But, by § 96, S(Vaẞ) y = Ta Tẞ sin 0 Sny. Here Tn 1, let & be the angle between 7 and y, then finally = η S.aßy-Ta TB Ty sin 0 cos p. But as n is perpendicular to a and B, Ty cos & is the length of the perpendicular from the extremity of y upon the plane of a, ß. And as the product of the other three factors is (§ 96) the area of the parallelogram two of whose sides are a, ß, we see that the magnitude of S.aẞy, independent of its sign, is the volume of the parallelepiped of which three cöordinate edges are a, ß, y; or six times the volume of the pyramid which has a, ß, y for edges. 101.] Hence the equation S.aẞy = 0, if we suppose a, B, y to be actual vectors, shews either that i.e. two of the three vectors are parallel, or all three are parallel to one plane. This is consistent with previous results, for if y = pß we have S.aßypS.aẞ2 = 0; and, if y be coplanar with a, ß, we have y = pa+qß, and 7 102.] This property of the expression S.aßy prepares us to find that it is a determinant. And, in fact, if we take a, ẞ as in § 83, and in addition y = ix" + jy′′ +kz′′, we have at once S.aßy=-x" (yz′ —zy')—y′′ (zx′ — xz′)—z′′ (xy′ —yx′), The determinant changes sign if we make any two rows change places. This is the proposition we met with before (§ 89) in the form S.aẞy S.Bay = S.ẞya, &c. If we take three new vectors a1 = ix+jx′ + kx”, B1 = iy + jy + ky", Y1 = iz +jz + kz′′, we thus see that they are coplanar if a, ß, y are so. That is, if 103.] We have, by § 52, If q = aß, we have Kq = ßa, and the formula becomes In Cartesian cöordinates this is (x2 + y2+z2) (x22 + y2 + z2) = : (xx′+yy +zz′)2 + (yz′ — zy′)2 + (zx′ — xz′)2 + (xy′ — yx')2. = (ww′ —xx' — yy —zz′)2 + (wx′ +w′x+yz — zy′)2 +(wy +w'y + zx′ — xz′)2 + (wz′ +w'z + xy' — yx^)2, a formula of algebra due to Euler. 104.] We have, of course, by multiplication, (a+ẞ)2 = a2 + aß + ßa + ß2 = a2 + 2 Saß + B2 (§ 86 (3)). Translating into the usual notation of plane trigonometry, this becomes c2 = a2-2 ab cos C+b2, the common formula. Again, (a+3) (a−B) = — Vaß + Vßa = -2 Vaß (§ 86 (2)). == βα Taking tensors of both sides we have the theorem, the parallelogram whose sides are parallel and equal to the diagonals of a given parallelogram, has double its area (§ 96). = and vanishes only when a2 = 82, or Ta TB; that is, the diagonals of a parallelogram are at right angles to one another, when, and only when, it is a rhombus. Later it will be shewn that this contains a proof that the angle in a semicircle is a right angle. 105.] The expression obviously denotes a vector whose tensor is equal to that of ß. But we have so that p is in the plane of a, B. Also we have so that ẞ and p make equal angles with a, evidently on opposite sides of it. Thus if a be the perpendicular to a reflecting surface and ẞ the path of an incident ray, p will be the path of the reflected ray. Another mode of obtaining these results is to expand the above expression, thus, § 90 (2), P = 2a-1Saß-B = 2a ̄1Saß-a-1(Saß + Vaß) = a ̄1(Saß-Vaß), so that in the figure of § 77 we see that if OA = a, and OB = ß, we have OD=p=aßa-1. Or, again, we may get the result at once by transforming the equation to UP a ប 106.] For any three coplanar vectors the expression ρ = αβγ is (§ 101) a vector. It is interesting to determine what this vector is. The reader will easily see that if a circle be described about the triangle, two of whose sides are (in order) a and ẞ, and if from the extremity of ẞ a line parallel to y be drawn again cutting the circle, the vector joining the point of intersection with the origin of a is the direction of the vector aßy. For we may write it in the form α p = aß2ß ̄1y = — (T3)2 aß−1y — — (TB)2 ~—- Y, a which shews that the versor () which turns 3 into a direction parallel to a, turns y into a direction parallel to p. And this expresses the long-known property of opposite angles of a quadrilateral inscribed in a circle. Hence if a, ẞ, y be the sides of a triangle taken in order, the tangents to the circumscribing circle at the angles of the triangle are parallel respectively to αβγ, βγα, and γαβ. Suppose two of these to be parallel, i. e. let αβγ = αβγα = λαγβ (§ 90), since the expression is a vector. Hence i. e. the triangle is right-angled. And geometry shews us at once that this is correct. Again, if the triangle be isosceles, the tangent at the vertex is parallel to the base. Here we have whence y2 = a2, or Ty Ta, as required. = As an elegant extension of this proposition the reader may prove that the vector of the continued product aßyd of the vector-sides of a quadrilateral inscribed in a sphere is parallel to the radius drawn to the corner (a, d). 107.] To exemplify the variety of possible transformations even of simple expressions, we will take two cases which are of frequent occurrence in applications to geometry. and thus that P is any point equidistant from two fixed points,] may be written whence (p+a)2 = (p − a)2, or p2+2 Sap + a2 = p2-2 Sap + a2 (§ 104), p = (p + a) ̄1a{p + a), T.(pa) (pa) = 2 TVap. All of these express properties of a sphere. They will be interpreted when we come to geometrical applications. 108.] We have seen in § 95 that a quaternion may be divided into its scalar and vector parts as follows: where is the angle between the directions of a and 3, and UVB α ß, e= is the unit-vector perpendicular to the plane of a and ẞ so situated that positive (i. e. left-handed) rotation about it turns a towards ẞ. Similarly we have (§ 96) aß = Saẞ+ Vaẞß = Ta Tẞ(cos + sin 0), and having the same signification as before. 109.] Hence, considering the versor parts alone, we have being the positive angle between the directions of y and ẞ, and e the same vector as before, if a, ß, y be coplanar. = cos cos 0-sin & sin 0 + € (sin cos 0+ cos o sin 0), from which we have at once the fundamental formulae for the cosine and sine of the sum of two arcs, by equating separately the scalar and vector parts of these quaternions. And we see, as an immediate consequence of the expressions above, that cos mo+e sin m0 = (cos 0+ sin 0)m if m be a positive whole number. For the left-hand side is a versor which turns through the angle mo at once, while the right-hand |