Engineering Calculus Notes 61

# Engineering - ab cos θ or 2 ab cos θ = a 2 b 2 − c 2(1.16 We can compute the right-hand side of this equation by substituting the expressions

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1.4. PROJECTION OF VECTORS; DOT PRODUCTS 49 How do we calculate this projection from the entries of the two vectors? To this end, we perform a theoretical detour. 9 Suppose −→ v = ( x 1 ,y 1 ,z 1 ) and −→ w = ( x 2 ,y 2 ,z 2 ); how do we determine the angle θ between them? If we put them in standard position, representing −→ v by −−→ O P and −→ w by −−→ O Q (Figure 1.29 ), then we have a triangle △O PQ O P ( x 1 ,y 1 ,z 1 ) Q ( x 2 ,y 2 ,z 2 ) a b c θ −→ v −→ w Figure 1.29: Determining the Angle θ with angle θ at the origin, and two sides given by a = | −→ v | = r x 2 1 + y 2 1 + z 2 1 b = | −→ w | = r x 2 2 + y 2 2 + z 2 2 . The distance formula lets us determine the length of the third side: c = dist( P,Q ) = R x 2 + y 2 + z 2 . But we also have the Law of Cosines (Exercise 13 ): c 2 = a 2 + b 2 2
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Unformatted text preview: ab cos θ or 2 ab cos θ = a 2 + b 2 − c 2 . (1.16) We can compute the right-hand side of this equation by substituting the expressions for a , b and c in terms of the entries of −→ v and −→ w : a 2 + b 2 − c 2 = ( x 2 1 + y 2 1 + z 2 1 ) + ( x 2 2 + y 2 2 + z 2 2 ) − ( △ x 2 + △ y 2 + △ z 2 ) . Consider the terms involving x : x 2 1 + x 2 2 − △ x 2 = x 2 1 + x 2 2 − ( x 1 − x 2 ) 2 = x 2 1 + x 2 2 − ( x 2 1 − 2 x 1 x 2 + x 2 2 ) = 2 x 1 x 2 . 9 Thanks to my student Benjamin Brooks, whose questions helped me formulate the approach here....
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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.

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