# Determine point of intersection or find the value of $z$ [closed]

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

## closed as off-topic by RRL, Riccardo.Alestra, Adrian Keister, Andrew, ShaunFeb 5 at 15:39

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• Have you made any effort whatsoever to solve this on your own? Please show your work. – amd Jan 17 at 21:47
• Yes I have, although last time I showed some effort on my question no one even bothered to respond so please don't judge prematurely – DreamVision2017 Jan 18 at 18:41
• You haven’t posted a question. You’ve posted a context-free problem statement and then made a demand, however politely. Taken by itself, this is not the sort of “question” that will be well-received on this site. That aside, this one is a minor variation of the exercises that you’ve asked about in previous questions. If you’re having trouble with this subject, you’d be well advised to review and seek help from your instructor/TA. – amd Jan 18 at 21:17

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 $$L_2 = \begin{pmatrix}-13 \\ -12 \\ 13\end{pmatrix} + t\begin{pmatrix}6 \\ 3 \\ -6\end{pmatrix}$$

• The other person who answered got $1$. At least one of you is wrong. – amd Jan 17 at 21:46

(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 $$13$$.

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

• @amd You are right. I made arith. error at end. Rest of calculation, including point of intersection, is correct. $P_1(z)$ corrected. – herb steinberg Jan 17 at 22:25