Linear Algebra: Line Intersection Let L1 be the line passing through $$P = (-1, 3, -2) and Q = (5, -3, 10)$$ Let L2 be the line passing through $$(4,0,3)$$ in the direction of v= $$\begin{bmatrix}
3 \\
-1 \\
1
\end{bmatrix}
\quad$$ Do these lines intersect? If so, find the point of intersection.
My work:
L1 = P x Q which should give a det matrix of: \begin{vmatrix}
i & -1 & 5 \\
j & 3 & -3 \\
k & -2 & 10
\end{vmatrix}
Which should give you: $$ i(30-6)-j(-10+10)+k(3-15)$$
which gives the equation: $$24x-12z=0$$

L2 = $$\begin{bmatrix}
4 \\
0 \\
3
\end{bmatrix}
\quad + \begin{bmatrix}
3 \\
-1 \\
1
\end{bmatrix}
\quad t$$
Which gives: $$x = 4+3t$$ $$ y = -t$$ $$z= 3+t$$

Now you have to sub in these values into L1:
$$ 24(4+3t) - 12(3+t) = 0$$ $$60 + 60t = 0$$ $$t = -1$$

Now sub t back into L2:
 $$x = 4+3(-1) = 1$$ $$ y = -t = 1$$ $$z= 3+t = 2$$

Is this the point of intersection? I don't know if I did this question correctly.
 A: Since $P=(-1,3,-2)$ and $Q=(5,-3,10)$, we can describe $L_1$ as 
$$L_1=\left\{(-1,3,-2)+t(6,-6,12)\mid t\in \mathbb{R}\right\}.$$ Indeed, the direction is simply determined by $P-Q$ (or $Q-P$ as I did here).
As you said, $$L_2=\left\{(4,0,3)+t(3,-1,1)\mid t\in \mathbb{R}\right\}.$$
Let $P=(x,y,z)\in L_1\cap L_2$, then $\exists t,s\in \mathbb{R}$ such that \begin{eqnarray}
(x,y,z) &=& (-1+4t,3-6t,-2+12t)\\
&=& (4+3s,0-s,3+s)\\
\end{eqnarray}Hence $$(-5+4t+3s,3-6t+s,-2+12t-s)=(0,0,0).$$
Now you can solve this linear system in the variables $s$ and $t$.
A: You will get the system
$$-1+6s=4-3t$$
$$3-6s=-t$$
$$-2+12s=3+t$$
adding the last two equations we obtain
$$1+6s=3$$ so $$s=\frac{1}{3}$$ and $$t=-1$$ plugging this in your first equation we have
$$1=1$$ so the intersection point is given by $$[x,y,z]=[7;1;2]$$
A: First line:
Direction vector: $\;\vec{PQ}=Q-P=(6,-6,12)\implies \text{direction vector is}\;\;(1,-1,2)\;$
Thus the line is 
$$\;\color{red}{r_1(t)=P+t\vec{PQ}=(-1,3,-2)+t(1,-1,2) =(t-1, -t+3, 2t-2)\;,\;\;t\in\Bbb R}\;$$
Second line:
Directly, the line is  
$$\;\color{red}{r_2(t)=(4,0,3)+t(3,-1,1)=(3t+4,-t,t+3)\;,\;\;t\in\Bbb R}\;$$
Both lines intersect iff there exist $\;t,\,s\in\Bbb R\;$ such that $\;r_1(t)=r_2(s)\;$ , which means:
$$(t-1,\,-t+3,\,2t-2) =(3s+4,\,-s,\,s+3)\iff\begin{cases}I&t-1=3s+4\\II&-t+3=-s\\III&2t-2=s+3\end{cases}$$
From $\;I-II\;$ we get $\;2=2s+4\implies s=-1\;$
and thus from $\;I:\;\;t-1=-3+4\implies t=2\;$
and now we substitute in $\;II\;$ in order to find whether things come together (and thus there's an intersection point)m or else we get a contradiction and there is no such a point:
$$III:\;\;2\cdot2-2=-1+3\iff2=2...\checkmark$$
and thus both lines do  intersect each other, at point $\;r_1(2)=r_2(-1)\;$
A: A vector that can 'represent' line connecting $P=(−1,3,−2)$ and $Q=(5,−3,10)$ is :
$$ v_{pq}  = [6, -6, 12]$$
The line $L_{1}$ is therefore can be represented as
$$L_{1} \rightarrow x = -1 + 6t, \: \: y = 3 -6t, \:\: z = -2 + 12t, \:\:\:\: t \in \mathbb{R} $$
For $L_{2}$ is clearly
$$L_{2} \rightarrow x = 4 + 3t', \: \: y =  -1t', \:\: z = 3 + t', \:\:\:\: t' \in \mathbb{R} $$
so for intersection, we must have :
$$ -1+6t = 4+3t' \implies 6t - 3t' = 5 \: \: \: \: (l_{1})$$
$$ 3-6t = -t' \implies-6t + t' = -3 \: \: \: \: (l_{2})$$
$$ -2+12t = 3 + t' \implies 12t - t' = 5 \: \: \: \: (l_{3})$$
hence if 2D lines $l_{1}, l_{2}, $ and $l_{3}$ have a point at which all three intersect at same time, then $L_{1}$ and $L_{2}$ intersect each other. 
$$ -6t + t' = -3 \implies t' = 6t - 3 $$
so
$$  6t - 3t' = 5  \implies t = 1/3 $$
$$ 12t - t' = 5 \implies  t = 1/3$$
so there is a solution, that is t $t=1/3$ and $t' = -1$. So $L_{1}$ and $L_{2}$ intersect each other at this point : $$x(t=1/3)=x(t' = -1)=..., $$
$$ y(t=1/3)=y(t'=-1)=..., $$
$$ z(t=1/3)=z(t' = -1)=...$$
