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So I have this problem (self learning linear algebra) and I try to solve it using the indicated steps. Question is: Am I right, is this a valid answer? Obviously for me makes sense but since I'm just starting with maths maybe there's something wrong:

If $\|\vec{v}\| = 5$ and $\|\vec{w}\| =3$, what are the smallest and largest values of $\|\vec{v}-\vec{w}\|$ and $\vec{v} \cdotp \vec{w}$?

EDIT

I'd moved my initial answer to a full answer. See below.

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Perfect. $\,\,\!\!$ –  Berci Apr 17 '13 at 16:37
    
@Berci Thanks :) Great to hear that. I have one question. Should I delete the question or can I let it in case any one found it valuable? –  Randolf R-F Apr 17 '13 at 16:39
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@Randolf definitely keep it around. Someone else doubtless will have a similar question and this can stand as a reference. –  Ian Coley Apr 17 '13 at 16:40
    
Intuitively, it should also be clear that if e.g. $\vec{v} = 5 \vec{e}_1$, then the choice of $\vec{w}$ minimizing the mutual distance is $\vec{w} = 3 \vec{e}_1$ (at distance $2$), and the choice of $\vec{w}$ maximizing the distance is $\vec{w} = -3 \vec{e}_1$ (at distance $8$). –  TMM Apr 17 '13 at 19:51

1 Answer 1

Using cosine laws for first part:

$$\begin{align}\|\vec{v}\|^2 + \|\vec{w}\|^2 &= \|\vec{v}-\vec{w}\|^2 + 2 \cdot \|\vec{v}\| \cdot \|\vec{w}\|\cos{\theta} \\ \\ &\equiv \\ \\ \|\vec{v}\|^2 + \|\vec{w}\|^2 - 2 \cdot \|\vec{v}\| \cdot \|\vec{w}\|\cos\theta &= \|\vec{v}-\vec{w}\|^2 \\ \\ &\equiv \text{(Given the fact that $\|\vec{v}\| = 5$ and $\|\vec{w}\| =3$)} \\ \\ 34 - 30\cos\theta &= \|\vec{v}-\vec{w}\|^2 \\ \\ &\equiv \\ \\ \sqrt{34 - 30\cos\theta} &= \|\vec{v}-\vec{w}\| \end{align}$$

And so the smallest value of $\|\vec{v}-\vec{w}\|$ will be 2 when $\cos\theta = 1$ (at $0$) and the largest value will be 8 when $\cos\theta = -1$ (at $\pi$).

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