Given $u=(-2,5,3)$, how to find unit vectors$u,w$ s.t $|u+v|$ is maximal and $|u-w|$ is minimal and another similar part I was stumbled with a basic calculus question by a friend.
The question first asks to find unit vectors $v,w$ s.t $|u+v|$ is
maximal and $|u-w|$ is minimal where $u=(-2,5,3)$.
Then the question asks to find unit vectors $v,w$ s.t $u\cdot v$
is maximal and $|u\cdot w|$ is minimal.
It's easy to write out the equations in all parts, for example for
the first part: denote $v=(x,y,z)$ then we wish to find the maximum
$$(z+3)^{2}+(y+5)^{2}+(x-2)^{2}$$ under $$x^{2}+y^{2}+z^{2}=1$$
and similarly for the second part with the minimum. But this doesn't
seem like the right way to go at this, since I only know to solve
such a question with Lagrange multipliers, and they didn't study this
(yet). 
Can anyone please help point me out in the right direction ?
 A: The right direction is to consider vectors $v$ and $w$ which are collinear with $u$.
For the second part of the question, observe that $u\cdot v=|u||v|\cos\left(\angle(u,v)\right)$. In your case, the absolute values of $u$ and $v$ are fixed, so the values of $u\cdot v$ and $|u\cdot w|$ depend only on the angle between these two vectors.
A: Given $u=(-2,5,3)$ consider the following function $f_\pm(v)=<u\pm v,u\pm v>$ which is the inner product of $u\pm v$ with itself. The critical point of $f_\pm$ coincide with the critical point of $|u\pm v|$ (why?).
Now $f_\pm(v)=<u\pm v,u\pm v>=<u,u>\pm 2<u,v>+<v,v>=|u|^2+|v|^2\pm 2<u,v>$, since $u=(-2,3,5)$ and $|v|=1$ we have that $f_\pm(v)=39\pm 2<u,v>$. Since $u$ is a fixed vector and $v$ is allowed to move on a circle , $f_\pm (v)$ will be the biggest when $u$ and $v$ are multiple of each other (observe that $f_\pm(v)=39\pm\sqrt{38}\cos\theta$ where $\theta$ is the angle between $u$ and $v$, so in order for $f_\pm(v)$ to be biggest $\cos\theta$ has to be biggest, that is $\theta=0$ or $\theta=\pi$) that is when $v=\dfrac{u}{|u|}$ or $v=-\dfrac{u}{|u|}$. 
This the reason why the previous suggestion
P.S. for each function you will have to check which of the two possible answers is the solution. 
