Pure and to Coordinate Geometry, &c. A number of illustrative examples will be worked in order to shew the manner in which problems should be attacked vectorially. The chapters on Rotors and Stress diagrams may however be taken before reading the remainder of the chapter. B 131. Let a and ẞ denote the adjacent sides of any parallelogram as vectors, then the diagonals are a+ B d-B A Fig. 86. Now a+ẞ and a- - ß. [a+B|a-B]=[aa]-[aß]+[Ba]-[BB] =2[Ba]. (Notice that unless the proper order of the vectors had been retained in the multiplication, the third term might have cancelled with the second.) This result says:--the area of any parallelogram is half that of the parallelogram having as adjacent sides its diagonals. 132. Taking the scalar product we have (a+Ba-B)=a2 — ß2 + (aß) — (aẞ) If then the two sides are equal in magnitude a2=82, and since neither a+ß nor a−ẞ is zero, the diagonals of a rhombus are perpendicular. Again (a+B)2=a2+2 (aß) + ß2 (a — B)2 = a2 — 2 (aß) +ß2. If then the figure is a rectangle, (aß)=0 and the magnitudes of the diagonals are equal. 133. Again (a+B)2+(a-B)2=2{a2-82). If then a and B are of constant length the right-hand side is constant. Now the end points A and B of a and ẞ if drawn from a common origin lie on concentric circles (or spheres). Hence in the figure BA2+BA,2 a constant. This says that if from any point of a circle (or sphere) lines be drawn to the extremities of any diameter of a concentric circle (or sphere), the sum of the squares on these lines is constant. = 134. The three perpendiculars from the vertices of a triangle to the opposite sides are concurrent. OAB is any triangle, let OA=a, OB=B, P the point of intersection of the perpendiculars from A and B, and OP=y. and Then remembering the meaning of a scalar product we see that (Ba)=(ya) (a3)=(yB). Now neither a − ß, i.e. BA, nor y, i.e. OP, is zero. Hence ОР 1 АВ and therefore the three perpendiculars meet in a point. 135. Four straight lines in a plane intersect in 6 points, these 6 points are called the vertices of the quadrilateral formed by the four lines, and the lines joining opposite vertices are called the diagonals. To prove that the mid-points of the diagonals of a quadrilateral are collinear. In the figure FC, FD, EB, ED are the lines, A, B, C, D, E, F the vertices, AC, BD, FE the diagonals. Denoting the sides by vectors a, ß, y, d, etc. we have The Vector from B to M1 the mid-point of BD is 1 a+B ....(1). 9 2 If the vector product of these vanishes, M1MM, are collinear. a=t1d+t, and ẞ=tza+ty; Now hence Hence [aẞ]=t1[88]=t1[ay]. 4 [M1M2| M1M2]=t1[a+y\d]+t1[ay] 2 = −[aß]+[aß], (for a+y=-ẞ-8) =0. 136. The straight line. A straight line is fixed when we Fig. 90. P know two points on it or when we know one point on it and the direction of the line. Thus in the figure the straight line AP is fixed relatively to an origin O when we know (a) two position vectors a and Ρ to two points A and P on the line, or (b) the position vector a and the ort ẞ giving the direction of the line. In the second case the position vector of any point P on the line would be given by a vector equation of the form p=a+ßt, where t denotes the length of AP. Whatever the value of t may be, p will always be the position vector of some point on the line, and the equation p=a+ẞt (with t a variable) is called the Vector Equation to the line. Multiply the last equation by a taking the scalar product (ap)=a2+t(aẞ). If now a be perpendicular to ß, then (aß)=0 and (ap)= a2. a being the position vector of a fixed point on the line, a2 is constant. The last equation tells us that the projection of p on a is always the same, hence for co-planar vectors (ap)=a constant says that the point whose position vector is p must lie on a straight line perpendicular to a. If we are not confined to a plane, then the point must lie in a plane perpendicular to a. For if we imagine the line AP in the figure to rotate round OA as an axis, then AP will generate a plane and any point on this plane is such that its projection on OA is equal to OA itself. A α Fig. 91. If κ be an ort, then (kp)=p says that the point whose position vector is p lies on a plane (or line) perpendicular to κ and the distance from the origin to the plane (or line) is p. 137. Strange as it may seem to a student of Euclid, it is only within recent years that mechanisms have been devised for drawing a straight line. Watts, Roberts, and Tchebicheff invented simple mechanisms for tracing approximate straight lines, but it was not until 1864 that Colonel Peaucellier discovered an exact method. Ten years later Mr Hart invented a simpler mechanism, and since then others have been devised by Professor Sylvester and Mr Kempe. On account of the great practical importance of the problem, a short investigation of the Peaucellier and Hart inventions may not be out of place here. 138. The mechanism due to Colonel Peaucellier consists of P 7 links connected by pin-joints. Four of equal length form a 6 rhombus BCPD, two of equal length OC, OD, join opposite points C, D, of the rhombus to a fixed point 0. The seventh link is from another fixed point A to B and is of length 40. If B describes a circular arc about A, D and C will describe circular arcs about 0 and we shall prove that P will describe a straight line perpendicular to OA. and α A Fig. 92. Let OA=a, AB=B, BC=Y From the symmetry of the figure we see that OBP must be a straight line and that OA and AB make equal angles with it as also do BC and BD. (The student should establish these properties direct from the equations a2=82 etc., the conditions for the equality of the lengths of OA and AB etc.) Hence P describes a straight line perpendicular to a. |