Given $u=(-2,5,3)$ find a unit vector $v$ s.t $|u\times v|$ is maximal, and then a unit vector $w$ s.t $|(u\times v)\cdot w|$ is minimal

Question

This is a similar question to the one I have posted before. The problem is as in the title:

Given $u=(-2,5,3)$ find a unit vector $v$ s.t $|u\times v|$ is maximal, and then a unit vector $w$ s.t $|(u\times v)\cdot w|$ is minimal

Again, I can write out the equations, but there should be an easier way to find the max/min then to solve the equations (since, as I stated in my previews question, they don't know how to solve for the max/min).

Any help on the problem is greatly appreciated!

Answer 1

Edit: for the second step, you can always take $w$ to be equal to $v$, as $|(u\times v)\cdot v|=0$ is obviously minimal. It is zero, of course, because $u\times v$ is orthogonal to $v$ by property of the cross-product. So I only leave my answer to part 2 in case you start, some day, from an arbitrary $u'$, and not in the form $u\times v$ with $v$ unit vector like here.

1- Maximization of the cross-product: if $\|v\|=1$ and if the angle formed by $u$ and $v$ is $\theta$ modulo $\pi$, then $$ \|u\times v\|=\|u\|\|v\||\sin\theta|=\sqrt{38}|\sin\theta|. $$ So the maximum is attained when $|\sin\theta|=1$, i.e. when $v$ is orthogonal to $u$. The solutions form a two-dimensional sphere. Here is the first one I that comes to my mind: $$ v=\left(\frac{5}{\sqrt{29}},\frac{2}{\sqrt{29}},0\right). $$ This yields $$ u':=u\times v=\left(-\frac{6}{\sqrt{29}},\frac{15}{\sqrt{29}},-\frac{29}{\sqrt{29}}\right). $$

2- Minimization of the dot product: now let $\|w\|=1$ and $\phi$ the angle formed by $u'$ and $w$. Then $$ |u'\cdot w|=\|u'\|\|w\||\cos\theta|=\sqrt{38}|\cos\theta| $$ is minimal when $\cos\theta=0$, i.e. when $w$ and $u'$ are orthogonal. Again, the solutions form a two-dimensional unit sphere. And the first choice I can think of is $$ w=\left(\frac{15}{\sqrt{261}},\frac{6}{\sqrt{261}},0\right)=\left(\frac{5}{\sqrt{29}},\frac{2}{\sqrt{29}},0\right)=v. $$

Answer 2

Hint: you can maximize the cross product by choosing a vector $v$ that is orthogonal to $u$ (there are infinitely many choices). You can minimize the inner product by choosing a vector $w$ that lies in the plane spanned by $u$ and $v$ (because then it's orthogonal to $u\times v$ and the inner product equals $0$).