Vectors in Three Dimensions - Point of Intersection Can anyone help me solve this question?
$r_1(t)=(-2.5, 13, 3) + t<1, 4, 2>$
$r_2(t)=(6.5, 7, -7) + t<3, 0, -2>$
Find the point of intersection, P, of the lines $r_1$ and $r_2$.
Any help is greatly appreciated! :)

The trick is that we have to change the name of the variable $t$ in one of the equations.
We are looking for a point $P$ such that $P=r_1(t)$ for some $t$ and $P=r_2(s)$ for some $s$. Equating them coordinatewise, we have to solve the following system of equations:
$$\begin{align}
-2.5+t &= 6.5+3s \\
13+4t &= 7\\
3+2t &= -7-2s
\end{align}$$
I would solve e.g. the last two equations and check whether its solution $(s,t)$ satisfy the first one. 
(If yes, you get the intersection point by pluggin in the solution $t$ in $r_1$, if not, that means that the lines are skew.)
