Find shortest distance between lines given by
$$\frac{x-2}{3}=\frac{y-6}{4}=\frac{z+9}{-4}$$and$$\frac{x+1}{2}=\frac{y+2}{-6}=\frac{z-3}{1}$$Is there any shortcut method for this problems?
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$\begingroup$So you have two lines defined by the points $\mathbf{r}_1=(2,6,-9)$ and $\mathbf{r}_2=(-1,-2,3)$ and the (non unit) direction vectors $\mathbf{e}_1=(3,4,-4)$ and $\mathbf{e}_2 =(2,-6,1)$.
The coordinates of all the points along the lines are given by
$$\begin{align} \mathbf{p}_1 & = \mathbf{r}_1 + t_1 \mathbf{e}_1 \\ \mathbf{p}_2 & = \mathbf{r}_2 + t_2 \mathbf{e}_2 \\ \end{align}$$
where $t_1$ and $t_2$ are two scalar values. To find the closest points along the lines you recognize that the line connecting the closest points has direction vector $$\mathbf{n} = \mathbf{e}_1 \times \mathbf{e}_2 = (-20,-11,-26)$$
If the two direction vectors $\mathbf{e}_1$ and $\mathbf{e}_2$ are parallel (not in this specific case), this method cannot be applied because the cross product is zero:
$$\mathbf{e}_1 \times \mathbf{e}_2 = 0$$
If the points along the two lines are projected onto the cross line the distance is found with one fell swoop
$$ d = \frac{ \mathbf{n}\cdot \mathbf{p}_1}{\|\mathbf{n}\|} - \frac{ \mathbf{n}\cdot \mathbf{p}_2}{\|\mathbf{n}\|} = \frac{ \mathbf{n} \cdot ( \mathbf{p}_1-\mathbf{p}_2)}{\| \mathbf{n} \|} = \frac{ \mathbf{n} \cdot ( \mathbf{r}_1-\mathbf{r}_2+t_1 \mathbf{e}_1 -t_2 \mathbf{e}_2)}{\| \mathbf{n} \|} $$
But since $\mathbf{n}\cdot \mathbf{e}_1 = \mathbf{n}\cdot \mathbf{e}_2 = 0$, the above is
$$ \boxed{ d = \frac{ \mathbf{n} \cdot ( \mathbf{r}_1-\mathbf{r}_2)}{\| \mathbf{n} \|} }$$
In this case $$ d = \frac{ (-20,-11,-26) \cdot (3,8,-12) }{3 \sqrt{133}} = 4.74020116673185 $$
$\endgroup$ 2 $\begingroup$Here is an approach that uses dot products instead of cross products.
This works in any number of dimensions, not just 3.
The skew lines are $L = a+bt, M=c+ds$.
The distance between two points on $L$ and $M$ is $D =(a+bt-c-ds)^2 =(e+bt-ds)^2 $ where $e = a-c$.
For this to be a minimum, taking partials, we want $D_s = D_t = 0$.
$D_s = -2d(e+bt-ds) $ and $D_t = 2b(e+bt-ds) $.
Therefore, with multiplication of vectors being dot product, $0 =d(e+bt-ds) =de+dbt-d^2s $ and $0 =b(e+bt-ds) =be+b^2t-bds) $.
These are two equations in the two unknowns $s$ and $t$:
$\begin{array}\\ de &= d^2s-dbt\\ be &= bds-b^2t\\ \end{array} $
The determinant is $A =-b^2d^2+(bd)^2 =-(b^2d^2-(bd)^2) $. By Cauchy-Schwarz, this is non-zero unless $b$ and $d$ are parallel (which is a good thing).
The solutions (by Cramer's rule) are $s =\dfrac{-(b^2)(de)+(be)(db)}{A} $ and $t =\dfrac{(d^2)(be)-(be)(db)}{A} $.
Putting these into $L = a+bt, M = c+ds, D =(e+bt-ds)^2 $ we get the endpoints of the closest line and the distance.
$\endgroup$ 3 $\begingroup$If anyone is interested on a implementation of the algorithm proposed by @John Alexiou using python there it is:
def distance_from_two_lines(e1, e2, r1, r2): # e1, e2 = Direction vector # r1, r2 = Point where the line passes through # Find the unit vector perpendicular to both lines n = np.cross(e1, e2) n /= np.linalg.norm(n) # Calculate distance d = np.dot(n, r1 - r2) return dHint:
write the equations of the two lines in the form $\vec x=\vec p+t\vec q$: $$ r_1) \qquad \begin{pmatrix} x\\y\\z \end{pmatrix}=\begin{pmatrix} 2\\6\\-9 \end{pmatrix}+t\begin{pmatrix} 3\\4\\-4 \end{pmatrix} $$ $$ r_2) \qquad \begin{pmatrix} x\\y\\z \end{pmatrix}=\begin{pmatrix} -1\\-2\\3 \end{pmatrix}+t\begin{pmatrix} 2\\-6\\1 \end{pmatrix} $$
than, noted the the two lines are not parallel nor intersecting, use the formula from here.
$\endgroup$ $\begingroup$I'm using a method similar to marty cohen's but without matrix/linear system
We first parametrize the two lines as:$$ \vec{L_0} = \vec{p_0}+s\vec{u}\\ \vec{L_1} = \vec{p_1}+t\vec{v} $$The distance between any points on line 1 to line 0 is$$ d^2 = \dfrac{\left[(\vec{p_1}-\vec{p_0}+t\vec{v})\times\vec{u}\right]^2}{\vec{u}\cdot \vec{u}} $$To find the minimum distance, we need to minimize the numerator$$ \dfrac{d\left[(\vec{p_1}-\vec{p_0})\times \vec{u}+t\vec{v}\times\vec{u}\right]^2}{dt}=\dfrac{d\left[2(\vec{p_1}-\vec{p_0})\times\vec{u}\cdot(t\vec{v}\times\vec{u})+t^2(\vec{v}\times\vec{u})^2\right]}{dt}\\ =(\vec{p_1}-\vec{p_0})\times\vec{u}\cdot(\vec{v}\times\vec{u})+t(\vec{v}\times\vec{u})^2=0 $$
Then we can get the solution of t$$ t = -\dfrac{(\vec{p_1}-\vec{p_0})\times\vec{u}\cdot(\vec{v}\times\vec{u})}{(\vec{v}\times\vec{u})^2} $$
Taking the value of $t$ back to the distance expression,$$ d^2=\dfrac{\left[(\vec{p_1}-\vec{p_0})\times\vec{u}\right]^2(\vec{v}\times\vec{u})^2-\left[((\vec{p_1}-\vec{p_0})\times\vec{u})\cdot(\vec{v}\times\vec{u})\right]^2}{\vec{u}\cdot \vec{u}(\vec{v}\times\vec{u})^2}\\ = \dfrac{\left[((\vec{p_1}-\vec{p_0})\times\vec{u})\times(\vec{v}\times\vec{u})\right]^2}{\vec{u}\cdot \vec{u}(\vec{v}\times\vec{u}^2}\\ = \dfrac{\left[\vec{u}\cdot((\vec{p_1}-\vec{p_0})\times\vec{v})\vec{u}\right]^2}{\vec{u}\cdot \vec{u}(\vec{v}\times\vec{u})^2}\\ = \dfrac{\left[\vec{u}\cdot((\vec{p_1}-\vec{p_0})\times\vec{v})\right]^2}{(\vec{v}\times\vec{u})^2}\\ = \dfrac{\left[(\vec{p_1}-\vec{p_0})\cdot(\vec{v}\times\vec{u})\right]^2}{(\vec{v}\times\vec{u})^2} $$The final solution$$ d=\dfrac{\left|(\vec{p_1}-\vec{p_0})\cdot(\vec{v}\times\vec{u})\right|}{|\vec{v}\times\vec{u}|} $$By simply switching the index, we can also get the solution of s$$ s = -\dfrac{(\vec{p_1}-\vec{p_0})\times\vec{v}\cdot(\vec{v}\times\vec{u})}{(\vec{v}\times\vec{u})^2}\\ $$
Then we can also get the direction of the line connecting the two point:$$ \vec{n}=\vec{p_0}+s\vec{u}-\vec{p_1}-t\vec{v} $$
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