Suppose I have a plane $$z=x+3$$ and sphere $$x^2 + y^2 + z^2 = 6z$$ what will be their intersection ?
I wrote the equation for sphere as
$$x^2 + y^2 + (z-3)^2 = 9$$ with center as (0,0,3) which satisfies the plane equation, meaning plane will pass through great circle and their intersection will be a circle.
However when I try to solve equation of plane and sphere I get
$$x^2 + y^2 + (x+3)^2 = 6(x+3)$$
which does not looks like a circle to me at all. What did I do wrong?
Best Answer
Center of circle: at $(0,0,3)$ , radius = $3$.
Plane $z = x +3$ passes through center.
Normal of plane: $(1,0,-1)$ .
Let vector $(a,b,c)$ be perpendicular to this normal:
$(a,b,c) \cdot (1,0,-1)$ = $0$ ; $a - c = 0$.
Yields 2 independent, orthogonal vectors perpendicular to the normal $(1,0,-1)$ of the plane:
1) $(1/√2)(1,0,1)$ and
2) $(0,1,0)$.
Let $\vec{s}$ = $\alpha (1/√2)(1,0,1) +\beta (0,1,0)$.
In analogy to a circle traced in the $x, y$ - plane:
$\vec{s} \cdot (1/√2)(1,0,1)$ = $3 cos(\theta)$ = $\alpha$.
$\vec{s} \cdot (0,1,0)$ = $3 sin(\theta)$ = $\beta$
Finally the parameter representation of the great circle:
$\vec{r}$ = $(0,0,3) + (1/√2)3cos(\theta)(1,0,1) + 3sin(\theta)(0,1,0)$