The Outlines of QuaternionsLongmans, Green, 1894 - 190 páginas |
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Página 64
... intersection of the two planes , PQ . Let E be any point in that line . Produce OB , and draw EF and EG parallel respectively to AB and CD . Then , OF OBOG OD OE - = OA ' OE OC Let another vector , OH , be drawn in the 64 EQUALITY AND ...
... intersection of the two planes , PQ . Let E be any point in that line . Produce OB , and draw EF and EG parallel respectively to AB and CD . Then , OF OBOG OD OE - = OA ' OE OC Let another vector , OH , be drawn in the 64 EQUALITY AND ...
Página 66
... intersect in , or are parallel to , a common line are said to be Collinear . For example , the quaternions OB . OA and OD . OC , fig . 19 , 42 ° , are collinear ; and OL . ON , OM . ON , fig . 18 , 35 ° , are also collinear , whatever ...
... intersect in , or are parallel to , a common line are said to be Collinear . For example , the quaternions OB . OA and OD . OC , fig . 19 , 42 ° , are collinear ; and OL . ON , OM . ON , fig . 18 , 35 ° , are also collinear , whatever ...
Página 95
... intersection . 80 ° . Let σ = Vaß , and V.oVẞy = ySoß – ßSyo . V. VaẞV By = yS . ( Vaß ) B - BS . yVaß = YSBVaẞ- BSyaß . 99 But SẞVaßo , because ẞ and Vaß are at right angles . Therefore , We have , therefore , 81 ° . Adding V. VaẞVẞy ...
... intersection . 80 ° . Let σ = Vaß , and V.oVẞy = ySoß – ßSyo . V. VaẞV By = yS . ( Vaß ) B - BS . yVaß = YSBVaẞ- BSyaß . 99 But SẞVaßo , because ẞ and Vaß are at right angles . Therefore , We have , therefore , 81 ° . Adding V. VaẞVẞy ...
Página 141
... intersection of the planes OAB and ODE ; or , L is the point in which the arcs BA and DE duced meet . The arcs AB and ED produced meet in L ' , the point upon the sphere diametrically opposite to L ; and CL ' is a quadrant . To find TA ...
... intersection of the planes OAB and ODE ; or , L is the point in which the arcs BA and DE duced meet . The arcs AB and ED produced meet in L ' , the point upon the sphere diametrically opposite to L ; and CL ' is a quadrant . To find TA ...
Página 155
... intersection of two circles is perpendicular to the join of their centres . Let the circles intersect in A and A ' ; let their centres be K and K ' ; let KK ' AA ' = 8. Then , and = ولا - ( KA — y ) 2 = K'A2 = K'A'2 = ( KA ' — y ) 2 , S ...
... intersection of two circles is perpendicular to the join of their centres . Let the circles intersect in A and A ' ; let their centres be K and K ' ; let KK ' AA ' = 8. Then , and = ولا - ( KA — y ) 2 = K'A2 = K'A'2 = ( KA ' — y ) 2 , S ...
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Términos y frases comunes
arcs axis bisects centre chord circle coinitial vectors collinear commutative law conic conjugate consequently coplanar coplanar quaternions Crown 8vo diameter differential diplanar direction equal equation geometric given Hence hyperbola Illustrations intersection latus rectum Let OA locus mean point mẞ Multiplying OABC operation opposite P₁ parabola parallel parallelopiped perpendicular plane AOB positive or negative reciprocal right angles right quaternion right versors rotation Saß Saßy scalar Scientific Works published Similarly sphere SrVq ẞa symbols tangent TaTẞ tensor three vectors triangle Tẞ TẞTa unit-vector variable Vaß vector versor vertex VqVr VrVq Vẞy Woodcuts α α αβ β α βα φρ ОА ов ос тв وو ᎢᏰ
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