Laws of VectorsDate: 4/10/96 at 13:31:28 From: Anonymous Subject: Geometry - Vectors Question - part 1: Use definitions I and II below to prove that k[ (a, b) + (c,d) ] = k(a,b) + k(c,d) I. Definition of Scalar Multiple k(a,b) = (ka, kb) II. Definition of Vector Addition (a,b) + (c,d) = (a + c, b + d) I came up with the following solution: To prove that k[ (a,b) + (c,d) ] = k(a,b) + k(c,d), first use definition I and apply it to the righthand side of the equation. Doing this results in the modified equation: k[ (a,b) + (c,d) ] = (ka, kb) + (kc,kd). Next apply definition II to the left side of the equation, forming the following equality: k(a + c, b + d) = (ka,kb) + (kc, kd). To further "simplify," I applied definition I to the left and came up with the equation: ( ka + kc, kb + kd) = (ka,kb) + (kc,kd). As a final step, I applied definition two to the righthand side of this equation. The result was: (ka + kc, kb + kd) = (ka + kc, kb + kd). This seems to prove what they asked me to prove, but doesn't match the solution in the solution key. Can you verify whether or not this is acceptable reasoning? Part 2 of the same question - Make a diagram illustrating what you proved in part 1. As to this part of the question, I have had absolutely no success. Any help available would be appreciated. * Note: I am homeschooled, and my teacher has given me permission to request a solution, as she too is uncertain as to go about this. Date: 6/13/96 at 14:52:54 From: Doctor Charles Subject: Re: Geometry - Vectors >Can you verify whether or not this is acceptable reasoning. No! It works in this case and there is a way of making it valid but as it stands I don't think that this is good reasoning. You are trying to prove that a=b so you rewrite this a number of times: a=b a=c d=c etc. and finally get: e=e and deduce (incorrectly) that it must be the case that a=b. Luckily, as all that you have done is rewritten each side of the equation independently it is easy to run the argument 'backwards' thus: e=e and e=d, d=a so e=a e=c, c=b so e=b hence a=b. This may seem to be pedantic but it is very easy to construct false proofs such as the following: We can prove 1=2. 1=2 0*1=0*2 0=0 this is true so it must be the case that 1=2 I agree that what you have done isn't this but it is getting slightly close to it. If you are writing out a formal proof of this you shouldn't use the equality that you are trying to use in your argument. I would reason as follows. k[(a,b)+(c,d)] = k[(a+b,c+d)] by definition II so k[(a,b)+(c,d)] = (k(a+b),k(c+d)) by definition I so k[(a,b)+(c,d)] = (ka+kb,kc+kd) (i) by standard algebra of numbers and k(a,b)+k(c,d) = (ka,kb)+(kc,kd) by definition I so k(a,b)+k(c,d) = (ka+kb,kc+kd) (ii) by definiton II hence by equating (i) and (ii) we have k[(a,b)+(c,d)] = k(a,b)+k(c,d) which is the result to be shown. >part 2 of the same question - Make a diagram illustrating what >you proved in part 1. For your diagram you need to plot the points (0,0) (a,b) (c,d) (a+c,b+d). These will form a parallelogram. You should then plot the points (ka,kb) (kc,kd) and (ka+kc,kb+kd). These, with (0,0), will form a similar paralellogram containing the smaller one. The vectors are represented by an arrowed line joining the origin to the corresponding point or a line of the same length and parallel to this line. Vector addition is show by putting two such lines 'head to tail', the first starting at the origin, the second starting were the last left off. The line joining the origin to the final point reached represents the vector which is the sum of the original two vectors. (kc,kd) (ka+kc,kb+kd) __________ / / / / (c,d) /__ / / / / /___/_____/ (0,0) (a,b) (ka,kb) Scalar multiplication gives a vector in the same direction as the orginal vector but k times as long. The diagram should show that the line joining (0,0) and (ka+kc,kb+kd) is in the same direction and k times as long as the one joining (a+c,b+d) to the origin. -Doctor Charles, The Math Forum |
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