plane at P to the tangent PT. Let the plane cut the axis of the cylinder in Q; let OA = a, OQ = x, a = PTT. Since T is the plane CD, т ولا p2=(cta + a'ẞ) (cta + a2ß) = — c2t2 + cta2 + 1ß + cta'ßa + a'ßaß, = c2t2 + cta' (aß + ẞa) + V2a'ß, In words, PQ is perpendicular to the axis of the cylinder, or is a normal to the cylinder. It follows that the locus of all the perpendiculars let fall from the helix upon the axis of the cylinder is a screw-surface, or helicoid, bounded by the surface of the cylinder, and containing the helix itself; its equation being of the form, p = cta + uaß, (12) where t and u are independent variables. Or, reverting to the conception of a moving circle, we may say that u is a function of the velocity of rotation of the circumference of the circle, and t is a function of the velocity of translation of its centre, the two velocities being absolutely independent. To gain a definite idea of the shape of this surface, we have only to imagine that the cylinder is upright, and that a corkscrew staircase is constructed round the axis. The (smooth) bottom surface of the staircase is a helicoid. SECTION 7 The Plane 197°. Three given coinitial vectors, a, ß, y, terminate in a plane; to find the perpendicular from the origin upon the plane, OD = d. Since SV (By+ya+aß) = a scalar, V (By + ya + aẞ) || S. 198°. To find the condition that four points shall be coplanar. Let the vectors of the four points be a, ß, y, p. If the point P lie in the plane passing through the points A, B, C, which are supposed to be non-collinear, then or, S. (ap) (B - p) (y — p) = 0, Sẞyp+Syap + Saßp – Saßy=0. The geometric meaning of this equation is obvious. As a verification, let p=xi +yj + zk ; a = x ̧i + y ̧j + z ̧k ; &c., &c. Then 199°. The intersection of two planes is a straight line.1 For let OA = a, OB = B lie in one plane, and OC=y, OD = 8 lie in the other. Then, 79°, the intersection of the two planes is V. VaßVyd, that is, a straight line. As the point of intersection of two lines is called their cross, I venture to suggest that the line of intersection of two planes might be called their cut. 200°. The condition that three planes shall intersect in one common straight line. Let the equations of the three planes, P1, P2, P3, be respectively, Sapp; Sẞp=q; Syp = r; and let σ be a vector along their common line of intersection. Then, since a is perpendicular to P1, it is perpendicular to σ. Similarly, ẞ and y are perpendicular to σ. Therefore, and Saßy=0, pV By + qVya + rVaß = o. This is the condition that the three planes should be parallel to the same straight line. 201°. To find the condition that four planes shall meet in a point. Let the equations of the given planes be Saph; Sẞp=1; Syp = m; S&p=n. For the point of intersection, the variables of the first three equations must have a common value, But since the fourth plane passes through the same point, this value of p must be also a value of its variable. Therefore, substituting this value of p in the fourth equation, we get 202°. To find the diameter of the sphere circumscribing a given tetrahedron. This will be given in the Section on the sphere, 211°. N 203°. If two pairs of opposite edges of a tetrahedron, OABC, be at right angles, the third pair will be also at right angles. Let OAL BC, OBCA. Then 204°. To find the condition that the perpendiculars from the corners of a tetrahedron, OABC, upon the opposite faces shall be concurrent. Let xVẞy and yVya, the respective perpendiculars from A and B upon the opposite faces, intersect in P. Then, since (Ba), Vẞy and Vya are coplanar, V. VẞyVya is perpendicular to ẞ- a. Therefore, ß Therefore, that is, S(Ba) V. Vẞy Vya= 0, 0 = Sya - Sẞy = Sy (a — B), . . . . OC LAB. (1) We get corresponding results in the other cases; therefore the sought condition is that the opposite edges shall be at right angles. 205°. If the opposite edges of a tetrahedron be at right angles, then the sums of the squares of the opposite edges are equal. For, by (1) of 204°, we have OB2 + CA2 = OA2 + BC2. We get corresponding results in the other cases; therefore, &c., &c. This is another form of the condition of 204°. 206°. The sum of the vector-areas of the faces of a tetrahedron, OABC, is zero. OBC Vẞy; OCA = Vya; OAB = {Vaß. = As we have taken the positive areas of these three faces as seen by an observer standing upon the point O, i.e. from the outside of the tetrahedron; we must take the positive area of the remaining face from a corresponding point of view -exterior to the solid. Accordingly, = ACB = V (y — a) (ẞ — a) — — 1V (By + ya + aß). Therefore, = OBC + OCA + OAB + ACB = {Vẞy + Vya + Vaß — 1V (By + ya + aß) = 0. In general, the sum of the vector-areas of the faces of any polyhedron is zero. is zero. Let O be any point within the polyhedron, and let it be divided into a number of tetrahedra by planes passing through O. Then the sum of the faces of each separate tetrahedron Adding, the sum of the faces of all the tetrahedra is zero, that is, the sum of the internal faces, plus the sum of the external faces, is zero. But the internal areas, taken two and two, cancel each other, since each two adjacent areas have the same vector-expressions with opposite signs. Therefore the sum of the external faces, that is, the sum of the vector-areas of the faces of the original polyhedron, is zero. SECTION 9 The Sphere 207°. If a quadrilateral be not inscriptible in a circle, then, whether it be plane or gauche, we can always circumscribe two circles, OAB, OBC, about the two triangles formed by drawing the diagonal OB, fig. 44. We then have, 96°, T |