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anonymous

  • one year ago

For a triangle ABC with |AB| = c, |AC| = b, |BC| = a, let M be the midpoint of AB. Find |CM|

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  1. ganeshie8
    • one year ago
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    |dw:1434172244507:dw|

  2. ganeshie8
    • one year ago
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    Say \(\angle AMC = \theta\), then \(\angle BMC = 180-\theta\) as they both are linear pair angles.

  3. ganeshie8
    • one year ago
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    Apply law of cosines in the left side triangle and get \[\large b^2 = |CM|^2+(c/2)^2-2|CM|(c/2)\cos(\theta)\tag{1}\] Apply law of cosines in the right side most triangle and get \[\large a^2 = |CM|^2+(c/2)^2-2|CM|(c/2)\cos(180-\theta)\tag{2}\]

  4. ganeshie8
    • one year ago
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    Adding them gives the desired result but im not so sure if you're allowed to use law of cosines

  5. anonymous
    • one year ago
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    I don't think it's not allowed. We were supposed to do it 2 ways. I'm sure something like this is fine for the first way, but for the 2nd way it had to be done with vectors.

  6. ganeshie8
    • one year ago
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    |dw:1434173370575:dw|

  7. ganeshie8
    • one year ago
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    we want to express the value of \(\large \|\frac{a+b}{2}\|\) in terms of \(\|a\|,~\|b\|,\|c\|\)

  8. ganeshie8
    • one year ago
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    |dw:1434173828264:dw|

  9. ganeshie8
    • one year ago
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    we may use the fact that dot product is distributive : \[\|a+b\|^2=\langle a+b,~a+b\rangle = \langle a,a\rangle + \langle a,b\rangle + \langle b,a\rangle + \langle b,b\rangle \tag{1}\] \[\|a-b\|^2=\langle a-b,~a-b\rangle = \langle a,a\rangle - \langle a,b\rangle - \langle b,a\rangle + \langle b,b\rangle \tag{2}\]

  10. ganeshie8
    • one year ago
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    Add them and we can conclude

  11. anonymous
    • one year ago
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    I guess I'm just really rusty on my vector properties, lol. So I'm trying to recall, magnitude u is \(\sqrt{u\cdot u}\) when thinking in terms of dot product?

  12. ganeshie8
    • one year ago
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    Yes, suppose \(u = \begin{pmatrix}x_1\\x_2\\x_3\\\vdots\end{pmatrix}\) then the dot product\(u\cdot u\) is given by \[u\cdot u=x_1^2+x_2^2+x_3^2+\cdots =\|u\|^2\]

  13. anonymous
    • one year ago
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    (not sure the command to type the magnitude bars really) Okay, so when we have \(||u+v||^{2}\), we're saying \((u+v)\cdot (u+v)\) I guess I'm trying to see what each part of what we're doing is done for. Like, drawing out the parallelogram idea for the addition of vectors lets me see why that CM length is \(||a+b||/2\) But then I'm not sure why we need \(||a-b||^{2}\). That and, from the same perspective as the parallelogram drawings, I would've assumed this: |dw:1434174991533:dw| I assume the position of a-b doen't matter since its magnitude is the same? Or does its position matter and I'm not imagining it properly?

  14. ganeshie8
    • one year ago
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    Without the second equation, how do you propose to eliminate below terms ? |dw:1434175327321:dw|

  15. anonymous
    • one year ago
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    RIght, I see that. And then we have: \(||a+b||^{2} + ||a-b||^{2} = 2<a,a> + 2<b,b>\) And then I assume this means: \(||a+b||^{2} + ||a-b||^{2} = 2||a||^{2} + 2||b||^{2}\) ?? \(||a+b||/2 = ||a||^{2} + ||b||^{2} - ||a-b||^{2}/2\) And that was the idea then, to get \(||a+b||/2\) correct?

  16. ganeshie8
    • one year ago
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    Yep, also notice that \(\|a-b\|\) can be replace by \(\|c\|\) so the final expression will be in terms of magnitudes of sides only

  17. ganeshie8
    • one year ago
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    you may use ` \| ` for norm

  18. anonymous
    • one year ago
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    Right, I see that. So, is c = a-b because we're allowed to move a vector around as long as we don't change its magnitude and direction? Because if I were to try and draw a-b, I would get a resultant vector in a different spot. I assume it has the same magnitude and direction and thus it doesn't matter, but I just want to be sure.

  19. ganeshie8
    • one year ago
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    In above diagram, we're fixing the vertex \(C\). \(a\) and \(b\) are position vectors, so we can't move them around. \(c\) is not a position vector, so we can translate it anywhere

  20. ganeshie8
    • one year ago
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    |dw:1434176429728:dw|

  21. anonymous
    • one year ago
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    Okay, that makes sense then. Is knowing that the length is \(\|a+b\|/2\) a property of some sort or just an obvious conclusion?

  22. ganeshie8
    • one year ago
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    It is easy to see that result

  23. ganeshie8
    • one year ago
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    follows trivially from vector addition

  24. ganeshie8
    • one year ago
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    Add these two vectors |dw:1434176693552:dw|

  25. ganeshie8
    • one year ago
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    \[\large b + \dfrac{a-b}{2} = ?\]

  26. anonymous
    • one year ago
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    Alright, cool. For some reason my brain isn't working with these things. I know some of this is more obvious than I'm making it seem, sorry, lol.

  27. ganeshie8
    • one year ago
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    np :)

  28. anonymous
    • one year ago
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    Yeah, thanks again :)

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