# Determine the value of $b$ so that the lines $L_1$ and $L_2$ intersect [closed]

$L_1$ is given by

and $L_2$ goes through the points $(−1, −1, 2)$ and $(1, b, 1)$, where $b$ is a constant.

1. Determine all the values of $b$ so that the lines $L_1$ and $L_2$ intersect

In order to find the intersection I have to put $L_1=L_2$ but how do I find the parameterization of $L_2$ from the given points?

Thank you!

## closed as off-topic by TheSimpliFire, José Carlos Santos, darij grinberg, Shailesh, Eevee TrainerJan 16 at 1:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, José Carlos Santos, darij grinberg, Shailesh, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.

The second line it's $(-1,-1,2)+s(2,b+1,-1)$ and we need $(-1,1,-1)\not||(2,b+1,-1)$.
Now, for the intersecting we need that there are $s$ and $t$ for which $$(2,2,-1)=t(-1,1,-1)+s(2,b+1,-1)$$ and solve the system. I got $t=0$, $s=1$ and $b=1$.
• How did you find the $s(2, b+1, -1)$ ? – J.Se Jan 4 '18 at 17:16
• It's $(1,b,1)-(-1,-1,2)$. If $B(1,b,1)$ and $A(-1,-1,2)$ then $\vec{AB}(2,b+1,-1)$. – Michael Rozenberg Jan 4 '18 at 17:17
To parametrise a line you need a point it passes through and its direction. You can get from the vector formed by the two points it passes through. Thus $L_2=(-1,-1,2)+t(2,b+1,-1).$L_1$is given by $$x=1-t$$$$y=1+t$$$$z=1-t$$ and$L_2$is given by $$x=-1+(1-(-1))s=-1+2s$$$$y=-1+(b-(-1))s=-1+(b+1)s$$$$z=2+(1-2)s=2-s$$ Now we solve $$1-t=-1+2s\implies2-t=2s$$ $$1-t=2-s\implies t=s-1$$ so using the equations for$x$and$z$, $$2-s+1=2s\implies \boxed{s=1}\implies \boxed{t=0}$$ Let's do the same for$y$. We solve $$1+t=-1+(b+1)s\implies 1+0=-1+(b+1)(1)\implies 1=-1+b+1=b$$ Hence intersection only occurs when $$\boxed{b=1}$$ In other words, for an intersection,$L_2$must go through the points$(-1, -1, 2)$and$(1, 1, 1)\$.