This preview shows page 1. Sign up to view the full content.
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 v A ( ABC ) = 1 2 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 + 3 2 + 3 2 k . Looked at dierently, the two sides of ABC emanating from vertex A are represented by the vectors V = AB = 2 + + k w = AC = + 2 k...
View Full
Document
 Fall '08
 ALL
 Calculus, Vectors

Click to edit the document details