Also let OB, and B, C, be the projections of B and y on a. (a + By)=(ay)+(By) ...... Form (ii). 126. The Vector Product. For this it is necessary to consider separately the two cases: (1) for coplanar vectors, (2) for space vectors. 126 a. Let a, ß, y be any three coplanar vectors, Proof. In the figure OA = a, AB=ß, OC=y and d=a+B. Now [a + B│y] = [dy]= OBEC (as a vector) [ay]=OADC= OGHC [By]= ABED=GBEH, the sense of each area being given by the order of the Imagine the (last) figure to represent a triangular prism, the area OBEC being in the plane of the paper while OADC and ABED are inclined to it. In this case [ay] and [By] are aspect-areas in different planes; we therefore require a definition for the sum of such aspect-areas. Definition :-If two areas with definite aspects be represented as vectors, then their sum is that area which is represented by the sum of the vectors. The aspect orts of [ay], [By], and [a+ẞy] all point away from the plane of the paper, or what is the same thing, those of the two former point from the faces outside the prism whilst that of the latter points from the face inside the prism. We have seen (§ 112) that the magnitude of a vector product does not depend on the shape of the area but only on its size, hence we may replace the areas ОADC, ABED and OBEC by rectangles of the same magnitude and sense. Through A and D draw planes perpendicular to y, we shall then get the right prism O'AB'E'C', as in figure, then as vectors The three vectors representing the areas are perpendicular to the sides C'D, DE', and E'C', their magnitudes are C'D. DA, DE'. DA, and C'E'. DA, that is, they are proportional to the lengths of C'D, DE', and C'E'. R [6] 8 ปี To find the sum of these vectors [By] and [a], set off from any point R, RQ proportional to the length of E'D and 1 to it, from Q set off QP proportional and [a] perpendicular to DC', then RP will be proportional and perpendicular to C'E'. For by construction PQR and C'DE' are similar triangles and since as vectors can always be considered as the sum of two vectors B and where σ = y + 8 + ε + .... Hence from the Distributive Law it follows that [a\B+y+d+e+...]=[aß]+[a│y+d+e+...] = = [ad] + [BS]+[yd] + [ae] + [Be] +[ye]. Exactly similar results hold for the Scalar Product. Since the Commutative Law does not hold for the Vector Product, care must be taken in such multiplications to preserve the proper order of the vectors, the vectors of the left-hand factor must always precede those of the right-hand factor (or else the sign must be changed). 128. In the formula let y = ma, then [a\B+y] = [aß] + [ay] [a]B + ma] = [aß] + m [aa] Hence a vector product remains unaltered if to one factor a multiple of the other be added. This shews that if the product and one factor are known, then the other factor is not uniquely determined; the division therefore belonging to this multiplication is not a unique operation. The geometrical meaning of the theorem is: Parallelograms on the same base and between the same parallels are equal. If the area and base are given, the other side may be any vector drawn between the parallels. The student should be careful to notice that from the equation which as we have seen is only true if either a = 0, or B = y, or B-ya. The latter says ẞ-y=ma as above. If however [a] = [ay] shall be true whatever a may be, then of necessity By, because By cannot be parallel to two different directions. The two equations [aß] = [ay], [a'ß] = [a'v], where a and a' are not parallel, can only be true if ß=y. Such reasoning enables us to dispense with the operation of Division by Vectors. This operation is complicated and will not be considered at all. It leads to the much more complicated Theory of Quaternions. 129. Summary of Results. 130. The remainder of this chapter will be devoted to various direct applications of the preceding theory to |