Find a parametrization of a half circle in $\Bbb R^3$

Question

Let $B$ be a sphere with centre $(0,0,0)$ and radius $a$. Let $C_1$ be the path over the half circle of $B$ from $(a,0,0)$ to $(-a,0,0)$, and through point $(0, \frac{a}{\sqrt2},\frac{a}{\sqrt2})$. Let $C_2$ be the path that goes in a straight line from $(-a,0,0)$ to $(a,0,0)$. Let $C=C_1 \cup C_2$.

How can I parametrize $C$? I'm having trouble because it's not in a $xy$ or $xz$ or $zy$ plane.

I found the parametrization of $C_2$, just having trouble with $C_1$.

Answer 1

We will parametrize $C_1$ by considering the "product" of two paths. Let $\alpha_1$ denote the linear path from $(a,0,0)$ to $(0,\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}})$ and $\alpha_2$ denote the linear path from $(0,\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}})$ to $(-a,0,0)$. Now set, $$\gamma_1(t)= \frac{\alpha_1(t)}{||\alpha_1(t)||} \ \ \ \ \ \text{and} \ \ \ \ \ \gamma_2(t)= \frac{\alpha_2(t)}{||\alpha_2(t)||} \ \ \ \ \forall \ t\in [0,1] $$ Then set $\gamma=\gamma_1*\gamma_2$.

Answer 2

If we find the plane containing all these three points, it has the equation $$y=z$$

So the parametrization of $C_1$ will be as the following: $$\begin{cases}x=a \cos t\\ y=a\frac{\sin t}{\sqrt 2}\\z=a\frac{\sin t}{\sqrt 2}\end{cases}$$ for $t\in [0, \pi]$