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I have a very basic question about geometry. The problem is: let $\vec{r}$, $\vec{u}$ and $\vec{v}$ be vectors in the plane such that $\vec{r} = \vec{u} + \vec{v}$ and such that $|\vec{r}|=10$. If the angle between $\vec{r}$ and $\vec{u}$ is $\pi/6$ find the angle $\theta$ between $\vec{r}$ and $\vec{v}$ such that $|\vec{v}|$ is minimum.

Well, my problem is: I cannot introduce any coordinate systems to write the vectors as elements of $\mathbb{R}^2$ and I cannot use calculus to find the minimum of the function $|\vec{v}|(\theta)$. The only things I can use is the law of cosines and the parallelogram rule.

My only attempt was try to write $|\vec{v}|$ as a quadratic function of $\cos \theta$, because then I could find the minimum without calculus: just finding the vertex. However I've failed with this (I've found the norm of $\vec{v}$ as an implicit function of $\theta$), and I didn't find any other way to do it.

My problem is that I get confused to attack problem with just this minimum framework. How can I solve this? I know that this is very basic, so sorry because of that, but I've found this problem in a basic analytic geometry book, tried it out, and as I didn't find a way to solve it I got curious.

Thanks in advance for your help, and sorry again for such a basic question.

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If $v$ is perpendicular to $u$ you get minimimum length. So the sides $u,v$ are legs of a right triangle with $r$ as the hypotenuse. And you know one of the non-right angles of this triangle... –  coffeemath Apr 15 '13 at 2:39
    
Are $\,u,v\,$ orthogonal? Is there any other info you may have missed? –  DonAntonio Apr 15 '13 at 2:48
    
Well, that's the only info the book gives. That's why I've got interested: how to solve this with so minimum info. I believe, however, that @coffeemath is right and the angle should really be $\pi/2$. I'm trying to prove it now. –  user1620696 Apr 15 '13 at 2:51
    
To prefix slightly on coffeemath's comment: The problem is equivalent to finding the distance between the fixed endpoint of r and the line extended (to infinity) along u from the start-point of r. This distance is found by dropping a perpendicular from the endpoint of r to the line through u, etc. as explained by @coffeemath. –  Pieter Geerkens Apr 15 '13 at 3:14
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You can do it directly with the law of cosines by completing the square. For short-hand I will just write $r$ for $|\vec{r}|$, etc., and $\phi$ for the angle between $\vec{r}$ and $\vec{u}$. $$ \begin{align} v^2 &=r^2+u^2-2ru\cos\phi \\ &=r^2+(u-r\cos\phi)^2-r^2\cos^2\phi \\ &\ge r^2(1-\cos^2\phi) = r^2\sin^2\phi \end{align} $$ with equality when $u=r\cos\phi$, i.e. when $r^2=u^2+v^2$.

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