Vector parametrization of a surface intersection

Question

How does one parametrize the following curve in 3-space to $\vec{g}(t): [a, b] \to \mathbb{R}^3$:

• the intersection of $x^2+y^2+z^2=a^2$ and $x+y+z=0$ ?

What I could come up with is as follows: $\vec{g}(t) = (a\cos(t), a\sin(t), -a(\sin(t)+\cos(t)))$. But I'm not so sure that this parametrization is correct.

Answer 1

Plugging in $x+y+z=0$ into the equation for the surface of the sphere.

$$x^2+xy+y^2=a^2/2$$

Completing the square

$$(x+y/2)^2+(\sqrt{3}y/2)^2=(a/\sqrt{2})^2$$

$$\sqrt{3}y/2=a/\sqrt{2}sin(t)$$

$$y(t)=\sqrt{2/3}a sin(t)$$

$$x+y/2=a/\sqrt{2}cos(t)$$

$$x(t)=\sqrt{1/6}asin(t)+a/\sqrt{2}cos(t)$$

$$z(t)=-x(t)-y(t)$$

Answer 2

Your suggested answer is

$$\vec{g}(t) = (a\cos(t), a\sin(t), -a(\sin(t)+\cos(t)))$$

which must satisfy $$x^2+y^2+z^2=a^2$$ However, here: $$x^2+y^2+z^2=a^2+a^2(1+2\sin{t}\cos{t})\neq a^2$$

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What I do is to use two variables for parametrizing the first surface:

$$x^2+y^2+z^2=a^2$$

is

$$\vec g(t,u)=(a \sin{t}\sin{u}, a \sin{t}\cos{u}, a\cos{t})$$ intersecting it with $x+y+z=0$ means:

$$a \sin{t}\sin{u} + a \sin{t}\cos{u} + a\cos{t}=0$$ or $$\sin{t}=\frac{\pm1}{\sqrt{1+(\sin{u}+\cos{u})^2}}$$

Here, I rewrite new $g$ which is the intersecting curve:

$$\vec g(u)=(\frac{\pm a \sin{u}}{\sqrt{1+(\sin{u}+\cos{u})^2}}, \frac{\pm a \cos{u}}{\sqrt{1+(\sin{u}+\cos{u})^2}}, \frac{\mp a (\sin{u}+\cos{u})}{\sqrt{1+(\sin{u}+\cos{u})^2}})$$