By Daniel Zelinsky and Samuel S. Saslaw (Auth.)

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4c). Thus all the algebraic operations and the concomitant properties of arrows in 3-space carry over to R"". We should Uke to call special attention to one of these vector spaces: the space of n-tuples with η = 1. An object in this vector space is a single number, [ a ] . Addition is just addi­ tion of numbers and multiphcation by a scalar is just ordinary number multiplication. Thus the set of all scalars is being con­ sidered as a vector space. You might say scalars are vectors! Hopefully this is not infinitely confusing, because we shall have much use for this vector space.

And the angle between ν and w is w = 61 v i + 62V2 + COS" &3V3, αΦι + α2&2 + azbz L iai' + a,' + az'V'Hbi' + + bz'y''] 6. 10. 39 Prove Schwarzas inequality: For every six numbers a i , a2, a s , 6i, 62, 63, we have (ai^ + ai" + as^) (bi^ + 62' + 63') > (ai6i + 0262 + azbzy. This one is difficult analytically, but easy geometrically: Consider the vectors A = [ a i , a2, a s ] and B = [61, 62, bz] and interpret both sides of the inequaUty in these terms (for example, ai^ + (h^ + az^ = |1 A H ^ ) .

The plane through the origin perpendicular to the 0-axis (that is, the a;2/-plane; here Po is (0, 0, 0) and ν = k ) has the vector equation k · r = 0, or the scalar equation Q{x - 0) + 0 ( 2 / - 0) + 1(0 - 0) = 0, which means 0 = 0 . Clearly we could have used for ν any vector parallel to k except 0. Such a vector is sk for some nonzero scalar s, and the resultant equation would have come out sk · r = 0 or S0 = 0^ which does indeed have the same graph. 3. The plane that is the perpendicular bisector of the line segment joining (1, 2, 3) to ( — 1, 2, 5) (here Po is the midpoint ( 0 , 2 , 4 ) and ν = (1 - ( - l ) ) i + (2 - 2 ) j + (3 — 5 ) k is the vector along the line segment) is 2 (a; — 0) + 0(1/ - 2) - 2 ( 0 - 4) = 0 , or, equivalently, rc - 0 + 4 = 0.

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