Engineering Calculus Notes 394

# Engineering Calculus Notes 394 - v 1 − 3 v 2 − 7 v 3 =...

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382 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION Substituting this into the second equation we have 2 v 1 + 3(4 v 1 + v 3 ) 3 v 3 = 0 or 10 v 1 = 0 . Thus v 1 = 0 and v 2 = v 3 . We fnd −→ u 2 = 1 2 (0 , 1 , 1) . Finally, λ = 6 leads to 4 v 1 2 v 2 +2 v 3 = 6 v 1 2 v 1 v 2 3 v 3 = 6 v 2 2 v 1 3 v 2 v 3 = 6 v 3 or 2 v 1 2 v 2 +2 v 3 = 0 2 v 1 7 v 2 3 v 3 = 0 2
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Unformatted text preview: v 1 − 3 v 2 − 7 v 3 = 0 . The frst equation says v 1 = − v 2 + v 3 and substituting this into the other two yields two copies o± − 5 v 2 − 5 v 3 = 0 or v 2 = − v 3 so v 1 = 2 v 3...
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## This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Spring '08 term at University of Florida.

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