The distance between two intersecting lines.

Let $L_{1} : \frac{x - x_{1}}{m_{1}} = \frac{y - y_{1}}{l_{1}} = \frac{z - z_{1}}{k_{1}}$ and $L_{2} : \frac{x - x_{2}}{m_{2}} = \frac{y - y_{2}}{l_{2}} = \frac{z - z_{2}}{k_{2}}$ be two intersecting lines. The distance $ρ (L_{1}, L_{2})$ between the lines $L_{1}$ and $L_{2}$ can be found using the following procedure:

1) Find the equation of the plane $P$ passing through the line $L_{1}$ and parallel to the line $L_{2}$ :

The plane $P$ passes through the point $M_{1} (x_{1}, y_{1}, z_{1})$ , perpendicular to the vector $\overset{―}{n} = [\overset{―}{s_{1}}, \overset{―}{s_{2}}] = (n_{x}, n_{y}, n_{z})$ , where $\overset{―}{s_{1}} = (m_{1}, l_{1}, k_{1})$ and $\overset{―}{s_{2}} = (m_{2}, l_{2}, k_{2})$ are the direction vectors of the lines $L_{1}$ and $L_{2}$ respectively. Therefore, the equation of the plane $P$ is $n_{x} (x - x_{1}) + n_{y} (y - y_{1}) + n_{z} (z - z_{1}) = 0$ .

2) The distance between the lines $L_{1}$ and $L_{2}$ is equal to the distance from any point on the line $L_{2}$ to the plane $P$ .

$ρ (L_{1}, L_{2}) = ρ (M_{2}, P) = | \frac{n_{x} x_{2} + n_{y} y_{2} + n_{z} z_{2}}{\sqrt{n_{x}^{2} + n_{y}^{2} + n_{z}^{2}}} | .$

Finding the common perpendicular of intersecting lines.

To find the common perpendicular of the intersecting lines $L_{1}$ and $L_{2}$ , it is necessary to find the equations of the planes $P_{1}$ and $P_{2}$ passing through the lines $L_{1}$ and $L_{2}$ respectively, perpendicular to the plane $P$ .

Let $P_{1} : A_{1} x + B_{1} y + C_{1} z + D_{1} = 0;$

$P_{2} : A_{2} x + B_{2} y + C_{2} z + D_{2} = 0.$

Then the equation of the common perpendicular has the form

${\begin{array}{lcl} A_{1} x + B_{1} y + C_{1} z + D_{1} = 0; \\ A_{2} x + B_{2} y + C_{2} z + D_{2} = 0. \end{array}$

Example.

2.214.

For the given lines $L_{1} : \frac{x + 7}{3} = \frac{y + 4}{4} = \frac{z + 3}{- 2}$ and $L_{2} : \frac{x - 21}{6} = \frac{y + 5}{- 4} = \frac{z - 2}{- 1}$ , it is required to:

a) prove that the lines do not lie in the same plane, i.e., they are intersecting;

b) write the equation of the plane passing through the line $L_{2}$ parallel to $L_{1}$ ;

c) calculate the distance between the lines;

d) write the equations of the common perpendicular to the lines $L_{1}$ and $L_{2}$ .

Solution.

a) If the lines $L_{1}$ and $L_{2}$ lie in the same plane, then their direction vectors $\overset{―}{s_{1}} (3, 4, - 2)$ , $\overset{―}{s_{2}} (6, - 4, - 1)$ , and the vector $\overset{―}{l}$ , connecting an arbitrary point on line $L_{1}$ and an arbitrary point on line $L_{2}$ , are coplanar. As such a vector $\overset{―}{l}$ , we can choose $\overset{―}{l} (x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1})$ . Let's check whether these vectors are coplanar.

$| \begin{matrix} 21 - (- 7) & - 5 - (- 4) & 2 - (- 3) \\ 3 & 4 & - 2 \\ 6 & - 4 & - 1 \end{matrix} | = | \begin{matrix} 28 & - 1 & 5 \\ 3 & 4 & - 2 \\ 6 & - 4 & - 1 \end{matrix} | =$ $= 28 \cdot 4 \cdot (- 1) + 3 \cdot (- 4) \cdot 5 + (- 1) \cdot (- 2) \cdot 6 -$ $- 6 \cdot 4 \cdot 5 - (- 4) \cdot (- 2) \cdot 28 - 3 \cdot (- 1) \cdot (- 1) =$

$- 112 - 60 + 12 - 120 - 224 + 3 = 501 \neq 0.$

Therefore, the vectors are not coplanar, and the lines do not lie in the same plane.

b) Let's write down the equation of the plane passing through the line $L_{2}$ parallel to $L_{1}$ . This plane passes through the point $M_{2} (21, - 5, 2)$ perpendicular to the vector $\overset{―}{n} = [{\overset{―}{s}}_{1}, {\overset{―}{s}}_{2}]$ .

$[s_{1}, s_{2}] = | \begin{matrix} i & j & k \\ 3 & 4 & - 2 \\ 6 & - 4 & - 1 \end{matrix} | = i | \begin{matrix} 4 & - 2 \\ - 4 & - 1 \end{matrix} | - j | \begin{matrix} 3 & - 2 \\ 6 & - 1 \end{matrix} | + k | \begin{matrix} 3 & 4 \\ 6 & - 4 \end{matrix} | =$ $= i (- 4 - 8) - j (- 3 + 12) + k (- 12 - 24) = - 12 i - 9 j - 36 k .$

Thus, the vector $\overset{―}{n}$ has coordinates $\overset{―}{n} (- 12, - 9, - 36)$ .

Let's find the equation of the plane: $P : - 12 (x - 21) - 9 (y + 5) - 36 (z - 2) = 0 \Rightarrow$ $\Rightarrow - 12 x - 9 y - 36 z + 252 - 45 + 72 = 0 \Rightarrow - 12 x - 9 y - 36 z + 279 = 0 \Rightarrow$ $\Rightarrow 4 x + 3 y + 12 z - 93 = 0.$

Answer: $4 x + 3 y + 12 z - 93 = 0.$

c) The distance between the lines $L_{1}$ and $L_{2}$ is equal to the distance from any point on line $L_{1}$ to the plane $P$ :

$ρ (L_{1}, L_{2}) = ρ (M_{1}, P) = | \frac{n_{x} x_{1} + n_{y} y_{1} + n_{z} z_{1}}{\sqrt{n_{x}^{2} + n_{y}^{2} + n_{z}^{2}}} |$ $ρ (M_{1}, P) = | \frac{4 (- 7) + 3 (- 4) + 12 (- 3)}{\sqrt{4^{2} + 3^{2} + 12^{2}}} | = | \frac{- 28 - 12 - 36}{\sqrt{16 + 9 + 144}} | =$ $= | \frac{- 76}{\sqrt{169}} | = \frac{76}{13} .$

Answer: $\frac{76}{13}$ .

g) Let's find the equations of planes $P_{1}$ and $P_{2}$ passing through lines $L_{1}$ and $L_{2}$ , respectively, perpendicular to the plane $P$ .

We have $M_{1} = (- 7, - 4, - 3) \in P_{1}$ ,

${\overset{―}{n}}_{1} = [{\overset{―}{s}}_{1}, \overset{―}{n}] = | \begin{matrix} i & j & k \\ 3 & 4 & - 2 \\ 4 & 3 & 12 \end{matrix} | = i | \begin{matrix} 4 & - 2 \\ 3 & 12 \end{matrix} | - j | \begin{matrix} 3 & - 2 \\ 4 & 12 \end{matrix} | + k | \begin{matrix} 3 & 4 \\ 4 & 3 \end{matrix} | =$ $= i (48 + 6) - j (36 + 8) + k (9 - 16) = 54 i - 44 j - 7 k .$

Thus, $P_{1} : 54 (x + 7) - 44 (y + 4) - 7 (z + 3) = 54 x - 44 y - 7 z + 378 - 176 - 21 =$ $= 54 x - 44 y - 7 z + 181 = 0.$

Similarly, we find $P_{2}$ :

We have $M_{2} = (21, - 5, 2) \in P_{2}$ ,

${\overset{―}{n}}_{2} = [{\overset{―}{s}}_{2}, \overset{―}{n}] = | \begin{matrix} i & j & k \\ 6 & - 4 & - 1 \\ 4 & 3 & 12 \end{matrix} | = i | \begin{matrix} - 4 & - 1 \\ 3 & 12 \end{matrix} | - j | \begin{matrix} 6 & - 1 \\ 4 & 12 \end{matrix} | + k | \begin{matrix} 6 & - 4 \\ 4 & 3 \end{matrix} | =$ $= i (- 48 + 3) - j (72 + 4) + k (18 + 16) = - 45 i - 76 j + 34 k .$

Thus, $P_{1} : - 45 (x - 21) - 76 (y + 5) + 34 (z - 2) = - 45 x - 76 y + 34 z + 945 - 380 - 68 =$ $= - 45 x - 76 y + 34 z + 497 = 0.$

Answer: ${\begin{array}{lcl} 54 x - 44 y - 7 z + 181 = 0; \\ - 45 x - 76 y + 34 z + 497 = 0. \end{array}$

Homework.

2.215.

For the given lines $L_{1} : \frac{x - 6}{3} = \frac{y - 3}{- 2} = \frac{z + 3}{4}$ and $L_{2} : \frac{x + 1}{3} = \frac{y + 7}{- 3} = \frac{z - 4}{8}$ , it is required to: