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

- anonymous

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

- jamiebookeater

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- ganeshie8

|dw:1434172244507:dw|

- ganeshie8

Say \(\angle AMC = \theta\), then \(\angle BMC = 180-\theta\) as they both are linear pair angles.

- 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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## More answers

- ganeshie8

Adding them gives the desired result but im not so sure if you're allowed to use law of cosines

- 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

|dw:1434173370575:dw|

- ganeshie8

we want to express the value of \(\large \|\frac{a+b}{2}\|\) in terms of \(\|a\|,~\|b\|,\|c\|\)

- ganeshie8

|dw:1434173828264:dw|

- 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

Add them and we can conclude

- 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

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

(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

Without the second equation, how do you propose to eliminate below terms ?
|dw:1434175327321:dw|

- 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

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

you may use ` \| ` for norm

- 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

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

|dw:1434176429728:dw|

- 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

It is easy to see that result

- ganeshie8

follows trivially from vector addition

- ganeshie8

Add these two vectors
|dw:1434176693552:dw|

- ganeshie8

\[\large b + \dfrac{a-b}{2} = ?\]

- 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

np :)

- anonymous

Yeah, thanks again :)

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