Introduction to Quaternions: With Numerous ExamplesMacmillan, 1882 - 232 σελίδες |
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Σελίδα 13
... intersection be respectively G ,, G2 , G3 . E and B C Retaining the notation of Ex . 4 , we have BD = 3a , CE = 3ẞ ; and .. BG = xBE = x ( 2a + 3B ) BG = BD + DG , 3 = 3a + yDA = 3a + y ( CA - CD ) = 3a + y ( 2ẞ - a ) ; .. 2x = 3 - y ...
... intersection be respectively G ,, G2 , G3 . E and B C Retaining the notation of Ex . 4 , we have BD = 3a , CE = 3ẞ ; and .. BG = xBE = x ( 2a + 3B ) BG = BD + DG , 3 = 3a + yDA = 3a + y ( CA - CD ) = 3a + y ( 2ẞ - a ) ; .. 2x = 3 - y ...
Σελίδα 26
... intersection of bisectors of the sides of a triangle from the opposite angles , the point of intersection of per- pendiculars on the sides from the opposite angles , and the point of intersection of perpendiculars on the sides from ...
... intersection of bisectors of the sides of a triangle from the opposite angles , the point of intersection of per- pendiculars on the sides from the opposite angles , and the point of intersection of perpendiculars on the sides from ...
Σελίδα 27
... intersect in O , then HA = bB - ba cos C , = - b ( B — a cos C ' ) , A and KB : = a ( a – ẞ cos C ' ) . - H B Now COCA + AO , and also = CB + BO , gives - BB + yb ( B – ca cos C ) = aa + xa ( a – B cos C ) , cos C ... ax = α b cos C sin ...
... intersect in O , then HA = bB - ba cos C , = - b ( B — a cos C ' ) , A and KB : = a ( a – ẞ cos C ' ) . - H B Now COCA + AO , and also = CB + BO , gives - BB + yb ( B – ca cos C ) = aa + xa ( a – B cos C ) , cos C ... ax = α b cos C sin ...
Σελίδα 30
... intersections of PQ , P'Q ' , & c . shall be in the angular points of a parallelogram EFGH constructed from PQRS as P'Q'R'S ' is constructed from ABCD . 5. The quadrilateral formed by bisecting the sides of a quadrilateral and joining ...
... intersections of PQ , P'Q ' , & c . shall be in the angular points of a parallelogram EFGH constructed from PQRS as P'Q'R'S ' is constructed from ABCD . 5. The quadrilateral formed by bisecting the sides of a quadrilateral and joining ...
Σελίδα 44
... intersection of OG , CD , and vector from 0 to the point of bisection of AF , as also to that of BE , and therefore to the intersection of AF , BE = 1/2 ( 8 + 1 ( 8+ a + B ) , hence vector which joins the points of intersection of ...
... intersection of OG , CD , and vector from 0 to the point of bisection of AF , as also to that of BE , and therefore to the intersection of AF , BE = 1/2 ( 8 + 1 ( 8+ a + B ) , hence vector which joins the points of intersection of ...
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
aßy axis bisects Cambridge centre Chemistry chord circle cone conjugate diameters constant Crown 8vo diagonals drawn ELEMENTARY TREATISE ellipse ellipsoid equal equation example Fcap find the equation find the locus given lines given point given straight lines gives Hence hyperbola intersection Lecturer middle points multiplication notation numerous Illustrations Owens College P. G. TAIT P₁ parabola parallelepiped parallelogram perpendicular PHYSICAL Professor quaternion right angles rotation Royal Sapa Saß scalar SCIENCE Second Edition second order semi-diameters shews Similarly simple shear sphere Spop squares ß² Tait tangent plane tetrahedron tion triangle unit vectors values Vaß vector parallel vector perpendicular Vẞy whence yẞ αβγ φρ
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