Determine if two straight lines given by parametric equations intersect Does
$[x,y,z] = [4,-3,2] + t[1,8,-3]$
intersect with
$[x,y,z] = [1,0,3] + v[4,-5,-9] ?$
Attempt
To find out if they intersect or not, should i find if the direction vector are scalar multiples? Clearly they are not, so that means they are not parallel and should intersect right? But the correct answer is that they do not intersect. How do you do this? Thanks!
 A: In 3 dimensions, two lines need not intersect. The two lines intersect if and only if there are real numbers $a$, $b$ such that $[4,-3,2] + a[1,8,-3] = [1,0,3] + b[4,-5,-9]$. We have the system of equations:
$$
\begin{aligned}
4+a &= 1+4b &(1) \\
-3+8a &= -5b &(2) \\
2-3a &= 3-9b &(3)
\end{aligned}
$$
$-(2)+(1)+(3)$ gives
$$
9-4a=4 \\ 
\Downarrow \\
a=5/4
$$
$(2)$ then gives
$$7 =-5b \\
\Downarrow \\
b = -7/5
$$
But in $(1)$ we then have
$$4+5/4=1-28/5$$
which is false. So no solution exists, and the lines do not intersect.
A: Vector equations can be written as simultaneous equations.
Thus
 [x,y,z] = [4,-3,2] + t[1,8,-3]

becomes
x =  4 +  t
y = -3 + 8t
z =  2 - 3t

Your two lines intersect if
[4,-3,2] + t[1,8,-3] = [1,0,3] + v[4,-5,-9]

or
 4 +  t = 1 + 4v
-3 + 8t = 0 - 5v
 2 - 3t = 3 - 9v

Thus, you have 3 simultaneous equations with only 2 unknowns, so you are good to go! If you can find a solution for t and v that satisfies these equations, then the lines intersect.
A: $\newcommand{\+}{^{\dagger}}%
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$$
\vec{A} + t\,\vec{B} = \vec{C} + v\,\vec{D}\quad\imp\quad
\left\lbrace%
\begin{array}{rcrcl}\quad
B^{2}\ t & - & \vec{D}\cdot\vec{B}\ v & = & \pars{\vec{C} - \vec{A}}\cdot\vec{B}
\\
\vec{B}\cdot\vec{D}\ t & - & D^{2}\ v & = & \pars{\vec{C} - \vec{A}}\cdot\vec{D}
\end{array}\right.\tag{1}
$$
If $\ds{0 \not= -B^{2}D^{2} + \pars{\vec{B}\cdot\vec{D}}^{2}
        = -B^{2}D^{2}\sin^{2}\pars{\angle\pars{\vec{B},\vec{D}}}
        = -\pars{\vec{B} \times \vec{D}}^{2}}$ which is equivalent to:
$$
\vec{B} \not= \vec{0}\quad\mbox{and}\quad\vec{D} \not= \vec{0}\quad\mbox{and}\quad
\vec{B} \not\parallel \vec{D},
$$
we can find the pair $\pars{t,v}$  from the pair of equations $\pars{1}$.
A: I think they are not on the same surface (plane).  Two vectors can be: (1) in the same surface in this case they can either (1.1) intersect (1.2) parallel (1.3) the same vector; and (2) not in the same surface. 
