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I have given 3 points, $p_0(x_0,y_0,z_0), p_1(x_1,y_1,z_1)$ and $p_2(x_2,y_2,z_2)$.

  1. How to find the point $p_v(x_v,y_v,z_v)$ which lies on the line $p_1p_2$ and is at perpendicular distance from point $p_0$?
  2. How to find point $p_vRev(x_r,y_r,z_r)$ which is obtained after rotating given point $p_v$ about point $p_0$ exactly $180$ degree in any axis? (since this point is same).
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    $\begingroup$ What is a perpendicular distance? A distance between a point and line is the length of the perpendicular segment from the point to the line, so it is always perpendicular in that meaning. At what distance from $p_0$ should $p_v$ be? $\endgroup$ – Roy Sht Jun 12 at 16:47
  • $\begingroup$ You can either use projection or simply the inner product to find point $p_0$. Also, you can find it by writing distance formula and minimizing the distance. $\endgroup$ – Heroshizen Jun 12 at 16:49
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    $\begingroup$ Perhaps you mean that the line $p_vp_0$ is perpendicular to $p_1p_2$? $\endgroup$ – TonyK Jun 12 at 17:03
  • $\begingroup$ @Heroshizen i have p0 as, said in question. i am trying to find pv and pvRev. Here pv is the point lying in Line segment p1 to p2 and is also at perpenducular distance from p0. Now in case of pvRev it is point got after rotating the p0pv line 180 degree in any direction. and pvrev is the end point of that rotated line segment. $\endgroup$ – Andrew Jun 12 at 17:15
  • $\begingroup$ What is a perpendicular distance? $\endgroup$ – user Jun 12 at 17:28
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Here is an idea, so correct me if I am wrong :

The line $p_1p_2$ can be written as :

$$\frac{x-x_1}{a } = \frac{y-y_1}{b} = \frac{z-z_1}{c} =\mu $$

$a,$ $b,$ and $c$ can be found by :

$$a = x_1-x_2$$

$$b= y_1-y_2$$

$$c=z_1-z_2$$

So, a point on the line can be expressed as : $(a\mu+x_1,b\mu+y_1,c\mu+z_1)$

Here $\mu$ is a parameter that can be varied to get any point on the line.

So the particular value of $\mu $ for $p_v$ can be found by solving : $(a\mu+x_1,b\mu+y_1,c\mu+z_1) \equiv (x_v,y_v,z_v)......(1)$

enter image description here

Now, the line $p_1p_v \perp p_op_v$

The vectors along these line can be found :

$$p_1p_v \ : \ a \hat{i} + b\hat{j} + c\hat{k} = \overrightarrow{x}$$

$$p_op_v \ : \ (x_o-x_1-a\mu)\hat{i} + (...)\hat{j} + (...)\hat{k} = \overrightarrow{y}$$

Now,

$$\overrightarrow{x} . \overrightarrow{y}=0$$

Solve and get $\mu$

Put this value of $\mu $ in ($1$) to get $p_v$

Apply distance formula to get the perpendicular distance.

Now, write the equation of $p_op_v$just like we wrote the equation of $p_1p_2$ and find a general point on this line (like with $p_v$) with a similar p

Then apply distance formula between this point and $p_v$ and equate this distance to perpendicular distance found before. You will get two solutions from this (2 points on two sides of $p_v$)

Apply similar distance formula relation with $p_o$ and the new point to get the actual one of the two obtained earlier. This will be $p_{v_{Rev}}$

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