Linear Algebra Solutions 94

# Linear Algebra Solutions 94 - A to plane BCD is | 1 × 1 +...

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= 1 6 | ( - A 1 A 2 × - A 1 A 3 ) · - A 1 A 4 | . 19. We have - CB = [1 , 4 , - 1] t , - CD = [ - 3 , 3 , 0] t , - AD = [3 , 0 , 3] t . Hence - CB × - CD = 3 i + 3 j + 15 k , so the vector i + j + 5 k is perpendicular to the plane BCD . Now the plane BCD has equation x + y +5 z = 9, as B = (2 , 2 , 1) is on the plane. Also the line through A normal to plane BCD has equation x y z = 1 1 5 + t 1 1 5 = (1 + t ) 1 1 5 . Hence x = 1 + t, y = 1 + t, z = 5(1 + t ). [We remark that this line meets plane BCD in a point E which is given by a value of t found by solving (1 + t ) + (1 + t ) + 5(5 + 5 t ) = 9 . So t = - 2 / 3 and E = (1 / 3 , 1 / 3 , 5 / 3).] The distance from
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Unformatted text preview: A to plane BCD is | 1 × 1 + 1 × 1 + 5 × 5-9 | 1 2 + 1 2 + 5 2 = 18 √ 27 = 2 √ 3 . To Fnd the distance between lines AD and BC , we Frst note that (a) The equation of AD is P = 1 1 5 + t 3 3 = 1 + 3 t 1 5 + 3 t ; (b) The equation of BC is Q = 2 2 1 + s 1 4-1 = 2 + s 2 + 4 s 1-s . 97...
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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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