anonymous
  • anonymous
For a triangle ABC with |AB| = c, |AC| = b, |BC| = a, let M be the midpoint of AB. Find |CM|
Mathematics
jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
|dw:1434172244507:dw|
ganeshie8
  • ganeshie8
Say \(\angle AMC = \theta\), then \(\angle BMC = 180-\theta\) as they both are linear pair angles.
ganeshie8
  • ganeshie8
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 \[\large a^2 = |CM|^2+(c/2)^2-2|CM|(c/2)\cos(180-\theta)\tag{2}\]

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ganeshie8
  • ganeshie8
Adding them gives the desired result but im not so sure if you're allowed to use law of cosines
anonymous
  • anonymous
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.
ganeshie8
  • ganeshie8
|dw:1434173370575:dw|
ganeshie8
  • ganeshie8
we want to express the value of \(\large \|\frac{a+b}{2}\|\) in terms of \(\|a\|,~\|b\|,\|c\|\)
ganeshie8
  • ganeshie8
|dw:1434173828264:dw|
ganeshie8
  • ganeshie8
we may use the fact that dot product is distributive : \[\|a+b\|^2=\langle a+b,~a+b\rangle = \langle a,a\rangle + \langle a,b\rangle + \langle b,a\rangle + \langle b,b\rangle \tag{1}\] \[\|a-b\|^2=\langle a-b,~a-b\rangle = \langle a,a\rangle - \langle a,b\rangle - \langle b,a\rangle + \langle b,b\rangle \tag{2}\]
ganeshie8
  • ganeshie8
Add them and we can conclude
anonymous
  • anonymous
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?
ganeshie8
  • ganeshie8
Yes, suppose \(u = \begin{pmatrix}x_1\\x_2\\x_3\\\vdots\end{pmatrix}\) then the dot product\(u\cdot u\) is given by \[u\cdot u=x_1^2+x_2^2+x_3^2+\cdots =\|u\|^2\]
anonymous
  • anonymous
(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?
ganeshie8
  • ganeshie8
Without the second equation, how do you propose to eliminate below terms ? |dw:1434175327321:dw|
anonymous
  • anonymous
RIght, I see that. And then we have: \(||a+b||^{2} + ||a-b||^{2} = 2 + 2\) 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?
ganeshie8
  • ganeshie8
Yep, also notice that \(\|a-b\|\) can be replace by \(\|c\|\) so the final expression will be in terms of magnitudes of sides only
ganeshie8
  • ganeshie8
you may use ` \| ` for norm
anonymous
  • anonymous
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.
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
|dw:1434176429728:dw|
anonymous
  • anonymous
Okay, that makes sense then. Is knowing that the length is \(\|a+b\|/2\) a property of some sort or just an obvious conclusion?
ganeshie8
  • ganeshie8
It is easy to see that result
ganeshie8
  • ganeshie8
follows trivially from vector addition
ganeshie8
  • ganeshie8
Add these two vectors |dw:1434176693552:dw|
ganeshie8
  • ganeshie8
\[\large b + \dfrac{a-b}{2} = ?\]
anonymous
  • anonymous
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.
ganeshie8
  • ganeshie8
np :)
anonymous
  • anonymous
Yeah, thanks again :)

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