Prove that all possible lines in function of m form a plane

1. May 30, 2014

powerof

1. The problem statement, all variables and given/known data
We are given a point A(1,1,1) and a vector v=(m-1,3m-5,2m-6). We are asked to write the parametric and continuous (I don't know if that's the appropriate term; in Spanish it's called "forma continua" but you'll see right away what I mean) equations of the line formed by these two.

Then prove that all the possible lines in function of m form a plane.

2. The equations mentioned above

$\left\{\begin{matrix} x=1+(m-1)\lambda \\ y=1+(3m-5)\lambda \\ z=1+(2m-6)\lambda \end{matrix}\right.$

$\frac{x-1}{m-1}=\frac{y-1}{3m-5}=\frac{z-1}{2m-6}$

3. The attempt at a solution

I tried to develop from the second equation the other form to write the line, that is, as an intersection of two planes. I wanted to see if the two normal vectors of the planes happened to be linearly dependent but I haven't had any luck so far. You get the following possible pair of planes:

$\left\{\begin{matrix}\pi'\equiv (2m-6)x+(1-m)z+5-m=0 \\ \pi''\equiv(3m-5)x+(1-m)y+4-2m=0 \end{matrix}\right.$

$\begin{matrix}\overrightarrow{u}_{\pi'}=(2m-6,0,1-m) \\ \overrightarrow{u}_{\pi''}=(3m-5,1-m,0) \end{matrix}$

I honestly don't like this way, it's way too messy in my opinion and I get lost easily. If possible, can I get a few pointers in another direction, to do it another way?

2. May 30, 2014

Saitama

I am not familiar with the method you are using. One thing you can do is consider the three lines with direction ratios $<m_i-1,3m_i-5,2m_i-6>$ for $i=1,2,3$ and show that the scalar triple product of these directions ratios is zero.

3. May 31, 2014

powerof

Yes, after a good night's sleep I realized my first attempt to do it doesn't make any sense so disregard it.

The normal vectors of the two planes don't have to be linearly dependent for all the possible lines to form a plane.

I did it the way you proposed and I got the correct solution:

$\begin{vmatrix} m_{1}-1 &3m_{1}-5 &2m_{1}-6 \\ m_{2}-1 &3m_{2}-5 &2m_{2}-6 \\ m_{3}-1&3m_{3}-5 & m_{3}-6 \end{vmatrix}=\begin{vmatrix} m_{1}-1 &3m_{1}-5 &2m_{1}-6 \\ m_{2}-m_{1} &3(m_{2}-m_{1}) &2(m_{2}-m_{1}) \\ m_{3}-m_{1}&3(m_{3}-m_{1}) & 2(m_{3}-m_{1}) \end{vmatrix}=(m_{3}-m_{1})(m_{2}-m_{1})\begin{vmatrix} m_{1}-1 &3m_{1}-5 &2m_{1}-6 \\ 1 &3 &2 \\ 1&3 & 2 \end{vmatrix}=0$

The first step I did was to subtract from the second and third rows the first one. After that the rest should be obvious at first sight.

Thank you very much for your help.

Last edited: May 31, 2014
4. May 31, 2014

Ray Vickson

There is a big difference between saying that the points "lie in a plane" vs. saying the points "form a plane". Certainly those points all lie in a certain plane $P$, but the issue is: do all the points in $P$ have the indicated form? In other words, given $(x,y,z)$ in the plane $P$, can we always find $\lambda$ and $m$ that satisfy your three equations? If that is true the points truly do form a plane; but if it is not true you can only say they lie in a plane, not that they "form" one.

Hint: they do NOT form a plane!

Last edited: May 31, 2014
5. Jun 1, 2014

Joffan

You could try taking the cross product of v(m) and v(m').

6. Jun 1, 2014

powerof

Thanks, that's another way, and even easier (I think). I checked it and it works. For any m and m' when you do the cross product you get a multiple of the same vector (if I'm not wrong it's (2,2,1) in it's simplest form). I'm too lazy to write it all in latex.