Unformatted text preview: − −→ (which is the one we used to calculate the signed area)–the triangle looks negatively oriented, so the oriented area should be p − 3 2 P ( − −→ ) = 3 2 −→ . This agrees with the geometric observation based on Figure 1.42 . We have seen that the oriented area v A ( △ ABC ) has projections proj −→ k v A ( △ ABC ) = 3 2 −→ k proj −→ ı v A ( △ ABC ) = − 1 2 −→ ı proj −→ v A ( △ ABC ) = 3 2 −→ . But these projections are simply the components of the vector, so we conclude that the oriented area v A ( △ ABC ) is v A ( △ ABC ) = − 1 2 −→ ı + 3 2 −→ + 3 2 −→ k . Looked at di±erently, the two sides of △ ABC emanating from vertex A are represented by the vectors −→ V = −−→ AB = 2 −→ ı + −→ + −→ k −→ w = −→ AC = −→ ı + 2 −→ − −→ k...
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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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