Unformatted text preview: − 2 y + 4 z = 0 , so any choice with y = 2 z will work, ±or example Q (4 , 2 , 1), or −→ v = −−→ PQ = 2 −→ + −→ k gives one such point. Un±ortunately, any third point given by this scheme will produce −→ w a scalar multiple o± −→ v , so won’t work. However, i± we set x = 0 we have − 2 y + 4 z = 12 , and one solution o± this is y = − 4 , z = 1 ,...
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
 Fall '08
 ALL
 Calculus, Vectors

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