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$L_1$ is given by enter image description here

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!

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The hint.

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 $b=1$.

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  • $\begingroup$ How did you find the $s(2, b+1, -1)$ ? $\endgroup$ – J.Se Jan 4 '18 at 17:16
  • $\begingroup$ It's $(1,b,1)-(-1,-1,2)$. If $B(1,b,1)$ and $A(-1,-1,2)$ then $\vec{AB}(2,b+1,-1)$. $\endgroup$ – Michael Rozenberg Jan 4 '18 at 17:17
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Jan 4 '18 at 17:19
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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).

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$L_1$ 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 $(1, 1, 1)$.

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