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76 Vector Multiplication
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Because real-number multiplication is commutative, we know that k+a = (qa,k+ra) = (qa,rak+) = a k+ As in the Cartesian case, we don t have to worry about whether we multiply on the left or the right The polar product of the vector and the positive scalar is the same either way
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An example In Fig 5-2, the polar vector (7p /4,3/2) is shown as a solid, arrowed line segment When we multiply this vector by 3 on the left, we get
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3 (7p /4,3/2) = [7p /4,(3 3/2)] = (7p /4,9/2) Multiplying by 3 on the right yields (7p /4,3/2) 3 = {7p /4,[(3/2) 3]} = (7p /4,9/2) This polar product vector is represented by a dashed, gray, arrowed line segment pointing in the same direction as the original vector, but 3 times as long
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Polar vector times negative scalar Again, consider our polar vector a = (qa,ra) Suppose that we want to multiply a on the left by a negative scalar k It s tempting to suppose that we can leave the angle the same and make
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Figure 5-2 Polar products of the scalars 3 and 3 with
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the vector (7p /4,3/2) Each radial division represents 1 unit
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the magnitude equal to k ra But that gives us a negative magnitude, which is forbidden by the rules we ve accepted for polar vectors The proper approach is to multiply the original vector magnitude ra by the absolute value of k In this situation, that s k Then we reverse the direction of the vector by either adding or subtracting p to get a direction angle that s nonnegative but smaller than 2p We define the result as the left-hand polar product of k and a, and write it as k a = [(qa + p ),( k ra)] if 0 qa < p, and k a = [(qa p ),( k ra)] if p qa < 2p Because k is negative, k is positive; therefore k ra is positive, which ensures that our scalar-vector product has positive magnitude If we multiply a on the right by k , we get the right-hand polar product of a and k , which is ak = [(qa + p ),ra( k )] if 0 qa < p, and ak = [(qa p ),ra( k )] if p qa < 2p As before, k is negative so k is positive; that means ra( k ) is positive, ensuring that our vector-scalar product has positive magnitude The commutative law assures us that for any negative scalar k and any polar vector a, it s always true that k a = ak As before, we can leave out the left-hand and right-hand jargon, and simply talk about the polar product of the vector and the scalar
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An example Look again at Fig 5-2 When we multiply the original polar vector (7p /4,3/2) by 3 on the left, we get
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3 (7p /4,3/2) = [(7p /4 p),(3 3/2)] = (3p /4,9/2) Multiplying the original polar vector by 3 on the right yields (7p /4,3/2) ( 3) = {(7p /4 p),[(3/2) 3]} = (3p /4,9/2) This result is shown as a dashed, gray, arrowed line segment pointing in the opposite direction from the original vector, and 3 times as long
78 Vector Multiplication
Are you confused
You ask, What happens when our positive scalar k+ is between 0 and 1 What happens when our negative scalar k is between 1 and 0 What do we get if the scalar constant is 0 If 0 < k+ < 1, the product vector points in the same direction as the original, but it s shorter If 1 < k < 0, the product vector points in the opposite direction from the original, and it s shorter If we multiply a vector by 0, we get the zero vector In all of these cases, it doesn t matter whether we work in the Cartesian plane or in the polar plane
Here s a challenge!
Prove that the multiplication of a Cartesian-plane vector by a positive scalar is left-hand distributive over vector addition That is, if k+ is a positive constant, and if a and b are Cartesian-plane vectors, then k+(a + b) = k+a + k+b
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