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# HW#3 - MA 116 Calculus II HW 3(Solutions Due March 4 2011 1...

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MA 116 – Calculus II HW 3 (Solutions) Due: March 4, 2011 1. Find the following by directly applying the definition of cross product, i.e., right-hand rule: a) ˆ ˆ ˆ i j k × = d) ˆ ˆ ˆ k j i = − × b) ˆ ˆ ˆ j k i × = e) ( ) 0 ˆ ˆ ˆ j j i × × = c) ( ) ˆ ˆ ˆ ˆ i j j i × × = f) ( ) 0 ˆ ˆ ˆ j j k × = 2. Let , v x y = and cos sin , sin cos v x y x y φ φ φ φ φ = + + . Show the angle between v and v φ is φ . Using the definition of dot product we have cos v v v v φ φ θ =     ( ) ( ) ( ) 2 2 2 2 (1) , cos sin , sin cos cos sin sin cos cos v v x y x y x y x xy xy y x y φ φ φ φ φ φ φ φ φ φ = + + = + + − + = +   2 2 (2) v x y = + ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 (3) cos sin cos sin cos sin cos sin v x y x y x y x y φ φ φ φ φ φ φ φ φ = + + − + = + + + = Putting (1), (2), and (3) into our definition of dot product we get cos cos . Since the dot product is only defined for 0 , cos is a one to one function for this given domain. Hence, = . θ φ θ π θ θ φ =

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