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I need to prove that a statement about vector products isn't true.

10 Answers

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Hi u/call_me_mistress99,

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Note: The wording is a little bit weird. If I understood correctly, what it's actually asked is to prove that a x b = a x c *does not imply* b = c.

Saying "it isn't true" is a little bit misleading, as it seems to say that it may never be true.

Another way to express the same, more standard, would be this:

*Prove that it's not true that if a x b = a x c then b = c.*

Note that we just need to change the place where we write "it's not true" for the phrase to make real sense.
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Remember that **a** x **b** = |**a**||**b**|sin(θ) where θ is the angle between the two vectors. Usually θ is between 0 and π. Is there two different values of θ such that sin(θ) is the same? What does this tell you about the two vectors?
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try parallel a, b, c
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That only implies that (b-c) is parallel to a, so b-c = k\*a, for which b = c is only one particular case.
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I assume that is just a summary of a question,  because b can equal c.


a cross b = a cross b.

It's a "doesn't imply that  b = c always".

Just write out the definition for both.

A B sin(theta1)

A C sin(theta2)

and keep in mind the direction of the result.
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Honestly I'm unsure where to start. I tried a = (1,0,1), b = (0,0,1), c = (1,2,0) and got a x b and a x c. Then I realized I had no idea what I'm doing.
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Which requirments need to be fullfilled so that the equation is true for a given a? Recall what the cross product does.
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Simply take **a** to be the zero vector, and let **b** and **c** be any distinct vectors. Then **a** x **b** and **a** x **c** are always equal (to the zero vector).
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Mathemaically speaking, this problem really depends in what space you are working in.

Looking however as the formulation of the question, i suppose this is school-level math and doesn't need any serious advanced math rigour.

I hence suppose that we are in R^3 (or 3D).

1. Solution 1 - Geometric proof: l = axb is the vector that is both orthogonal to a and b. Imagine you have a sheet of paper, if you were to place a pencil on perpendicular to the paper, you will be able to find more than 2 vectors a and b on the piece of paper that are orthogonal of the pencil. I can use concepts of differential geometrie here but they might be too advances.

2. Solution 2: Define a = [a1, a2, a3], do the same for b and show that axb = d can be solved for example when a = -a and b=b or not.

3. Solution 3: Use that formula: axb=|a||b|cos(a and b) and show that since this is not unique to a and b.

4. Some actual technics of analytical geometrie could be used here.

Technically speaking, the x operator is a inner product and the question answers itself.

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