Linear Algebra Solutions 86

# Linear Algebra Solutions 86 - 11 and P =-2 1 3 2 11 3 1 1...

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(ii) cos BAC = ( - AB · - AC ) / ( AB · AC ) , where - AB = [1 , 2 , 3] t and - AC = [5 , - 4 , 1] t . Hence cos BAC = 5 - 8 + 3 14 42 = 0 . Hence ABC = π/ 2 radians or 90 . (iii) cos ACB = ( - CA · - CB ) / ( CA · CB ) , where - CA = [ - 5 , 4 , - 1] t and - CB = [ - 4 , 6 , 2] t . Hence cos ACB = 20 + 24 - 2 42 56 = 42 42 56 = 42 56 = 3 2 . Hence ACB = π/ 6 radians or 30 . 7. By Theorem 8.5.2, the closest point P on the line AB to the origin O is given by P = A + t - AB , where t = - AO · - AB AB 2 = - A · - AB AB 2 . Now A · - AB = - 2 1 3 · 3 1 1 = - 2 . Hence t = 2
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Unformatted text preview: / 11 and P = -2 1 3 + 2 11 3 1 1 = -16 / 11 13 / 11 35 / 11 and P = (-16 / 11 , 13 / 11 , 35 / 11). Consequently the shortest distance OP is given by s µ-16 11 ¶ 2 + µ 13 11 ¶ 2 + µ 35 11 ¶ 2 = √ 1650 11 = √ 15 × 11 × 10 11 = √ 150 √ 11 . 89...
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## This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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