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Engineering Calculus Notes 92

# Engineering Calculus Notes 92 - R 3 We can think of the...

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80 CHAPTER 1. COORDINATES AND VECTORS Proof. If two vectors are linearly dependent, we can write both of them as scalar multiples of the same vector, and Δ ( r −→ v,s −→ v ) = rs Δ ( −→ v, −→ v ) = Δ ( s −→ v,r −→ v ) = Δ ( r −→ v,s −→ v ) where the last equality comes from skew-symmetry. So Δ ( r −→ v,s −→ v ) equals its negative, and hence must equal zero. To prove the reverse implication, write −→ v i = x i −→ ı + y i −→ , i = 1 , 2, and suppose Δ ( −→ v 1 , −→ v 2 ) = 0. This translates to x 1 y 2 x 2 y 1 = 0 or x 1 y 2 = x 2 y 1 . Assuming that −→ v 1 and −→ v 2 are both not vertical ( x i negationslash = 0 for i = 1 , 2), we can conclude that y 2 x 2 = y 1 x 1 which means they are dependent. We leave it to you to show that if one of them is vertical (and the determinant is zero), then either the other is also vertical, or else one of them is the zero vector. Of course, Corollary 1.6.3 can also be proved on geometric grounds, using Proposition 1.6.1 (Exercise 7 ). Oriented Areas in Space Suppose now that A , B and C are three noncollinear points in
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Unformatted text preview: R 3 . We can think of the ordered triple ( A < B < C ) as deFning an oriented triangle, and hence associate to it a “signed” area. But which sign should it have—positive or negative? The question is ill-posed, since the words “clockwise” and “counterclockwise” have no natural meaning in space: even when A , B and C all lie in the xy-plane, and have positive orientation in terms of the previous subsection, the motion from A to B to C will look counterclockwise only when viewed from above the plane; viewed from underneath , it will look clockwise . When the plane containing A , B and C is at some cockeyed angle, it is not at all clear which viewpoint is correct. We deal with this by turning the tables: 15 the motion, instead of being inherently “clockwise” or “counterclockwise”, determines which side of the 15 No pun intended! :-)...
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