Calculate points of parallel plane?

Question

If I have $3$ points $(x,y,z)$ that define a plane and want a parallel plane with the same amount of points, just offset by $2$ units, what would be a good start point regarding the math/algorithm?

Answer 1

Let you have $3-points A$, $B$ and $C$

and the equation of plane for the above 3 points is $ax+by+cz=d$

Which you can also write as: $\frac{ax+by+cz}{\sqrt{a^2+b^2+c^2}} = \frac{d}{\sqrt{a^2+b^2+c^2}} = say(D)$

Then, Your equation of the plane which is $2$ unit offset will be $\frac{ax+by+cz}{\sqrt{a^2+b^2+c^2}} = D±2$

• How it works:

  You are not changing the direction of normal.

Answer 2

Hint...Assuming you want the two planes parallel and exactly $2$ units apart, if you know how to find the normal $\underline{n}$ to the plane, then calculate the unit vector in this direction, namely, $$\underline{\hat{n}}=\frac{\underline{n}}{|\underline{n}|}$$.

Now add (or subtract) $2\underline{\hat{n}}$ to the position vectors of the three points you have already.

If you only need the equation of the plane, not all three points, then use this process to work out the new coordinates for one of the points, and then use $$\underline{r}\cdot\underline{n}=\underline{a}\cdot\underline{n}$$