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Given the lines:

$l_1: (1,-2,5)+t(2,1,-1)$

$l_2: (3,4,1)+s(2,1,-1)$.

Need to find the equation of the plane that is equally distance from those two lines, and Perpendicular to a plane that those lines creates.

My attempt:

I can see that the lines are parallels. i tried to find the equation of the plane that the two lines form but i don't know how to continue.

2 Answers 2

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Let's rewrite the two lines as follows $$ \begin{gathered} l_{\,1} :\mathbf{a} + t\,\mathbf{v} = \left( {1, - 2,5} \right) + t\left( {2,1, - 1} \right) \hfill \\ l_{\,2} :\mathbf{b} + t\,\mathbf{v} = \left( {3,4,1} \right) + s\left( {2,1, - 1} \right) \hfill \\ \end{gathered} $$ so that it is clear the definition of the two points $\mathbf a$ and $\mathbf b$ and vector $\mathbf v$.
Then as you correctly noted, being the vector $\mathbf v$ the same, the lines are parallel.
Consider now the segment joining the points $\mathbf a$ and $\mathbf b$ and the relevant difference vector $$ \mathbf{c} = \mathbf{b} - \mathbf{a} = \left( {2,6, - 4} \right) = 2\left( {1,3, - 2} \right) $$ The plane $\pi _{\,//} $ containing the two lines, shall contain both points and be parallel to $\mathbf v$, i.e. it shall pass by one of the points and be normal to the vector $$ \mathbf{n} = \mathbf{c} \times \mathbf{v} = \left( { - 2, - 2, - 4} \right) = - 2\left( {1,1,2} \right) $$ and since the vector $\mathbf n$ is not null, the two lines are distinct.

Now the required plane $\pi _{\, \bot } $ shall clearly be:
- orthogonal to $\pi _{\,//} $, i.e. parallel to $\mathbf n$
- parallel to the lines, i.e. parallel to $\mathbf v$
- equidistant from the lines, i.e. passing through the mid-point of the segment $\mathbf b - \mathbf a$

which means:
- orthogonal to the vector $\mathbf{m} = \mathbf{n} \times \mathbf{v} = \left( { - 3,5, - 1} \right)$
- passing through the point $\left( {\mathbf{b} + \mathbf{a}} \right)/2 = \left( {2,1,3} \right)$

So we get
$$ \pi _{\, \bot } :\left( {x - 2,y - 1,z - 3} \right) \cdot \left( { - 3,5, - 1} \right) = 0 $$

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    thank you! i think i got it. the plane is $8x-11y+5z-20=0$ – 2017-01-17
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    @mila: uhm, does not look correct. If you develop the dot product in my reply you get $3x-5y+z-4=0$. Where did you get your equation from ? – 2017-01-17
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    for the point on the new line i did average of $(1,-2,5), (3,4,1)$ so i got: $(2,1,3)$. that line as the same direction as two lines so the line is: $(2,1,3)+k(2,1,-1)$, or $(2+2k,1+k,3-k)$. then i subtract the point $(1,-2,5)$ so i got : $u=(2+2k-1,1+k+2,3-k-5)$. then i used the dot product: $u \cdot (2,1,-1)=0$, which gave me: $(8,-11,5)$. so $8x-11y+5z+d=0$, then i put:$(2,1,3)$ and got $8x-11y+5z-20=0$ – 2017-01-18
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    @mila: "Then I subtract the point $ (1,-2,5)$" what does that mean ? – 2017-01-18
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    for the direction – 2017-01-18
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Hint the equation of plane containing two lines is $(r-a_1). (b_1×b_2) $ where $a_1$ can be any of the two points. And $b_1,b_2$ are two vectors. Now find the point on the required plane which is at same distance from both lines. This can be done like $x-1=|x-3|$ same with $y,z $ to get the point as $-1,-3,3$ now this plane is perpendicular to other plane containing the lines which means the normals are perpendicular so let drs of normal of unknown plane be $l,m,n $ we can get drs of normal of other plane ie plane containing lines from the above mentioned equation to get $ll_1+mm_1+nn_1=0$ get ratios in terms of $l$ of $l,m,n $ . Equation of a plane is also $lx+my+nz=d $ where $l,m,n $ are drs and $d $ is a constant . This plane passes through $(-1,-3,3) $ to get the required plane.