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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
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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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434172244507:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Say \(\angle AMC = \theta\), then \(\angle BMC = 180\theta\) as they both are linear pair angles.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Apply law of cosines in the left side triangle and get \[\large b^2 = CM^2+(c/2)^22CM(c/2)\cos(\theta)\tag{1}\] Apply law of cosines in the right side most triangle and get \[\large a^2 = CM^2+(c/2)^22CM(c/2)\cos(180\theta)\tag{2}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Adding them gives the desired result but im not so sure if you're allowed to use law of cosines

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434173370575:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3we want to express the value of \(\large \\frac{a+b}{2}\\) in terms of \(\a\,~\b\,\c\\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434173828264:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3we may use the fact that dot product is distributive : \[\a+b\^2=\langle a+b,~a+b\rangle = \langle a,a\rangle + \langle a,b\rangle + \langle b,a\rangle + \langle b,b\rangle \tag{1}\] \[\ab\^2=\langle ab,~ab\rangle = \langle a,a\rangle  \langle a,b\rangle  \langle b,a\rangle + \langle b,b\rangle \tag{2}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Add them and we can conclude

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, suppose \(u = \begin{pmatrix}x_1\\x_2\\x_3\\\vdots\end{pmatrix}\) then the dot product\(u\cdot u\) is given by \[u\cdot u=x_1^2+x_2^2+x_3^2+\cdots =\u\^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(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 \(ab^{2}\). That and, from the same perspective as the parallelogram drawings, I would've assumed this: dw:1434174991533:dw I assume the position of ab doen't matter since its magnitude is the same? Or does its position matter and I'm not imagining it properly?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Without the second equation, how do you propose to eliminate below terms ? dw:1434175327321:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0RIght, I see that. And then we have: \(a+b^{2} + ab^{2} = 2<a,a> + 2<b,b>\) And then I assume this means: \(a+b^{2} + ab^{2} = 2a^{2} + 2b^{2}\) ?? \(a+b/2 = a^{2} + b^{2}  ab^{2}/2\) And that was the idea then, to get \(a+b/2\) correct?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yep, also notice that \(\ab\\) can be replace by \(\c\\) so the final expression will be in terms of magnitudes of sides only

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you may use ` \ ` for norm

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, I see that. So, is c = ab 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 ab, 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3In 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434176429728:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, that makes sense then. Is knowing that the length is \(\a+b\/2\) a property of some sort or just an obvious conclusion?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3It is easy to see that result

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3follows trivially from vector addition

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Add these two vectors dw:1434176693552:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\large b + \dfrac{ab}{2} = ?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, thanks again :)
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