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The line $r$ has the following equation: $ \begin{cases} x-y+2=0 \\ 2x-z+1=0 \end{cases} $

What's the equation of the line perpendicular to $r$ and going through $P(0,0,-1)$, written in the same form as $r$?

I started solving the system of equations and got $\begin{cases} x=-\dfrac{1}{2} + \dfrac{1}{2} t \\ y = \dfrac{3}{2} + \dfrac{1}{2}t \\ z=t\end{cases}$ but I'm not sure what to do from here.

Thanks in advance.

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4 Answers 4

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Now you have a parametric equation $R: \mathbb R\to\mathbb R^3$ for $r$.

Then solve for the $t$ such that $P-R(t)$ is perpendicular to the direction of $r$.

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  • $\begingroup$ I implemented your method. This seems to lead to a quadratic in t. The solutions are t=0 and t=-1 (which agrees with mine). How do you get rid of the spurious solution (t=0)? $\endgroup$ Aug 16, 2018 at 4:34
  • $\begingroup$ I can see the problem with t=0. It comes from the fact that the direction vector for r is a multiple of t, so its dot product will be 0 at the point where t=0. $\endgroup$ Aug 16, 2018 at 4:41
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    $\begingroup$ I'm overthinking! The r direction vector doesn't need the t, so the dot product is a linear equation with solution t=-1. $\endgroup$ Aug 16, 2018 at 5:01
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With $p = (x,y,z)$ given the planes

$$ \Pi_1\to < p-P_1, \vec n_1 > = 0\\ \Pi_2\to < p-P_2, \vec n_2 > = 0 $$

and the point $P_0$ we need a line

$$ L\to p = P_0 + \lambda \vec v $$

such that $< \vec v, \vec n_1 \times \vec n_2 > =0$ for $\vec v \ne 0$ and also

$$ \Pi_1 \cap L = \Pi_2 \cap L = p^* $$

or

$$ < p^*-P_1,\vec n_1 > = 0\\ < p^*-P_2,\vec n_2 > = 0\\ <P_0-p^*, \vec n_1\times\vec n_2> = 0 $$

those three conditions give $p^* = (x^*, y^*, z^*)$ the sought point.

NOTE

$<\cdot,\cdot >$ represents the scalar product of two vectors and $\cdot\times\cdot$ represents the vectorial product. Also

$$ \vec n_1 = (1,-1,0)\\ P_1 = (0,2,0)\\ \vec n_2 = (2,0,-1)\\ P_2 = (0,0,1) $$

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  • $\begingroup$ Your answer looks more complicated than the question! $\endgroup$ Aug 16, 2018 at 0:54
  • $\begingroup$ @herbsteinberg I'm sorry for you. Orthogonality is a simple idea. $\endgroup$
    – Cesareo
    Aug 16, 2018 at 6:18
  • $\begingroup$ I agree with statement about orthogonality. My comment was about your presentation - it was overwhelming in notation. Henning Makholm said it much better. $\endgroup$ Aug 16, 2018 at 15:20
  • $\begingroup$ @herbsteinberg Good for Henning but I prefer my way to tell. There are tastes for everything. $\endgroup$
    – Cesareo
    Aug 16, 2018 at 15:29
  • $\begingroup$ @Cesaeo I feel that clarity is most important. Your presentation is extremely difficult to wade through, starting with using symbols without defining them. $\endgroup$ Aug 16, 2018 at 18:54
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The perpendicular will be at the shortest distance between P and r. Distance (squared) is $(\frac{t-1}{2}-0)^2+(\frac{t+3}{2}-0)^2+(t+1)^2$ Minimum is $t=-1$. Therefor point on the line is $P_1=(-1,1,-1)$ and the line you want is $P+s(P_1-P)$.

As a check $(P_1-P)\cdot r=0$

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  • $\begingroup$ Since $z = t$, $P_1 = (-1, 1, -1)$. $\endgroup$
    – Maxim
    Aug 21, 2018 at 14:22
  • $\begingroup$ It's unlocked. You can now edit again. $\endgroup$
    – quid
    Aug 21, 2018 at 16:50
  • $\begingroup$ @Maxim You are right. It has been corrected. $\endgroup$ Aug 21, 2018 at 18:06
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You can take the plane normal to $r$ and going through $P$ and the plane going through $r$ and through $P$ as two planes defining the line $r_\perp$.

You know that the vector $\mathbf r =(1, 1, 2)$ is parallel to the line $r$, therefore $\mathbf r$ is a normal to the first plane and $\mathbf r \times (P_r - P)$ is a normal to the second plane, where $P_r$ is any point on $r$. Then $$r_\perp = \cases{ x + y + 2 z + 2 = 0 \\ x + y - z - 1 = 0}.$$

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