Show that exists on the sphere a derivable curve that connects A and B.

Question

Let $A$ and $B$ two points on the unit sphere. Suppose that $A \neq B$. Show that exists on the sphere a derivable curve that connects $A$ and $B$.

I tried to use the line $R(x)=A+x(B-A)$ with $0 \leq x \leq 1$, but I do not know how to start.

Answer 1

You were close, but you chose a straight line segment. You want each point to be a unit distance from the origin. So, provided $A\ne-B$ you should divide each vector by its magnitude.

$$R(x)=\frac{A+x(B-A)}{\Vert A+x(B-A)\Vert}$$

If $A=-B$ the straight line will pass through $(0,0,0)$ and you would be dividing by $0$.

Answer 2

Since

$A \ne B, \tag 1$

the only way $A$ and $B$ can be collinear is if

$A = -B; \tag 2$

in this case $A$ and $B$ are antipodal points and any plane containing both $A$ and $B$ will pass through $(0, 0,0)$; such a plane will intersect the sphere in a great circle containing $A$ and $B$, and we are done.

In the event that

$A \ne B \ne -A, \tag 3$

then the plane passing through $(0, 0, 0)$ and normal to $A \times B$ will again cut the sphere in a great circle containing $A$ and $B$, which then forms a differentiable curve in said sphere connecting $A$ and $B$; this is explained in greater detail, including explicit formulas for said curve, in my answer to this question.