Distance between two lines $L_1:\> x+y+z=6,\> x-2z=-5$ and $L_2:\> x+2y=3,\> y+2z=3 $

Question

Find the distance between the two lines defined by : $$\mathbb L_{1}= \begin{cases} x+y+z=6 & \\ x-2z=-5 & \\ \end{cases}$$ $$\mathbb L_{2}= \begin{cases} x+2y=3 & \\ y+2z=3 & \\ \end{cases}$$

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I know that if we have two lines : $\mathbb L_{1}=P_1+tv_1$ and $\mathbb L_{2}=P_2+tv_2$,then the distance is given by : $$d(\mathbb L_{1},\mathbb L_{2})=\frac{\left|\left(P_{2}-P_{1}\right)\cdot\left(v_{1}\times v_{2}\right)\right|}{\left|v_{1}\times v_{2}\right|}$$

But the problem is that thegiven equations are not in the mentioned form,and I 'm not sure even if they are line (the equations seems to be plane). So how to start?

Answer 1

Parametrize the two lines as follows:

For $\mathbb L_{1}$, let $z=t$. Then, $x= -5+2t$ and $y = 11-3t$, which leads to $P_1=(-5,11,0)$ and $v_1=(2,-3,1)$.

For $\mathbb L_{2}$, let $z=s$. Then, $x= 3-2s$ and $y = -3+4s$, which leads to $P_2=(-3,3,0)$ and $v_2=(4,-2,1)$.

Thus, $ v_{1}\times v_{2} = (-1,2,8) $ and the distance is

$$d(\mathbb L_{1},\mathbb L_{2})=\frac{\left|\left(P_{2}-P_{1}\right)\cdot\left(v_{1}\times v_{2}\right)\right|}{\left|v_{1}\times v_{2}\right|} = \frac{\left|(2,-8,0)\cdot(-1,2,8)\right|}{\left|(-1,2,8)\right|} = \frac{18}{\sqrt{69}} $$

Answer 2

Sketch of a general solution :

1. step one : find $P_1$ and $P_2$ by solving the two systems of equations. In the general case the plane equations have three variables (this is why their intersection is a line meaning an infinity of points) so you need to set for example $z=0$

2. step two : in order to find the directional vectors $v_1$ and $v_2$ we start with the two pairs of normal vectors to the planes whose intersections are the lines we’re studying. So for example with the first line the two normal vectors are $n_1=(1,1,1)^T$ and $n_1’=(1,0,-2)^T$ and a directional vector of the first line is orthogonal to the two normal vectors so for example we can take $v_1=n_1\times n_1’$ the vector product of the two normal vectors. Similarly $v_2=n_2\times n_2’$

3. step three : With all this in hand we can apply the formula

Answer 3

This answer was written mostly to convince myself that this can be brute-forced by Lagrange multipliers. Anyway, we are trying to minimize $$f(x,y,z,x',y',z') = (x-x')^2+(y-y')^2+(z-z')^2$$ with respect to constraints $$ \begin{cases} g_1(x,y,z,x',y',z') = x+y+z-6=0 & \\ g_2(x,y,z,x',y',z') = x-2z+5 =0& \\ \end{cases} \qquad \begin{cases} g_3(x,y,z,x',y',z') = x'+2y'-3 & \\ g_4(x,y,z,x',y',z') = y'+2z'-3 & \\ \end{cases}$$ The point $(x,y,z,x',y',z')$ which minimizes $f$ will satisfy $$\nabla f \in \operatorname{span}\{\nabla g_1, \nabla g_2,\nabla g_3,\nabla g_4\}.$$ The gradients are $$\nabla f (x,y,z,x',y',z') = 2(x-x',y-y',z-z',x'-x,y'-y,z'-z)$$ $$\nabla g_1 (x,y,z,x',y',z') = (1,1,1,0,0,0)$$ $$\nabla g_2 (x,y,z,x',y',z') = (1,0,-2,0,0,0)$$ $$\nabla g_3 (x,y,z,x',y',z') = (0,0,0,1,2,0)$$ $$\nabla g_4 (x,y,z,x',y',z') = (0,0,0,0,1,2)$$ so denoting $(\mathbf{x}, \mathbf{y}, \mathbf{z}) = (x-x',y-y',z-z')$ it follows that $$(\mathbf{x}, \mathbf{y}, \mathbf{z}) \in \operatorname{span}\{(1,1,1),(1,0,-2)\} \cap \operatorname{span}\{(1,2,0),(0,1,2)\}.$$ This intersection can be seen to be $\operatorname{span}\{(-1,2,8)\}$ since $$(-1,2,8) = 2(1,1,1)-3(1,0,-2) = -(1,2,0)+4(0,1,2).$$ Therefore $(\mathbf{x}, \mathbf{y}, \mathbf{z}) = t(-1,2,8)$ for some $t \in \Bbb{R}$. We also know $$ \begin{cases} x+y+z=6 \,\mid \cdot \,2& \\ x-2z=-5 \,\mid \cdot \,(-3) & \\ \end{cases} \qquad \begin{cases} x'+2y'=3 \,\mid \cdot \,(-1) & \\ y'+2z'=3 \,\mid \cdot \,4 & \\ \end{cases}$$ so $$-x+2y+8z=27, \qquad -x'+2y'+8z'=9.$$ Subtracting gives $$18 = -\mathbf{x}+2\mathbf{y}-8\mathbf{z} = t\|(-1,2,8)\|^2 = 69t \implies t = \frac{18}{69}.$$ Now we have $$f(x,y,z,x',y',z') = \|(\mathbf{x}, \mathbf{y}, \mathbf{z})\|^2 = t^2\|(-1,2,8)\|^2 = \frac{18^2}{69}$$ so finally $$d = \frac{18}{\sqrt{69}}.$$