Introduction to quaternions, by P. Kelland and P.G. Tait1873 |
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Σελίδα 13
... intersection be respectively G1 , G. , G ̧ . E and B F G ' Retaining the notation of Ex . 4 , we have BD 3a , CE = 3ẞ ; = and .. BG2 = xBE = x ( 2a + 3ẞ ) . BG2 = BD + DG3 = 3a + yDA = 3a + y ( CA - CD ) = 3a + y ( 2ẞ - a ) ; .. 2x = 3 ...
... intersection be respectively G1 , G. , G ̧ . E and B F G ' Retaining the notation of Ex . 4 , we have BD 3a , CE = 3ẞ ; = and .. BG2 = xBE = x ( 2a + 3ẞ ) . BG2 = BD + DG3 = 3a + yDA = 3a + y ( CA - CD ) = 3a + y ( 2ẞ - a ) ; .. 2x = 3 ...
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
Philip Kelland. 2o . Let AH , BK perpendiculars on the sides intersect in 0 , then HA = bß – ba cos C , = b ( B - a cos C ' ) , KB = a ( a - ẞ cos C ) . 1 Now COCA + AO , and also = CB + BO gives - bB + yb ( B − aa cos C ) = aa + xa ( a ...
Philip Kelland. 2o . Let AH , BK perpendiculars on the sides intersect in 0 , then HA = bß – ba cos C , = b ( B - a cos C ' ) , KB = a ( a - ẞ cos C ) . 1 Now COCA + AO , and also = CB + BO gives - bB + yb ( B − aa cos C ) = aa + xa ( a ...
Σελίδα 30
... intersections of PQ , P'Q ' , & c . shall be in the angular points of a parallelogram EFGH constructed from PQRS as P'QR'S ' is con- structed from ABCD . 5. The quadrilateral formed by bisecting the sides of a quadri- lateral and ...
... intersections of PQ , P'Q ' , & c . shall be in the angular points of a parallelogram EFGH constructed from PQRS as P'QR'S ' is con- structed from ABCD . 5. The quadrilateral formed by bisecting the sides of a quadri- lateral and ...
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 + a + B ) , hence vector which joins the points of intersection of diagonals 1 ...
... 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 + a + B ) , hence vector which joins the points of intersection of diagonals 1 ...
Άλλες εκδόσεις - Προβολή όλων
Introduction to Quaternions, by P. Kelland and P. G. Tait Philip Kelland Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2013 |
Introduction to Quaternions, by P. Kelland and P.G. Tait Philip Kelland Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2018 |
Introduction to Quaternions, by P. Kelland and P. G. Tait Philip Kelland Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2015 |
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 plane given point given straight lines gives Hence hyperbola latus rectum line of intersection line which joins mean point meet middle points multiplication notation nẞ operating P₁ parabola parallelepiped parallelogram prove quadrilateral Quaternions right angles rotation Sapa Saß scalar second order semi-diameters shews sides Similarly simple shear Spop 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 φρ
Δημοφιλή αποσπάσματα
Σελίδα 8 - Any two sides of a triangle are together greater than the third side.
Σελίδα 52 - Thus, for" example, he to whom the geometrical proposition, that the angles of a triangle are together equal to two right angles...
Σελίδα 89 - A point moves so that the sum of the squares of its distances from the points (0, 0), (1, 0) is constant.
Σελίδα 40 - To express the cosine of an angle of a triangle in terms of the sides. Let ABC be a triangle ; and retaining the usual notation of Trigonometry, let CB = a, CA=ß; then (vector AB)' =(a- ß)' = a'-2Saß + ß
Σελίδα 68 - Ex. 5. To find the locus of a point such that the ratio of its distances from a given point and a given straight line is constant — all in one plane. Let S be the given point, DQ the given straight line, SP = ePQ the given relation. Let vector SD = a,SP = p, DQ = yy, y being the unit vector along DQ, then eT(PQ), 5—2 gives p~ = e'PQ', where PQ is a vector, = i'(a»)" = eVa'. = a+yy; . : Sap + xa' = a', for Say = 0; and a;V = (a...
Σελίδα 72 - P is a surfaco of the second order. 3. Prove that the section of this surface by a plane perpendicular to the lin.e to which the generating lines are drawn perpendicular is a circle. 4. Prove that the locus of a point whose distances from two given straight lines have a constant ratio is a surface of the second order. 5. A straight line moves parallel to a fixed plane and is terminated by two given straight lines not in one plane ; find the locus of the point which divides the line into parts which...
Σελίδα 9 - 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 90 - 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 32 - G : shew that FG is parallel to CD. 346. From any point in the base of a triangle straight lines are drawn parallel to the sides : shew that the intersection of the diagonals of every parallelogram so formed lies in a certain straight line. 347. In a triangle ABC a straight...
Σελίδα 90 - IP 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