Engineering Calculus Notes 90

Engineering Calculus Notes 90 - 2 Δ −→ v 1 −→ v 2...

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78 CHAPTER 1. COORDINATES AND VECTORS b A b B b C b D + b A b D b C b B Figure 1.39: Oriented Quadrilaterals two nonzero vectors −→ v 1 = x 1 −→ ı + y 1 −→ −→ v 2 = x 2 −→ ı + y 2 −→  . Then the determinant using these rows Δ( −→ v 1 , −→ v 2 ) = v v v v x 1 y 1 x 2 y 2 v v v v . can be interpreted geometrically as follows. Let P ( x 1 ,y 1 ) and Q ( x 2 ,y 2 ) be the points with position vectors −→ v 1 and −→ v 2 , respectively, and let R ( x 1 + x 2 ,y 1 + y 2 ) be the point whose position vector is −→ v 1 + −→ v 2 (Figure 1.40 ). Then the signed area of △O PQ equals 1
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Unformatted text preview: 2 Δ( −→ v 1 , −→ v 2 ); but b O b P b R b Q −→ v 1 −→ v 2 Figure 1.40: Proposition 1.6.1 note that the parallelogram O PRQ has the same orientation as the triangles △O PQ and △ PRQ , and these two triangles △O PQ and △ PRQ are congruent, hence have the same area. Thus the signed area of the parallelogram O PRQ is twice the signed area of △O PQ ; in other words...
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