Find all points $C$ such that area of $ABC$ is $3$

Question

In an exercise I'm asked the following:

Consider the points $A=(0,0,0)$ and $B=(1,0,0)$. What are all the points $C$ such that the triangle $ABC$ has area 3?

I did the following:

let $C=(a,b,c)$, then the triangle has area equal to:

$$\frac{1}{2} ||B\times C||$$

so we have that $$\frac{1}{2}\sqrt{c^2 +b^2}=3$$

Giving:

$$c= \pm \sqrt{36 -b^2}$$

So all the points $C$ are:

$$C=(t,\lambda, \pm \sqrt{36 -\lambda^2} ), \ \ t,\lambda \in \mathbb R$$

My question is: Is this correct? It seems a little bit odd that there are infinite points that satisfy this condition.

Answer 1

Think about this possibility: the points $A$ and $B$ create the base of your triangle along the x-axis. An obvious starting point would be $C = (\frac{1}{2}, 6, 0)$ which creates an isosceles triangle that has an area of 3 (draw and check). Rotating this coordinate around the x-axis would leave you to create infinitely many possibilities for coordinates that create triangles with an area of 3 (provided in this case we are working in $\mathbb{R}$).

SO YES, it is possible to have infinitely many points.

The above example does NOT give ALL possible points which is where we need other tools, enter Linear Algebra.

Answer 2

Let C be in plane $z=0$

$$\dfrac12 \cdot1 \cdot h = 3 \to h=6 $$

So rotating by radius $ \sqrt{y^2+z^2} =6$ around x-axis gives all points on a torus as the locus required.