An Elementary Treatise on Quaternions |
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Page xv
... Hence the equations just written are true of any set of rectangular unit - vectors i , j , k . § 72 . The product , and the quotient , of two vectors at right angles to each other is a third perpendicular to both . Hence Κα = -α , and ...
... Hence the equations just written are true of any set of rectangular unit - vectors i , j , k . § 72 . The product , and the quotient , of two vectors at right angles to each other is a third perpendicular to both . Hence Κα = -α , and ...
Page 5
... Hence any expression , as AB , denoting a line considered with reference to direction as well as length , contains implicitly three numbers , and all lines parallel and equal to AB depend in the same way upon the same three . Hence ...
... Hence any expression , as AB , denoting a line considered with reference to direction as well as length , contains implicitly three numbers , and all lines parallel and equal to AB depend in the same way upon the same three . Hence ...
Page 6
... Hence these cöordinates enter linearly into the expression for a vector . 20. ] But we also see that if C and A coincide ( and C may be any point ) AC = 0 , for no vector is then required to carry A to C. Hence the above relation may be ...
... Hence these cöordinates enter linearly into the expression for a vector . 20. ] But we also see that if C and A coincide ( and C may be any point ) AC = 0 , for no vector is then required to carry A to C. Hence the above relation may be ...
Page 8
... Hence and the equation y = aa + BB . P = ( x + za ) a + ( y + zb ) ß , @ = ( § + $ a ) a + ( n + 56 ) ß , p = w now requires only the two numerical conditions x + za = & + 5a , y + zb = n + 5b . - 27. ] The Commutative and Associative ...
... Hence and the equation y = aa + BB . P = ( x + za ) a + ( y + zb ) ß , @ = ( § + $ a ) a + ( n + 56 ) ß , p = w now requires only the two numerical conditions x + za = & + 5a , y + zb = n + 5b . - 27. ] The Commutative and Associative ...
Page 11
... Hence Aa + Bb + Cc = 3 ( AB + BC + CA ) = 0 ; which ( § 21 ) proves the proposition . Also Aa = AB + BC = AB — ≥ ( CA + AB ) = ≥ ( AB — CA ) = { ( AB + AC ' ) , results which are sometimes useful . They may be easily verified by ...
... Hence Aa + Bb + Cc = 3 ( AB + BC + CA ) = 0 ; which ( § 21 ) proves the proposition . Also Aa = AB + BC = AB — ≥ ( CA + AB ) = ≥ ( AB — CA ) = { ( AB + AC ' ) , results which are sometimes useful . They may be easily verified by ...
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Common terms and phrases
a₁ axes axis Cartesian centre of inertia Chapter circle cloth cone conjugate constant cöordinates coplanar curvature curve developable surface diameters differential direction drawn easily ellipsoid envelop equal evidently expression Extra fcap extremity fcap Find the equation Find the locus given equation given line given vectors gives Hamilton Hence hodograph integral intersection last section length linear and vector locus normal obviously once operator origin osculating osculating plane P₁ parabola parallel perpendicular properties quaternion radius rectangular represents result right angles rotation S.aßy Saß scalar scalar equations second order self-conjugate shew solution sphere spherical straight line suppose surface surface of revolution tangent plane Taylor's Theorem tensor theorem three vectors triangle unit-vectors Vaß vector function versor w₁ write written Τρ φρ