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Laws of Vectors

Date: 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 

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 
and finally get:
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' 
                 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.

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 
   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

Associated Topics:
High School Linear Algebra

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