Linear Algebra Solutions 84

Linear Algebra - AB In fact A is between C and B with AC = AB 4 The points P on the line AB which satisfy AP = 2 5 PB are given by P = A t-AB where

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Section 8.8 1. The given line has equations x = 3 + t (13 - 3) = 3 + 10 t, y = - 2 + t (3 + 2) = - 2 + 5 t, z = 7 + t ( - 8 - 7) = 7 - 15 t. The line meets the plane y = 0 in the point ( x, 0 , z ), where 0 = - 2+5 t , or t = 2 / 5. The corresponding values for x and z are 7 and 1, respectively. 2. E = 1 2 ( B + C ), F = (1 - t ) A + t E , where t = AF AE = AF AF + FE = AF/FE ( AF/FE ) + 1 = 2 3 . Hence F = 1 3 A + 2 3 µ 1 2 ( B + C ) = 1 3 A + 1 3 ( B + C ) = 1 3 ( A + B + C ) . 3. Let A = (2 , 1 , 4) , B = (1 , - 1 , 2) , C = (3 , 3 , 6). Then we prove - AC = t - AB for some real t . We have - AC = 1 2 2 , - AB = - 1 - 2 - 2 . Hence - AC = ( - 1) - AB and consequently C is on the line
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Unformatted text preview: AB . In fact A is between C and B , with AC = AB . 4. The points P on the line AB which satisfy AP = 2 5 PB are given by P = A + t-AB , where | t/ (1-t ) | = 2 / 5. Hence t/ (1-t ) = ± 2 / 5. The equation t/ (1-t ) = 2 / 5 gives t = 2 / 7 and hence P = 2 3-1 + 2 7 1 4 5 = 16 / 7 29 / 7 3 / 7 . 87...
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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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