1
$\begingroup$

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.

enter image description here

$\endgroup$
  • 2
    $\begingroup$ Hint: $PP_1$ and $PP_2$ are parallel to the normal to the plane. $\endgroup$ – Yves Daoust Jul 26 '17 at 7:47
  • $\begingroup$ @YvesDaoust I see, but I don't know coordinates of $P_1$ and $P_2$, I only know the length of $PP_1$ and the fact that it's parallel to $(A,B,C)$. Maybe I need coordinates of a normal passing through $P$ $\endgroup$ – AnatoliySultanov Jul 26 '17 at 8:22
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You are welcome. $\endgroup$ – Peter Szilas Jul 27 '17 at 16:21
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Intersection with a sphere, although not incorrect, seems an unnecessary complication. Much simpler to use a unit normal, as Peter Szilas does. $\endgroup$ – amd Jul 26 '17 at 18:24
  • $\begingroup$ Interesting solution, thanks for the idea! $\endgroup$ – AnatoliySultanov Jul 27 '17 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.