How to find the coordinates of points on a line perpendicular to a given plane

Question

I am given a plane equation $Ax+By+Cz+D=0$ and coordinates $(x,y,z)$ of a point $P$ lying on a plane. I need to determine coordinates $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ of two points $P_1$ and $P_2$ which are located on a line that is passing through $P$ perpendicular to the plane. Furthermore I need $P_1$ and $P_2$ to be located at the same distance $d$ from the plane.

The illustration below shows everything better than I explained. Is it possible to find out these coordinates? I've always been bad with analytic geometry so I hope someone could put this to me simply. Thanks in advance.

Answer 1

Plane: $Ax + By + Cz + D = 0$.

Normal vector to this plane: $(A,B,C)$.

Normalized:

$\vec n = \frac {1}{(A^2 + B^2 + C^2)^{1/2}} (A,B,C)$;

Consider a straight line passing through the point $(x,y,z)$ on the plane with direction vector $\vec n$ :

$\vec r(t) = (x,y,z) + t \vec n$.

Since $\vec n$ is normalized:

Pick $t_1 = d$ and $t_2 = - d$ to find $P_1$ and $P_2$,

$(x_1,y_1,z_1) = (x,y,z) + d \vec n$, and

$(x_2,y_2,z_2) = (x,y,z) - d \vec n$.

Helps?

Answer 2

I'll give you an example, hoping it will be better understood

Let the plane be $\pi:\;x+2 y+3 z+4=0$ and the point on it $P(1;\;2;-3)$

The line $r$ passing through $P$ and perpendicular to $\pi$ has parametric equation $r=(1 + t,\; 2 + 2 t,\; -3 - 3 t)$

To find a pair of points $P_1$ and $P_2$ having distance $d=\sqrt{14}$ from $P$ you write the equation of the sphere with centre in $P$ and radius $\sqrt{14}$

$(x-1)^2+(y-2)^2+(z+3)^2=14$

and find the intersection with $r$ plugging $x=1+t;\;y=2+2t;\;z=-3-3t$ in the previous equation

$14 t^2=14\to t=\pm1$

therefore the points are $P_1(0;\;0;\;0),\;P_2(2;\;4;\;-6)$

Hope this helps