Determine point of intersection or find the value of $z$

Question

Let $L_1$ be the line passing through the points $Q_1=(4, −2, −4)$ and $Q_2=(5, −1, −5)$ and let $L_2$ be the line passing through the point $P_1=(−13, −12, 6)$ with direction vector $\underline{d}=[6, 3, −6]^T$. Determine whether $L_1$ and $L_2$ intersect. If so, find the point of intersection $Q$. If not, find a value for the $z$-coordinate of $P_1$ so the resulting lines do intersect.

Please tell me what the result is

Answer 1

Find the equation for both lines:

$$L_1 = \begin{pmatrix}4 \\ -2 \\ -4\end{pmatrix} + t\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}\;L_2 = \begin{pmatrix}-13 \\ -12 \\ 6\end{pmatrix} + s\begin{pmatrix}6 \\ 3 \\ -6\end{pmatrix}$$ Set $L_1=L_2$, to form a set of linear equations in s and t:

$$\begin{pmatrix}4 \\ -2 \\ -4\end{pmatrix} + t\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix} = \begin{pmatrix}-13 \\ -12 \\ 6\end{pmatrix} + s\begin{pmatrix}6 \\ 3 \\ -6\end{pmatrix}$$ Rearranging: $$\begin{pmatrix}17 \\ 10 \\ -10\end{pmatrix}=s\begin{pmatrix}6 \\ 3 \\ -6\end{pmatrix}-t\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}$$

Now $$17=6s- t,\ 10=3s-t\\\implies t=-3,s=\frac{7}{3}$$ Verify with the final equation: $$-10\neq-6(\frac{7}{3})+1(-3)$$ Since this does not hold, we must change either $L_1$ or $L_2$, e.g. $$-4-(t)=x-6(s)\\ \implies -4-(-3)=x-6(\frac{7}{3})\\ \implies x = 13.$$ So $$L_2 = \begin{pmatrix}-13 \\ -12 \\ 13\end{pmatrix} + t\begin{pmatrix}6 \\ 3 \\ -6\end{pmatrix}$$

Answer 2

(Check my arithmetic). $L_1=Q_1+(Q_2-Q_1)s=(4,-2,-4)+(1,1,-1)s$ and $L_2=P_1+d=(-13,-12,6)+(6,3,-6)T$. Match $x$ and $y$ coordinates to get simultaneous equations. $-13+6T=4+s,-12+3T=-2+s$. Solve and get $T=7/3, s=-3$. The $z$ coordinate for $L_1$ is$-1$ while the coordinate for $L_2$ is $4$. To get them to intersect change $z$ coordinate of $P_1$ to $13$.

$Q=(1,-5,-1)$.