Linear Algebra Solutions 90

# Linear Algebra Solutions 90 - where =(1 r)s =(1 r(1 s r(1 t...

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where α = (1 - r ) s, β = (1 - r )(1 - s ) + r (1 - t ) , γ = rt. Then 0 < α < 1 , 0 < γ < 1 , 0 < β < (1 - r ) + r = 1. Also α + β + γ = (1 - r ) s + (1 - r )(1 - s ) + r (1 - t ) + rt = 1 . 13. The line AB is given by P = A + t [3 , 4 , 5] t , or x = 6 + 3 t, y = - 1 + 4 t, z = 11 + 5 t. Then B is found by substituting these expressions in the plane equation 3 x + 4 y + 5 z = 10 . We Fnd t = - 59 / 50 and consequently B = µ 6 - 177 50 , - 1 - 236 50 , 11 - 295 50 = µ 123 50 , - 286 50 , 255 50 . Then AB = || - AB || = || t 3 4 5 || = | t | p 3 2 + 4 2 + 5 2 = 59 50 × 50 = 59 50 . 14. Let A = ( - 3 , 0 , 2) , B = (6 , 1 , 4) , C = ( - 5 , 1 , 0). Then the area of triangle ABC is 1 2 || -
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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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