given vectors are the concurrent edges through one end of that diagonal. Conversely, if we are given p we can find the components in three given directions. For let OP=p be the given vector, and OA, OB, and OC, the given directions. Draw through P a line parallel to OC, this will cut the plane containing OA and OB in some point S. Through S draw lines SA and SB parallel to OB and OA respectively cutting OA and OB in A and B. Then by moving the parallelogram OASB parallel to itself so that S moves along y to P, O, A and B will move along y to C, R, and Q respectively, and a parallelopiped will be generated, of which three concurrent edges will be OA, OB, and OC, the required components of p. We might of course have started the decomposition by drawing PR or PQ parallel to B or a, and then proceeded as before; in whichever way we start, it is seen that the equations (i) to (iii) hold and therefore we always get the same parallelopiped. The resolution is therefore unique. 51. If a, ẞ, y be three independent vectors, then any other vector & can be expressed in terms of them so that 8 = aa+bB+cy, where a, b, c, are definite scalars. For 8 can be decomposed in the directions of a, B, and y, calling the components a', B', y' we have d = a' + B' + y'. But a and a', B and B', y and y are like vectors and therefore a'aa, ß' = bß, y' = cy, which ment. 52. proves the state Coordinates in Space. If the lengths of OA, OB, and OC are x, y, and z, then x, y, z are called the coordinates of the point P and the three directions through O the axes of coordinates. Given the origin and the directions of the three axes, then the position vector of a point determines its coordinates and conversely the coordinates determine the position vector of a point. Let к, λ, μ, be the orts (called reference orts) in the directions Ox, Oy, and Oz, and x, y, z the coordinates, then By means of the relation (i) or (ii) any vector equation can be transformed into three scalar equations; for instance if p = 2a + 3B, α = α141 + A2l2 + A33 B = b12+b2l2 + b313, then x = 2α1 + 3b1 y= 2α2+3b2 2= 2a+3b3. One vector equation is therefore equivalent to three scalar equations. Theorem. Since the resolution into three components is unique, it follows that equal vectors have equal components, and vectors having equal components in any three directions in space are themselves equal. 53. In the particular case of rectangular orts, as in the figure, R x the components of any given vector can be easily calculated. By construction PS is 1 to the plane OASB, and PSO a right angle, hence Fig. 32. where 01= POA, and 02=POB. ...... ...(iii) ...(iv), ...... cos 01, cos 2, cos 03 are called the direction cosines of OP because they determine its direction. or If we square and add (ii), (iii), and (iv) we get x2+y2+22=r2(cos2 01+cos2 02+cos2 03), so that 01, 02, 0, are not independent. It will be seen that the only angles we really require to determine the components are COP and AOS, for The distance OP (denoted by r), the angle COP (denoted by 0), and the angle AOS (denoted by p) are usually called the Polar Coordinates of P. The angles corresponding to and are called in Astronomy and Navigation the co-latitude and the longitude. Conversely, given the rectangular coordinates of a point, the vector from the origin to the point can be found in magnitude, sense and direction. and the direction and sense by the three direction cosines. Since x, y, and z may be either positive or negative we get eight different arrangements of signs. The planes xOy, yOz, and z0x divide all space into 8 parts called octants, each octant having its own arrangement of coordinate signs. Hence when we know the signs of the coordinates of a point, we know in which octant the point is situated. and 54. If a and ẞ be the position vectors of two points A and B, then AB=(b1-a1)+(b2−ɑ2) 12 + (b3−Ag) 43. Comparing this with (1) of § 53 we see that if c denote the length of AB b1-a1 is the x, b2-ɑ, is the y, b2-a, is the z of the preceding section. Hence the direction cosines of the vector p-a are (1) The four diagonals of a parallelopiped meet and bisect each other. If a, B, y represent three concurrent edges, then the diagonals are given by a +B+y, B+y-a, and a+y-B. The position vectors of the mid-points are (2) The lines joining the mid-points of opposite edges bisect the diagonals and one another. (3) a, B, y, d being the position vectors of four points in space, shew that in whatever order the points be joined to form a quadrilateral, the lines joining the mid-points of adjacent sides form a parallelogram. (4) a, B, y being three independent vectors, if we join their end points we obtain a four-faced figure called a tetrahedron. It has 4 vertices, and 6 edges. Non-intersecting edges are called opposite edges. Find as vectors the lines joining the mid-points of opposite edges. (6) In figure to (4) suppose OA, OB, and OC produced to A1, B1, and C1 where OA1=20A, OB1=30B, ÓC1=40C. Shew that the points where AB and Д1B1, BC and B1C1, CA and C11 intersect are collinear. (7) Extend (5) to the case where A1B1 C1 are any points on the three concurrent edges. (8) A rectangular room is 13 ft. high, 20 ft. long and 18 ft. wide, find the lengths of the lines joining one corner to all the other corners. (9) A point in a rectangular room is distant 3 ft. from the floor, 7 ft. from a side wall and 5 ft. from the end wall, find the distance of the point from the corner where the floor and the two walls meet. Find also the angles the line joining the corner to the point make with the lines of intersection of floor and walls. (10) A point in the room (question 9) is distant 11 ft. from the corner, the line joining it to the corner makes angles 25° and 40° with the intersection of the floor with the end and side walls. Find the distance of the point from floor and walls. (11) The rectangular coordinates of two points are 3, 1, 2, 1, 3, 5; find as vectors the lines joining these points to the origin and to one another, and find the distance between the points. |