Two impossible crossproduct problems

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The discussion revolves around two vector problems. The first problem requires proving the identity |UxV|² = |U|² + |V|² - (U·V)², with suggestions to use geometric interpretations and trigonometric identities involving the angle between vectors. The second problem involves finding a vector C such that AxB = AxC, where C is not equal to B, with the solution indicating that C can be expressed as B plus any vector V that is parallel to A. The participants emphasize the importance of understanding vector properties and relationships without relying heavily on algebra. Overall, the conversation highlights methods to approach vector problems using both algebraic and geometric perspectives.
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Hi, I really need some help here.. Vectors remind me why I hate geomtery.

problem 1: Prove that |UxV|2 = |U|2+|V|2 - (U*V)2

How can I prove that these two are equal without spending 1 hour using algebra? Maybe there is some geometry quirk that I'm not seeing?

problem 2: We have two vectors A=[1,1,1] and B=[1,2,3]

Find a vector C so that AxB = AxC, where C =! B.

I tried using algebra on this but I just ended up with crazy expressions for Cx, Cy and Cz where each of them were dependent on the others.

So.. Is there some other way? All help is appreciated =)
 
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I would also like to point out that I haven't learned yet about things like the ratio between sine(x) and cosine(x), so pls if possible tell me this can be solved by simple algebra or by using geometry?
 
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For the second one, since it just asks for a vector C not equal to B, the easiest thing to do would be to choose C = B + V, where V is some vector such that A x V = 0. Do you know what kinds of vectors have that property?
 
Nikitin said:
problem 1: Prove that |UxV|2 = |U|2+|V|2 - (U*V)2

How can I prove that these two are equal without spending 1 hour using algebra? Maybe there is some geometry quirk that I'm not seeing?
Do you know how to rewrite U·V and |U×V| in terms of |U|, |V| and the angle between U and V. You need to use those identities, and also \cos^2x+\sin^2x=1.
 
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Mute said:
For the second one, since it just asks for a vector C not equal to B, the easiest thing to do would be to choose C = B + V, where V is some vector such that A x V = 0. Do you know what kinds of vectors have that property?

ahh, so AxC = Ax(B + V) = AxB + AxV where AxV=0.

Very cunning. Yes, V equals any vector which is pararell with A. Thank you 4 the help!

Fredrik said:
Do you know how to rewrite U·V and |U×V| in terms of |U|, |V| and the angle between U and V. You need to use those identities, and also \cos^2x+\sin^2x=1.

Sure, |U|2*|V|2 - |U|2*|V|2*cos(x)2= |U|2*|V|2(1-cos(x)2)=|U|2*|V|2(sine(x)2)=|UxV|2

correct? tho we haven't learned about the 1=cos(x)^2 + sine(x)^2 trick in my maths class so maybe there is another way to prove it?

thank u very much 4 the help anyways
 
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