Equation of plane passing through 3 points
Given three points $a, b, c\in \mathbb R^3$, how do we find the equation of the plane passing through these three points?
Original Question : Let $p : \mathbb{R} \to \mathbb{R}^3$ be a twice differentiable parametric curve, which is nowhere straight. For arbitrary h find the equation of the plane that goes through the three points $p(t)$, $p(t+h)$, $p(t−h)$. Then find the equation of the limit plane that one obtains from letting $h\to 0$. Verify that the tangent and normal unit vectors at $p(t)$ lie on that plane. I just want to know how to form the equation of the plane using the three coordinates provided. (Please no complete answer to the above question.)
$\endgroup$ 52 Answers
$\begingroup$The parametric equation of the plane passing through the three points $p(t)$, $p(t+h)$, $p(t−h)$ is $$Q_{t,h}(r,s):=p(t)+\frac{p(t+h)-p(t)}{h}\cdot r+\frac{p(t-h)-p(t)}{-h}\cdot s$$
$\endgroup$ 7 $\begingroup$First, assume that the 3 points are not co-linear. The normal to the plane is then nontrivial: $\mathbf n = (\mathbf b - \mathbf a) \times (\mathbf c - \mathbf a)$.
Let $\mathbf r$ be a point on the plane. The equation of the plane is:
$$ (\mathbf r - \mathbf a)\cdot \mathbf n = 0, $$
or explicitly:
$$ (\mathbf r - \mathbf a)\cdot \big( \mathbf (\mathbf b - \mathbf a) \times (\mathbf c - \mathbf a) \big)= 0. $$
Intuition: $(\mathbf r - \mathbf a)$ is a vector parallel to the plane, thus its dot product with the plane's normal has to vanish.
$\endgroup$ 1
