Vectors in Three Dimensions - Point of Intersection

Question

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! :)

Answer 1

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.)