Minimizing and maximizing $ax+by+cz$ for unit vectors $(x,y,z)$? Suppose you have a unit vector $(x,y,z)$ and you want to minimize or maximize some linear relation $ax+by+cz$. Of course, one could do this with lagrance multipliers. 
Is there an alternative way to do this just with linear algebra in general? If so, why does it work?
I'm curious because earlier I had to do a homework problem minimizing $x+2y+3z$ on the unit sphere. 
 A: Hint: Using Cauchy Schwarz inequality is one way:
$$|X\cdot Y| \le ||X|| \, ||Y|| \iff -||X|| \, ||Y|| \le X\cdot Y \le ||X|| \, ||Y||, $$
where $|X\cdot Y| = ||X|| \, ||Y||$ iff $\{X,Y\}$ is linearly dependent.
A: This can be done using just linear algebra.
The expression you want to maximize ($ax + by +cz$) is equivalent to the dot product $(a, b , c) \cdot (x, y, z)$
From the identity $A\cdot B = |A||B|\cos\theta$, where $\theta$ is the angle between $A$ and $B$, and the fact that $\cos\theta$ has a maximum whene $\theta = 0$, we then know that, to maximize $(a, b , c) \cdot (x, y, z)$, we need the vector $(x,y,z)$ to point in the direction of  of $(a,b,c)$.
The rest is just a scalar multiplication.
A: You may try a geometric approach.
The equation $x + 2y+3z = c$ gives a a plane. Which $c$ are achievable?
the ones that $c$ defines a plane so that it has a non empty intersection with the sphere. 
For $c=0$ of course we the sphere and $x+2y+3z=0$ have a non empty intersection, actual the cut is a great circle.
As $c$ becomes more negative then imagine the plane going lower and lower, the crucial point is when the plane and the sphere have only one point in common. And that would be the minimum $c$. Since we would be able to pick any smaller.
The plane is then tangent to the sphere. The normal vector of the plane and $x+2y+3z=(1,2,3)\cdot (x,y,z)$ is $(1,2,3)$ is the same as the normal vector of the sphere. Therefore the point where the sphere and the plane meet is at the intersection of the line $l(t)  = t(1,2,3)$ and the sphere.
Substituning to the equation of the sphere we get $14 t^2 = 1$, now for that $t_0<0$
we the $\min   (x+2y +3z) =c =t_0+2t_0 +3t_0 =6t_0 $
