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9. What conditions must be satisfied with regard to a number of given lines in space that it may be possible to draw through each of them a plane in such a way that these planes may intersect in a common line?

10. Find the equation of the locus of a point the sum of the squares of whose distances from a number of given planes is con

stant.

11. Substitute "lines" for "planes" in (10).

12. Find the equation of the plane which bisects, at right angles, the shortest distance between two given lines.

Find the locus of a point in this plane which is equidistant from the given lines.

13. Find the conditions that the simultaneous equations

Sap = a,

Sβρ

Spp = b,

may represent a line, and not a point.

Syp = c,

14. What is represented by the equations

(Sap)2 = (SBp)2 = (Syp)2,

where a, ß, y are any three vectors?

15. Find the equation of the plane which passes through two given points and makes a given angle with a given plane.

16. Find the area of the triangle whose corners have the vectors

a, B, y.

Hence form the equation of a circular cylinder whose axis and radius are given.

17. (Hamilton, Bishop Law's Premium Ex., 1858).

(a.) Assign some of the transformations of the expression

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where a and ẞ are the vectors of two given points A and B. (6.) The expression represents the vector y, or OC, of a point C in the straight line AB.

(c.) Assign the position of this point C.

18. (Ibid.)

(a.) If a, B, y, d be the vectors of four points, A, B, C, D, what is the condition for those points being in one plane?

(b.) When these four vectors from one origin do not thus terminate upon one plane, what is the expression for the volume of the pyramid, of which the four points are the corners?

(c). Express the perpendicular & let fall from the origin O on the plane ABC, in terms of a, B, Y.

19. Find the locus of a point equidistant from the three planes

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20. If three mutually perpendicular vectors be drawn from a point to a plane, the sum of the reciprocals of the squares of their lengths is independent of their directions.

21. Find the general form of the equation of a plane from the condition (which is to be assumed as a definition) that any two planes intersect in a single straight line.

22. Prove that the sum of the vector areas of the faces of any polyhedron is zero.

CHAPTER VII.

THE SPHERE AND CYCLIC CONE.

216.] AFTER that of the plane the equations next in order of simplicity are those of the sphere, and of the cone of the second order. To these we devote a short Chapter as a valuable preparation for the study of surfaces of the second order in general.

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denotes that the length of p is the same as that of a given vector a, and therefore belongs to a sphere of radius Ta whose centre is the origin. In § 107 several transformations of this equation were obtained, some of which we will repeat here with their interpretations. Thus

S (pa) (pa) = 0

shews that the chords drawn from any point on the sphere to the extremities of a diameter (whose vectors are a and -a) are at right angles to each other.

T(pa) (pa) = 2TVap

shews that the rectangle under these chords is four times the area of the triangle two of whose sides are a and p.

p = (p+a)1a (p+a) (see § 105)

shews that the angle at the centre in any circle is double that at the circumference standing on the same arc. All these are easy consequences of the processes already explained for the interpretation of quaternion expressions.

218.] If the centre of the sphere be at the extremity of a the equation may be written

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in which case the origin is a point on the surface of the sphere, this becomes

From this, in the form

p2 - 2 Sap = 0.

Sp (p-2a) = 0

another proof that the angle in a semicircle is a right angle is derived at once.

219.] The converse problem is-Find the locus of the feet of perpendiculars let fall from a given point (p=ß) on planes passing through the origin.

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be one of the planes, then (§ 208) the vector-perpendicular is

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[This is an example of a peculiar form in which quaternions sometimes give us the equation of a surface. The equation is a vector one, or equivalent to three scalar equations; but it involves the undetermined vector a in such a way as to be equivalent to only two indeterminates (as the tensor of a is evidently not involved). To put the equation in a more immediately interpretable form, a must be eliminated, and the remarks just made shew this to be possible.]

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so that, as is evident, the locus is the sphere of which ẞ is a diameter.

220.] To find the intersection of the two spheres

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square the equations, and subtract, and we have

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which is the equation of a plane, perpendicular to a―a1 the vector

joining the centres of the spheres. whether the spheres intersect or not. their Radical Plane.

This is always a real plane
It is, in fact, what is called

221.] Find the locus of a point the ratio of whose distances from two given points is constant.

Let the given points be 0 and A, the extremities of the vector a. Also let P be the required point in any of its positions, and OP=p. Then, at once, if n be the ratio of the lengths of the two lines, T(p-a) nTp.

This gives

=

p2-2Sap+ a2 = n2 p2,

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na

Thus the locus is a sphere whose radius is 7(2), and whose

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T( 1

, a definite point in the line OA.

222.] If in any line, OP, drawn from the origin to a given plane, OQ be taken such that OQ.OP is constant, find the locus of Q.

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be the equation of the plane, a vector of the required surface. Then, by the conditions,

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Substituting in the equation of the plane, we have

aw2+b2Saw = 0,

which shews that the locus is a sphere, the origin being situated on it at the point farthest from the given plane.

223.] Find the locus of points the sum of the squares of whose distances from a set of given points is a constant quantity. Find also the least value of this constant, and the corresponding locus. Let the vectors from the origin to the given points be ......a, and to the sought point p, then

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Σα

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i.e.

n

the equation of a sphere the vector of whose centre is whose centre is the mean of the system of given points. Suppose the origin to be placed at the mean point, the equation becomes

p2

c2 +

(a2)

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n

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