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In a $3$-dimensional Euclidean space, we are given the points $A(1,0,-2)$ and $B(3,-1,1)$. Find two lines $a$ and $b$ such that $A\in a$ , $B\in b$ , and the distance $d(a,b)$ between the lines is $2$.

Clayton
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M.M.
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  • How is the distance between two lines defined? – Clayton Feb 14 '18 at 23:55
  • The distance between two lines is defined as the minima of the distances from Q to P, where Q is a point of the first line and P is a point of the second one. – M.M. Feb 14 '18 at 23:59
  • Do you know about skew lines and how to find the shortest Euclidean distance between any two skew lines? – Prasun Biswas Feb 15 '18 at 00:03
  • I know the formula to calculate the distance between two skew lines, but I don't know how to obtain the equations for the 2 lines. My idea was to use parametric equations, imposing that the line $a$ passes through $A$ and $b$ passes through $B$, but I don't know how to find the direction vectors for these lines. – M.M. Feb 15 '18 at 00:15
  • you can reduce this problem to 2d: pick a third point not collinear with A. B and you can find equation of a plane. On that plane, draw a circle of radius 2 and use A the center. Then draw a tangent line from B to that circle. The tangent point will be $\sqrt{10}$ from B. Finally, draw line parallel to that tangent line that goes thru A. – Vasili Feb 15 '18 at 00:50

1 Answers1

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You form first the vector $\vec{AB}=2\vec i-\vec j+3\vec k$. In order to use the formula $$d=\frac{|\vec{AB}\cdot(\vec u_a\text{ x } \vec u_b)|}{|\vec u_a\text{ x } \vec u_b|}$$ you need to choose the director vectors $\vec u_a$ and $\vec u_b$ of the searched lines $a$ and $b$. For this do you have many choices (your "two lines" are not unique).

Put, for example, $\vec u_a=\vec i+\vec j+\vec k$ and $\vec u_b=\vec i+2\vec j+t\vec k$ where $t$ is to be calculated (for this values $t$ is real and the choice is good but for many other values $t$ could be non-real so the choice could be bad).

ligne $a$: $\dfrac{x-1}{1}=\dfrac{y-0}{1}=\dfrac{z+2}{1}$

ligne $b$: $\dfrac{x-3}{1}=\dfrac{y+1}{2}=\dfrac{z-1}{t}$

$\vec{u_a}\text{ x }\vec{u_b}=\left|\begin{array}{cc}\vec i & \vec j&\vec k\\1 & 1&1\\1&2&t\end{array}\right|=(t-2)\vec i-(t-1)\vec j+\vec k$

$|\vec{u_a}\text{ x }\vec{u_b}|=\sqrt{(t-2)^2+(t-1)^2+1}=\sqrt{2t^2-6t+6}$

Now $$d=\dfrac{|\vec{AB}\cdot(\vec u_a\text{ x } \vec u_b)|}{|\vec u_a\text{ x } \vec u_b|}=\dfrac{|(2,-1,3)\cdot(t-2,-(t-1),1)|}{\sqrt{2t^2-6t+6}}=\dfrac{|3t-2|}{\sqrt{2t^2-6t+6}}=2$$ This gives the equation $t^2+12t-20=0$ and the values $$\color{red}{t=-6\pm\sqrt{56}}$$ each of them gives an answer for the lines $a$ and $b$.

► A good exercise for you could be to find another answer using the cartesian form of the lines

$$\begin{cases}x=az+p\\y=bz+q\end{cases}\space \space \space \begin{cases}x=a'z+p'\\y=b'z+q'\end{cases}$$ and the formula $$d=\frac{(p-p')(b-b')-(q-q')(a-a')}{\sqrt{(a-a')^2+(b-b')^2+(ab'-ba')^2}}$$ (You already know you can take suitable arbitrary values to get an answer).

Piquito
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  • How did you choose $\vec u_a$ and $\vec u_b$? Did you choose them randomly? – M.M. Feb 15 '18 at 14:53
  • You have freedom to choose. Y have chosen first (1,1,1) and (1,1,t) but this was a bad choice giving non real solutions for t. With (1,2,t) this choice worked. I have chosen simple values of course. Mi intention was just to help you to solve your problem. – Piquito Feb 15 '18 at 15:39