Linear Algebra Solutions 93

Linear Algebra Solutions 93 - Hence the shortest distance...

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Unformatted text preview: Hence the shortest distance between lines AC and BD is equal to 0 −6 −2 · 1 E 1 5 3 | AB ·(X × Y )| √ = =√ . ||X × Y || 62 62 18. Let E be the foot of the perpendicular from A4 to the plane A1 A2 A3 . Then 1 vol A1 A2 A3 A4 = ( area ∆A1 A2 A3 ) · A4 E. 3 Now E E 1 area ∆A1 A2 A3 = || A1 A2 × A1 A3 ||. 2 Also A4 E is the length of the projection of A1 A4 onto the line A4 E . (See figure above.) E Hence A4 E = | A1 A4 ·X |, where X is a unit direction vector for the line A4 E . We can take E E A1 A2 × A1 A3 X= . E E || A1 A2 × A1 A3 || Hence E vol A1 A2 A3 A4 = = E E E E 1 | A1 A4 ·(A1 A2 × A1 A3 )| || A1 A2 × A1 A3 || E E 6 || A A × A A || E E E 1 | A1 A4 ·(A1 A2 × A1 A3 )| 6 96 1 2 1 3 ...
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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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