A problem involving vectors I don't know how to approach
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Otis
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Hi nombreuso. I think it has more to do with definition of equivalent vector arrangements than with the dot product.
Did you draw a picture? Label vectors A,B,C. Rotate your diagram 90 degrees counterclockwise. Look for a consistent relationship between vectors A,B,C before rotation and the triangle sides after rotation. Also think about what you proved in part a).
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I guess I have no other option
Otis
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I assume by "it" you're talking about an arrangement of vectors A,B,C. That doesn't form just "a" triangle. ABC is identical to PQR, rotation notwithstanding.
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Otis
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So, since they all rotate to the same direction, the entire triangle is rotated. How can I express this rotation?
Otis
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You just made a good effort, but let's not use pronouns, and it isn't that the original vectors each rotate to point in the same direction -- it's that they each rotate by the same amount (90 degrees counter-clockwise):
You could finish with an explicit comment about why proving part b) would basically be a repeat of part a). In other words, justify why you've already proved the vector equation/relationship in part b).
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Ok. Maybe this is not exactly what you meant, but could I say that the sum of all the vectors rotated 90 degrees counterclockwise is the same as the first sum rotated 90 degrees counterclockwise, which is 0 rotated counterclockwise, which since doesn't have any direction, is still 0?
Otis
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Hi nombrueso. I can't tell what you're thinking above, but I hope you're not confusing a measurement of 0 with the zero vector.
I don't think you need to say any more about rotation. Your first statement about the rotated vectors (the statement that I'd edited in post #9) is good enough, I believe. Maybe you can get away with simply saying, "ABC is the same as PQR". You decide.
Otherwise, you've already proved in part a) that a sum of vectors forming a closed figure (like a triangle) represents the zero vector. In other words, you've proved that, after combining such vectors, we end up right back where we started. That is what the zero vector means, in such a sum. No net displacement.
Rotating the figure or changing the vectors' names does not affect the part a) proof. Noting that fact in a comment could look like this:
Since vectors PQ,QR,RP all rotate by the same amount, the entire triangle is rotated, and that rotation (or changing vector names to A,B,C) does not alter the vector sum, so the proofs in part a) and part b) are the same.
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Ok, thank you.Hi nombrueso. I can't tell what you're thinking above, but I hope you're not confusing a measurement of 0 with the zero vector.
I don't think you need to say any more about rotation. Your first statement about the rotated vectors (the statement that I'd edited in post #9) is good enough, I believe. Maybe you can get away with simply saying, "ABC is the same as PQR". You decide.
Otherwise, you've already proved in part a) that a sum of vectors forming a closed figure (like a triangle) represents the zero vector. In other words, you've proved that, after combining such vectors, we end up right back where we started. That is what the zero vector means, in such a sum. No net displacement.
Rotating the figure or changing the vectors' names does not affect the part a) proof. Noting that fact in a comment could look like this:
Since vectors PQ,QR,RP all rotate by the same amount, the entire triangle is rotated, and that rotation (or changing vector names to A,B,C) does not alter the vector sum, so the proofs in part a) and part b) are the same.
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