Introduction to Quaternions, with Numerous ExamplesMacmillan, 1873 - 227 σελίδες |
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Αποτελέσματα 1 - 5 από τα 75.
Σελίδα 10
... hence vector CB CA + AB = ß + a , = and = CD + DB = ma + nß ; ••• a + ẞ ́ = ma + nß . Hence ( Art . 6 ) m = 1 , n = 1 , i.e. the opposite sides of the paral- lelogram are equal . Again , as vectors , AO + OB = AB = CD = = CO + OD ; And ...
... hence vector CB CA + AB = ß + a , = and = CD + DB = ma + nß ; ••• a + ẞ ́ = ma + nß . Hence ( Art . 6 ) m = 1 , n = 1 , i.e. the opposite sides of the paral- lelogram are equal . Again , as vectors , AO + OB = AB = CD = = CO + OD ; And ...
Σελίδα 11
... hence ( Art . 4 ) BA is parallel to DE , and equal to 2DE . Again , BG + GABA F G B D Ꮆ = 2DE = 2 ( DG + GE ) . Now vector BG is along GE , and vector GA along DG . .. ( Art . 6 ) BG = 2GE , GA = 2GD , D whence the same is true of the ...
... hence ( Art . 4 ) BA is parallel to DE , and equal to 2DE . Again , BG + GABA F G B D Ꮆ = 2DE = 2 ( DG + GE ) . Now vector BG is along GE , and vector GA along DG . .. ( Art . 6 ) BG = 2GE , GA = 2GD , D whence the same is true of the ...
Σελίδα 12
... hence CG is in the same straight line with GF , and equal to 2GF . Ex . 5. When , instead of D and E being the middle points of the sides , they are any points whatever in those sides , it is required to find G and the point in which CG ...
... hence CG is in the same straight line with GF , and equal to 2GF . Ex . 5. When , instead of D and E being the middle points of the sides , they are any points whatever in those sides , it is required to find G and the point in which CG ...
Σελίδα 14
... hence BG2 = BA + AG2 = 2a + 2ẞ + AG ̧ ; 3 .. AG1 = 2 ( 2ẞ − a ) ; 2 - line AG == line DA and similarly of the others . 3 = 2DG ,, 3 Ex . 7. The middle points of the lines which join the points of bisection of the opposite sides of a ...
... hence BG2 = BA + AG2 = 2a + 2ẞ + AG ̧ ; 3 .. AG1 = 2 ( 2ẞ − a ) ; 2 - line AG == line DA and similarly of the others . 3 = 2DG ,, 3 Ex . 7. The middle points of the lines which join the points of bisection of the opposite sides of a ...
Σελίδα 16
... hence the line PQ is parallel to BC . The method pursued in this example leads to the solution of all similar problems . It consists , as we have already stated , in reach- ing the points P and Q respectively by two different routes ...
... hence the line PQ is parallel to BC . The method pursued in this example leads to the solution of all similar problems . It consists , as we have already stated , in reach- ing the points P and Q respectively by two different routes ...
Άλλες εκδόσεις - Προβολή όλων
Introduction to Quaternions, with Numerous Examples Philip Kelland,Peter Guthrie Tait Πλήρης προβολή - 1873 |
Introduction to Quaternions, with Numerous Examples. Philip Kelland Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2010 |
Introduction to Quaternions, with Numerous Examples P. Kelland,P. G. Tait Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2017 |
Συχνά εμφανιζόμενοι όροι και φράσεις
ABCD aßy axis centre chord circle cone conjugate diameters constant diagonals drawn ellipse ellipsoid equal example find the equation find the locus given lines given point given straight lines gives Hence hyperbola latus rectum line of intersection line which joins m₁ mean point meet middle points multiplication notation nẞ operating parabola parallelepiped parallelogram prove quadrilateral Quaternions right angles rotation Sapa Saß scalar second order semi-diameters shews sides Similarly simple shear squares ß² ß³ strain subtraction Tait tangent plane tensor tetrahedron three vectors triangle unit vectors values Vaß vector parallel vector perpendicular Vẞy whence William Rowan Hamilton yẞ αβγ γαβ δαβ φρ
Δημοφιλή αποσπάσματα
Σελίδα 9 - Any two sides of a triangle are together greater than the third side.
Σελίδα 90 - A point moves so that the sum of the squares of its distances from the points (0, 0), (1, 0) is constant.
Σελίδα 10 - FG [Hypothesit. and joined towards the same parts by the straight lines BE, CH. But straight lines which join the extremities of equal and parallel straight lines towards the same parts are themselves equal and parallel.
Σελίδα 91 - Thus a parabola is the locus of a point which moves so that its distance from a fixed point is equal to its distance from a fixed straight line (see fig.
Σελίδα 71 - Find the locus of a point whose distances from two given straight lines are in a given ratio.
Σελίδα 67 - Find the locus of a point such that the ratio of its distances from two fixed points is constant.
Σελίδα 6 - Hamilton extended algebra to space : ' ' He had done a considerable amount of good work, obstructed as he was, when about the year 1843, he perceived clearly the obstruction to his progress in the shape of an old law, which prior to that time, had appeared like a law of common sense. The law in question is known as the commutative law of multiplication. Presented in its simplest form it is nothing more than this : ' five times three is the same as three times five' ; more generally, it appears under...
Σελίδα 91 - IF we define a conic section as "the locus of a point which moves so that its distance from a fixed point bears a constant ratio to its distance from a fixed straight line
Σελίδα 153 - a where p is the perpendicular from the centre on the tangent plane, r the distance from the focus, and A, B the constants of integration.
Σελίδα 7 - A vector is the representative of transference through a given distance, in a given direction.