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Explain the vector statement in a geometric way.

Let a, b and c be different vectors in 3D space, where a + b = c.
 
Given the statement: a x b = c x b
 
Question:
 
Explain this statement geometrically using the definition of vector addition & cross product, in anyway.
 
 

Comments

I am leaving this a comment so it won't show answered and maybe someone else will come by.
The following is the case for the AXB = CXB statement.
I need to more about think about what adding the A+B=C to that means.
 
Upper case is vector, the lower case = scalar
A X B = C X B implies C • A - C • B = c | A - B|
(A-B) X C => 0 => A - B || C
=> (A - B) • C = c | A-B | => C • A - C • B = c |A - B|
I could also say:
A + B = C  =>
A X B = C X B  =  (A + B) X B = (A X B ) + (B X B) = (A X B) + 0 = A X B
not very helpfull
 
A + B = C  =>  B = C - A  =>
A X B = C X B = C X (C - A) = (C X C) - (C X A) = 0 -(C X A)
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1 Answer

OK Brian,
I'll have a crack at it, without a lot of vector math. First of all, if A + B = C, that means that A, B, and C are lying all in some common plane in space. It doesn't matter which plane for the discussion which follows, so we might as well pick a convenient one for discussion, such as the x-y plane.
In fact, you can arrange these 3 vectors in the form of the sides of a triangle (why not, that's what the vector addition says).
When you think of a triangle, and the lengths of its sides, if any two things multiply a third thing to give identical products, they must have been the same magnitude themselves originally. Thus, legs A and C are isoceles (so to speak), whilst B can be different (or can be the same) magnitude.
Now, since all these three vectors are in the same plane, the perpendicular to that plane figures in (is the vector direction of) any cross product. So AXB is some vector (from the origin, or from the point of intersection of A and B, it really doesn't matter, it's a vector, it has a length and a direction) along that perpendicular direction, as is CXB.
The other piece of a cross-product definition, the multiplier sin(θ), is identical for the two identical angles of the isoceles triangle; it's maximized to 1 for an infinitely thin isosceles slice (|B|->0), but the |cross product| is then also ->0. Likewise, as |B| -> 2|A| (i.e. as the isoceles wedge widens and flattens to zero height), the multiplier sin(θ) minimizes to 0, making the cross product ->0. Where, then (= at what geometry of A, B, and C), does the maximum for the cross product lie? You should have some geometrical idea about this from the respective triangles -- but can you prove it? You may want to look up a formula for area of an isosceles triangle from its sides -- or you may want to use a different triangle geometry Law?
Hope this helps,
-- S.

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