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1 We ve been given the Cartesian vectors a = (5, 5) and b = ( 5,5)
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When we multiply a on the left by 4, we get 4a = 4 (5, 5) = {(4 5), [4 ( 5)]} = (20, 20) When we multiply b on the left by 4, we get 4b = 4 ( 5,5) = {[ 4 ( 5)],( 4 5)} = (20, 20) 2 The Cartesian vector a has the coordinates xa = 5 and ya = 5, so it terminates in the fourth quadrant The direction angle for the polar form of a can be found using the conversion formula for a vector in the fourth quadrant, giving us qa = 2p + Arctan ( 5/5) = 2p + Arctan ( 1) = 2p + ( p /4) = 7p /4 The magnitude of a is found by the distance formula ra = [52 + ( 5)2]1/2 = (25 + 25)1/2 = 501/2 Therefore, the polar version of a is a = (7p /4,501/2) The Cartesian version of b has xb = 5 and yb = 5 It terminates in the second quadrant Using the conversion formula for the direction angle of a vector in that quadrant, we get qb = p + Arctan [5/( 5)] = p + Arctan ( 1) = p + ( p /4) = 3p /4 The magnitude of b is rb = [( 5)2 + 52]1/2 = (25 + 25)1/2 = 501/2 Therefore, the polar version of b is b = (3p /4,501/2) When we multiply a on the left by 4, we get 4a = 4 (7p /4,501/2) = (7p /4,8001/2) When we multiply b on the left by 4, we get 4b = 4 (3p /4,501/2) = (3p /4, 8001/2) = (7p /4,8001/2)
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In the last step in the equation for 4b, we must take the absolute value of the negative magnitude coordinate, because we can t allow a vector to have negative magnitude We do this by reversing the direction, in this case by adding p to the angle 3 We want to prove that positive-scalar multiplication is right-hand distributive over vector subtraction in the Cartesian xy plane Let s start with (a b)k+ where a = (xa,ya), b = (xb,yb), and k+ is a positive real number Expanding the vectors into their ordered pairs in our initial expression, we get (a b)k+ = [(xa xb),( ya yb)]k+ The definition of right-hand scalar multiplication tells us that we can morph this equation to obtain (a b)k+ = {[(xa xb)k+],[( ya yb)]k+} The right-hand distributive law for real numbers allows us to transform the equation further, getting (a b)k+ = [(xak+ xbk+),( yak+ ybk+)] Let s put this equation aside for moment We ll come back to it! Now, instead of the product of the vector difference and the constant, let s start with the difference between the products ak+ bk+ We can expand the individual vectors into ordered pairs to get ak+ bk+ = (xa,ya)k+ (xb,yb)k+ By the definition of right-hand scalar multiplication, we have ak+ bk+ = (xak+,yak+) (xbk+,ybk+) When we add the elements of the ordered pairs individually to get a new ordered pair, we obtain ak+ bk+ = [(xak+ xbk+),( yak+ ybk+)] The right-hand side of this equation is the same as the right-hand side of the equation we put aside a minute ago That equation, once again, is (a b)k+ = [(xak+ xbk+),( yak+ ybk+)]
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Taken together, the above two equations show us that (a b)k+ = ak+ bk+ 4 We ve been given the Cartesian vectors a = (4,4) and b = ( 7,7) We can define the coordinate values as xa = 4, xb = 7, ya = 4, and yb = 7 The Cartesian dot product of a and b, in that order, is therefore a b = xaxb + yayb = 4 ( 7) + 4 7 = 28 + 28 = 0 The Cartesian dot product of b and a, in that order, is b a = xbxa + ybya = 7 4 + 7 4 = 28 + 28 = 0 5 The Cartesian vector a has the coordinates xa = 4 and ya = 4, so it terminates in the first quadrant The direction angle for the polar form of a is therefore qa = Arctan (4/4) = Arctan 1 = p /4 The magnitude of a is ra = [42 + 42]1/2 = (16 + 16)1/2 = 321/2 so the polar form of a is a = (p /4,321/2) The Cartesian vector b has xb = 7 and yb = 7 It terminates in the second quadrant Using the conversion formula for the direction angle of a vector in the second quadrant, we get qb = p + Arctan [7/( 7)] = p + Arctan ( 1) = p + ( p /4) = 3p /4 The magnitude of b is rb = [( 7)2 + 72]1/2 = (49 + 49)1/2 = 981/2 Therefore, the polar version of b is b = (3p /4,981/2)
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