# How do I find the line perpendicular to the intersection of two planes and going through a certain point?

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.

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 $r$.

• 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)? Aug 16, 2018 at 4:34
• 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. Aug 16, 2018 at 4:41
• I'm overthinking! The r direction vector doesn't need the t, so the dot product is a linear equation with solution t=-1. Aug 16, 2018 at 5:01

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

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

• Your answer looks more complicated than the question! Aug 16, 2018 at 0:54
• @herbsteinberg I'm sorry for you. Orthogonality is a simple idea. Aug 16, 2018 at 6:18
• I agree with statement about orthogonality. My comment was about your presentation - it was overwhelming in notation. Henning Makholm said it much better. Aug 16, 2018 at 15:20
• @herbsteinberg Good for Henning but I prefer my way to tell. There are tastes for everything. Aug 16, 2018 at 15:29
• @Cesaeo I feel that clarity is most important. Your presentation is extremely difficult to wade through, starting with using symbols without defining them. Aug 16, 2018 at 18:54

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 $(P_1-P)\cdot r=0$

• Since $z = t$, $P_1 = (-1, 1, -1)$. Aug 21, 2018 at 14:22
• It's unlocked. You can now edit again.
– quid
Aug 21, 2018 at 16:50
• @Maxim You are right. It has been corrected. Aug 21, 2018 at 18:06

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 $$r_\perp = \cases{ x + y + 2 z + 2 = 0 \\ x + y - z - 1 = 0}.$$