Find a point on the same alignment of normal vector of a plane

Question

I need to find a point(x,y,z) that is - distance 2 from a known point P (x1,y1,z1) - on the same alignment of normal vector for plane A - P is on the plane A

the same question as:

Find a point that trace a line with another known point P on a plane, and this line are perpendicular to the plane, the distance between the point an P must be 2

   . A point I want to known (x,y,z)
   | 
   |  -> distance 2

------P----- -> Plane

   |  -> distance 2
   |  
   . A' point I wanto to known (x',y',z')

P is point (x1,y1,z1)

Answer 1

Let the plane be $ax+by+cz=d$, $P=(x_1,y_1,z_1)$. Then we have

$$x=x_1+at\\y=y_1+bt\\z=z_1+ct$$

for some parameter $t$. Since the distance is $2$, we have

$$\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}=\sqrt{(a^2+b^2+c^2)t^2}=2$$

Solving this gives us

$$t=\frac{2}{\sqrt{a^2+b^2+c^2}}$$

So the new point can be found using this value of $t$.