Distance of two parametrized lines

Question

Consider the four points $A = (2, 4, 0), B = (3, 1, 1), C = (1, 1, 3), D = (0, 5, 1)$. Find the distance between the lines $(AB)$ and $(CD)$, i.e. the distance between the closest points on these two lines, by using this method:

Find parametric equations of the lines $(AB)$ and $(CD)$, and find the times at which the line segment connecting a point $P_1$ on $(AB)$ to a point $P_2$ on $(CD)$ is perpendicular to both lines. The length of this segment is then the distance between the lines.

The parametric equations of both lines are:

$$x = t+2$$ $$x = -3t+4$$ $$x = t$$

and

$$x = -t_1+ 1$$ $$x = 4t_1+1$$ $$x = -2t_1+3$$

the line segment connecting them would be:

$$ABCD = (-2,7,-3)$$

how can I find when the segment is perpendicular to both lines?

note $$AB \times CD = (2,1,1)$$

Answer 1

The first line is

$ \ell_1 = P_1 + t d_1 $

with $P_1 = (2, 4, 0) $ and $ d_1 = (1 , -3 , 1 ) $

While the second line is given by

$ \ell_2 = (1,1,3) + s (-1, 4, -2) $

The vector extending from a point on $\ell_1$ to a point on $\ell_2$ is given by

$ v = P_2 + s d_2 - (P_1 + t d_1 ) = (P_2 - P1) + s d_2 - t d_1 \\ = (1,1,3) - (2, 4, 0) + s (-1, 4, -2) - t (1, -3, 1) \\ = (-1, -3, 3) + s (-1, 4, -2) - t (1, -3, 1) \\ = (-1 - s - t , - 3 + 4 s + 3 t , 3 - 2 s - t ) $

We want this vector to be perpendicular to $d_1$ and $d_2$, so we get the following two equations

$ v \cdot d_1 = 0 $, i.e. $(-1 - s - t, -3 + 4s + 3 t , 3 - 2 s - t) \cdot (1, -3, 1) = 0 $

and

$ v \cdot d_2 = 0 $, i.e. $(-1 -s -t , -3 +4s + 3 t, 3 -2 s - t) \cdot (-1, 4, -2) = 0 $

These two equations, after carrying out the dot product simplify to

$11 - 15 s - 11 t = 0$

$-17 + 21 s + 15 t = 0 $

Re-arranging, we get

$ 11 t + 15 s = 11 $

$ 15 t + 21 s = 17 $

Solving these two equations using Cramer's rule gives

$ t = \dfrac{ (11)(21) - (17)(15) }{ (11)(21) - (15)^2 } = -4 $

$ s = \dfrac{ (11)(17) - (15)(11) }{ (11)(21) - (15)^2 } = \dfrac{11}{3} $

Therefore, the first point on the first line is

$ P = P_1 + t d_1 = (2, 4, 0) - 4 (1, -3, 1) = (-2, 16, - 4) $

And the second point on the second line is

$ Q = P_2 + s d_2 = (1, 1, 3) + \dfrac{11}{3} (-1, 4, -2) = \dfrac{1}{3} (-8, 47, - 13 ) $

Therefore, the distance between the two lines is just the length $\overline{PQ}$

$\overline{PQ} = \dfrac{1}{3} \sqrt{ (-6 + 8)^2 + (48 - 47)^2 + (-12 + 13)^2 } = \dfrac{\sqrt{6}}{3} = \sqrt{ \dfrac{2}{3} } $

To verify this result, very quickly, with a more efficient method, calculate $d_1 \times d_2$:

$ N = d_1 \times d_2 = (1, -3, 1) \times (-1, 4, -2) = ( 2, 1 , 1 ) $

So the plane containing the first line and parallel to the second line is

$ \pi_1: N \cdot (r - P_1) = 0 $ , i.e. $2 (x- 2) + (y - 4) + z = 0$

(Note that $r = (x,y,z) $ )

Now use the distance formula, to find the distance of $P_2$ from $\pi_1$

$d = \dfrac{ | N \cdot (P_2 - P_1) | }{\sqrt{ N \cdot N }} = \dfrac{ |2 (-1) + (-3) + 3 |}{\sqrt{ 6} } = \dfrac{2}{\sqrt{6}} = \sqrt{ \dfrac{2}{3} } $

So the answers match.

Answer 2

Hint: we can find component forms of both lines. Hence, smallest distance between two lines given in component form is: $$d=\frac{\pm\begin{vmatrix} x_1-x_2 & y_1-y_2 & z_1-z_2\\ l_1 & m_1 & n_1\\ l_2 & m_2 & n_2 \end{vmatrix}}{\sqrt{\begin{vmatrix} l_1 & m_1 \\ l_2 & m_2 \end{vmatrix}^2+\begin{vmatrix} m_1 & n_1 \\ m_2 & n_2 \end{vmatrix}^2+\begin{vmatrix} n_1 & l_1 \\ n_2 & l_2 \end{vmatrix}^2}}.$$

If the determinant in the numerator is equal to zero, the lines intersect each other.