to a and B. [The disappearance of x and y in the elimination instructively illustrates the note above.]

EXAMPLES TO CHAPTER VI.

1. What propositions of Euclid are proved by the mere form of the equation p = (1-x) a+xß,

which denotes the line joining any two points in space?

2. Shew that the chord of contact, of tangents to a parabola which meet at right angles, passes through a fixed point.

3. Prove the chief properties of the circle (as in Euclid, III) from the equation p = a cos 0+ẞ sin 0;

where Ta TB, and Saß = 0.

4. What locus is represented by the equation

intersect? If this is not satisfied, what is the shortest distance between them?

6. Find the equation of the plane which contains the two parallel lines Va (p-B) = 0, Va(p-B1) = 0.

7. Find the equation of the plane which contains

8. Find the equation of a straight line passing through a given

point, and making a given angle with a given plane.

Hence form the general equation of a right cone.