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### Laws of Vectors

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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
this.
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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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Associated Topics:
High School Linear Algebra

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