Engineering Calculus Notes 98

Engineering Calculus Notes 98 - (which is the one we used...

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86 CHAPTER 1. COORDINATES AND VECTORS at the yz -plane from the direction of the positive x -axis, then we see a “clockwise” triangle, so the oriented area is 1 2 −→ ı —it points in the direction of the negative x -axis. Finally, the projection onto the xz -plane has vertices A (2 , 4), B (4 , 5), and C (3 , 3). We sketched this in Figure 1.38 , and calculated a negative signed area of 3 / 2. Note, however, that if we look at our triangle from the direction of the positive y -axis, we see a counterclockwise triangle. Why the discrepancy? The reason for this becomes clear if we take into account not just the triangle, but also the axes we see. In Figure 1.38 , we sketched the triangle with positive abcissas pointing “east” and ordinates pointing “north”, but in Figure 1.46 the positive x -axis points “west” from our point of view. In other words, the orientation of the x -axis and z -axis (in that order) looks counterclockwise only if we look from the direction of the negative y -axis. From this point of view–that is, the direction of
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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...
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