139. Mr Hart employed only 5 links to obtain the straight line motion. Four of the links form a contra-parallelogram and it is on the properties of this figure that the theory of the mechanism depends. B In the figure the bars OA and BC of equal length are joined crosswise by two other bars OC and AB also of equal length. C Fig. 93. and ẞ is parallel to y-a. Hence PQR is a straight line. Similarly PQS can be shewn to be a straight line. (ii) PQ. PR is constant, for which depends only on the lengths of the bars and on the ratio of their divisions. (iii) Let the point P be fixed, and by aid of another link QO1, describe a circle about 01 passing through P. make Let Then since PQR is a straight line and PQ × QR is constant, 141. Many of the properties of the circle are deduced with great ease from this equation. (1) The angle in a semicircle is a right angle. (2) The tangent at any point is perpendicular to the radius at that point. Taking the definition of the tangent at any point A as the limiting position of any secant AP when P approaches and ultimately coincides with A, then AP coincides with 414 and the theorem follows at once. (3) If two straight lines intersect within or without a circle, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other. Let AB, CD be the lines intersecting at P and let O be the centre of the circle. Let PO=y, then since the radii are all equal, (4) If two circles intersect, the line joining the centres is perpendicular to the line joining the points of intersection (i.e. to the radical axis). Let a and ẞ be the vectors from the centre of one circle to the points of intersection, and y the vector to the centre of the second circle. Then a-y, and B-y are the vectors from the centre of the second circle to the points of intersection. Then a2=82 and (y-a)2=(y—B)2. By squaring out the terms of the second equation we get at once (a-Bly)=0, whence y is perpendicular to a - B. γ PRODUCTS OF VECTORS AND THEIR CONNECTION WITH COORDINATES. 142. Area of a triangle. Let a and ẞ be the position vectors of two points whose rectangular coordinates are x, y and x1, y1, then Hence a=xi1+Yl2 [aß]=[x12+y2X14+Y14] =XX1[44]+YX1 [424]+xY1 [42]+YY1 [422] Now the magnitude of [aß] is twice the area of the triangle OAB, hence the area OAB is given by It should be noticed that the area is taken with its boundary having a counter-clockwise sense; if the sense be reversed then the area will change sign. 143. Angle between two straight lines. Expressing a and B as in the previous paragraph we get for the scalar product (aß)=xX1(4141)+YX1 (1241)+XY1 (4142)+YY1 (1242) Hence, using the definition of the cosine of an angle given in § 117, we have, if 0 denote the angle between a and ß, 144. If a be the position vector of a point A in space whose rectangular coordinates are a1 α,ɑ ̧, then which gives the angle between any two coinitial lines in terms of the coordinates of their end points. 145. By taking the vector product of a and ß, we get |