find $m$ based on $p$ and $q$

Question

Two lines are given, $p:\frac{x-1}{3}=\frac{y-2}{m}=\frac{z}{-1}$ and $q:\frac{x+3}{1}=\frac{y+1}{3}=\frac{z-3}{-2}$. Find the value of $m$ so that $p$ and $q$ belong to the same plane $\alpha$. I have tried some things, but it led me nowhere. I found the normal vector of $\alpha$ and tried using that, but could not get to the solution. Any help is appreciated.

Answer 1

Direction vector of lines $p$ and $q$ are:

$$\vec{p} = 3 \vec{i} + m \vec{j} - \vec{k}$$ $$\vec{q} = \vec{i} + 3 \vec{j} - 2 \vec{k}$$

These lines are not parallel because there is no scalar $a$ such that:

$$\vec{p} = a \vec{q}$$

So these lines will have to interesect in order to belong to the same plane. Let's find the intersection point:

$$\frac{x-1}{3}=\frac{y-2}{m}=\frac{z}{-1}=b$$ $$\frac{x+3}{1}=\frac{y+1}{3}=\frac{z-3}{-2}=c$$

From the last two equations we have:

$$x = 3b + 1, y = mb+2, z=-b$$ $$x = c - 3, y = 3c-1, z=-2c+3$$

Or:

$$ 3b+1 = c - 3, mb + 2 = 3c - 1, -b=-2c + 3$$

From the first and the third equation you get:

$$b = -1, c= 1$$

...and from the second you get:

$$m=0$$

(which seems weird but it's a perfectly valid solution)

Answer 2

$p:(1,2,0)+t(3,m,-1)$ and $q:(-3,-1,3)+s(1,3,-2)$.

Thus, we need $(1+3,2+1,0-3)=t(3,m,-1)+s(1,3,-2)$,

which gives: $3t+s=4$ and $-t-2s=-3$.

From here we obtain: $t=1$ and $s=1$.

Thus, $3=1\cdot m+1\cdot3$ or $m=0$.