Cross product distributive proof

[MB]

Hi

Could someone please give me a proof for distributive law of the cross
product:

a x (b + c) = a x b + a x c

But there's a catch! I am assuming the we have defined the cross
product as (|a||b|sinO)n, etc. Using the above proof (that is, taking
the above as true) THEN I will use the above 'truth' to derive the
rectangular form of the cross product. So I need the above proved
without resorting to the rectangular form/defn of the cross product.

Thanx in advance
MB

[saibot]

Proof it for the canonical unit vectors e1, e2, e3. This is easy,
there are only a few cases to consider.
Then set a = a1*e1 + a2*e2 + a3*e3 where a1, a2, a3 are the coodinates
of a. The same for b and c and see what you can get. Probably it is
useful to do the case first where a is one of the canonical unit
vectors and only b and c are arbitrary.

[Ken Pledger]

In article <4d1c6aa9.03101...@posting.google.com>,
mar...@webmail.co.za (MB) wrote:

> ....

> Could someone please give me a proof for distributive law of the cross
> product:
>
> a x (b + c) = a x b + a x c
>
> But there's a catch! I am assuming the we have defined the cross

> product as (|a||b|sinO)n, etc....

When teaching this several years ago, I used a proof slightly
simplified from that of Seymour Schuster, "Elementary Vector Geometry,"
(Wiley, 1962), pp.139-140. The trick is to work ahead (without assuming
the distributive law) to the scalar triple product. You need to show that
its dot and cross can be interchanged, and also to have proved the
distributive law for the _dot_ product (which can also be a little
delicate). After that, here's the proof. Please pretend every variable
is in bold type. :-)

Let d = a x (b + c) - a x b - a x c

so it is required to prove that d = 0.

d^2 = d.d

= d.(a x (b + c) - a x b - a x c)

= d . a x (b + c) - d . a x b - d . a x c

= d x a . (b + c) - d x a . b - d x a . c

= d x a . (b + c) - d x a . (b + c)

= 0.

Therefore d = 0, so a x (b + c) = a x b + a x c.

Then you can multiply by -1 and use the anticommutative law to get
the other distributive law.

Schuster didn't use d^2 but I think it makes the proof clearer.

Ken Pledger.

[mulkey...@gmail.com]

Holy Cross me up! I wish I could. Still a bit of a newbie as I'm takin' Remedial Math/Algebra. The answer I came up with wasn't a rectangle. It was a Sphere. What did I do wrong? Plz help. Thanks in advance! :)

LMH

[Peter Percival]

mulkey...@gmail.com wrote:
> Holy Cross me up! I wish I could. Still a bit of a newbie as I'm takin' Remedial Math/Algebra. The answer I came up with wasn't a rectangle. It was a Sphere. What did I do wrong? Plz help. Thanks in advance! :)

Are you seeking a proof of this:

(A + B) x C = A x C + B x C

for vectors A, B, C in R^3?

If D is any vector in R^3 then

((A + B) x C).D = (A + B).(C x D)
= A.(C x D) + B.(C x D)
= (A x C).D + (B x C).D
= ((A x C) + (B x C)).D
so
(A + B) x C = A x C + B x C.

That requires

A.(C x D) = (A x C).D

which can be deduced by considering the volume of the parallelepiped
with side A, C, D; and

if E.D = F.D for all(*) D then E = F

which may be proved by considering

E.D = F.D => (E - F).D = 0
and if E - F were not 0 it would be orthogonal to D,
but D is arbitrary...

Question: does that go through in R^7 where x is also definable?


(*) Or just three linearly independent D's.
--
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