Problem 2 - Geometry

The following are two lines in the 3-dimensional $(x,y,z)$-space which are assumed to be non-parallel:

$$g_i: \frac{x-x_i}{l_i} = \frac{y-y_i}{m_i} = \frac{z-z_i}{n_i} \quad (i=1,2)$$

1.

What is the minimum distance between these two lines?

2.

What is the common perpendicular line of these two lines?

1.

For $g_1$:

$$g_1: \frac{x-x_1}{l_1} = \frac{y-y_1}{m_1} = \frac{z-z_1}{n_1}$$

$$g_1: \begin{tabular}{l} $m_1x - l_1y = m_1x_1 - l_1y_1$ \\ $n_1x - l_1z = n_1x_1 - l_1z_1$ \end{tabular}$$

The following vector indicates the direction of $g_1$:

$$u_1 = \left( \begin{tabular}{c} $m_1$ \\ $-l_1$ \\ 0 \end{tabul... ...eft( \begin{tabular}{c} $l_1^2$ \\ $l_1m_1$ \\ $l_1n_1$ \end{tabular} \right)$$

For $g_2$:

$$g_2: \frac{x-x_2}{l_2} = \frac{y-y_2}{m_2} = \frac{z-z_2}{n_2}$$

$$g_2: \begin{tabular}{l} $m_2x - l_2y = m_2x_2 - l_2y_2$ \\ $n_2x - l_2z = n_2x_2 - l_2z_2$ \end{tabular}$$

The following vector indicates the direction of $g_2$:

$$u_2 = \left( \begin{tabular}{c} $m_2$ \\ $-l_2$ \\ 0 \end{tabul... ...eft( \begin{tabular}{c} $l_2^2$ \\ $l_2m_2$ \\ $l_2n_2$ \end{tabular} \right)$$

Thus, the following vector indicates the direction of the common perpendicular line:

$$v=\left( \begin{tabular}{c} $l_1^2$ \\ $l_1m_1$ \\ $l_1n_1$ \en... ..._2m_1-n_1m_2$ \\ $l_2n_1 - l_1n_2$ \\ $l_1m_2 - l_2m_1$ \end{tabular} \right)$$

$M_1=(x_1,y_1,z_1)$ and $M_2=(x_2,y_2,z_2)$ are respectively parts of $g_1$ and $g_2$. If we project the vector $\overrightarrow{M_1M_2}$ on a normalized $v$ vector, we obtain the minimum distance between $g_1$ and $g_2$:

$$d = \frac{\sqrt{ l_1 l_2\left( \begin{tabular}{c} $n_2m_1-n_1m_2$... ..._2$ \\ $y_1-y_2$ \\ $z_1-z_2$ \end{tabular} \right)}}{\vert\vert v\vert\vert}$$

2.

We are seeking for a line $\Delta$ (of direction $v$) such that it belongs to both a plan $P_1$ that contains $g_1$ and is orthogonal to $g_2$ and a plan $P_2$ that contains $g_2$ and is orthogonal to $g_1$.

An equation of $P_1$ has the following form:

$$\lambda_1(m_1x-l_1y-m_1x_1+l_1y_1) + \mu_1(n_1x-l_1z-n_1x_1 + l_1z_1) = 0 \quad (\lambda_1,\mu_1)\neq(0,0)$$

$$x(m_1\lambda_1 + n_1\mu_1) + y(-\lambda_1 l_1) + z(-\mu_1 l_1) + (-\lambda_1 m_1 x_1 + \lambda_1 l_1 y_1 -\mu_1 n_1 x_1 + \mu_1 l_1 z_1) = 0$$

and, since $P_1$ is parallel to $v$, we have moreover:

$$\left( \begin{tabular}{c} $n_2m_1 - n_1m_2$ \\ $l_2n_1-l_1n_2$ \... ..._1 + n_1\mu_1$ \\ $-\lambda_1 l_1$ \\ $-\mu_1 l_1$ \end{tabular} \right) = 0$$

$$\lambda_1 (m_1^2n_2 - n_1m_1m_2 - l_1l_2n_1 + l_1^2n_2) + \mu_1 (n_1n_2m_1 - n_1^2m_2 - l_1^2m_2 + l_1 l_2 m_1) = 0$$

$$\lambda_1 A_1 + \mu_1 B_1 = 0$$

Provided that $A_1$ and $B_1$ are not equal to 0, we can choose $\lambda_1 = \frac{1}{A_1}$ and $\mu_1 = -\frac{1}{B_1}$, and we have an equation of the particular $P_1$ we are looking for.

Similarly, an equation of $P_2$ is given by:

$$\lambda_2(m_2x-l_2y-m_2x_2+l_2y_2) + \mu_2(n_2x-l_2z-n_2x_2 + l_2z_2) = 0 \quad (\lambda_2,\mu_2)\neq(0,0)$$

under the condition that:

$$\lambda_2 (m_2 n_2 m_1 - n_1 m_2^2 - l_2^2 n_1 + l_1 l_2 n_2) + \mu_2 (n_2^2m_1 - n_1n_2m_2 - l_1l_2m_2 + l_2^2m_1) = 0$$

$$\lambda_2 A_2 + \mu_2 B_2 = 0$$

Provided that $A_2$ and $B_2$ are not equal to 0, we can choose $\lambda_2 = \frac{1}{A_2}$ and $\mu_2 = -\frac{1}{B_2}$, and we have an equation of the particular $P_2$ we are looking for.

Then, an equation of the line $\Delta$ is:

$$\Delta: \begin{tabular}{l} $\frac{1}{A_1}(m_1x-l_1y-m_1x_1+l_1y_1... ...y-m_2x_2+l_2y_2) - \frac{1}{B_2}(n_2x-l_2z-n_2x_2 + l_2z_2) = 0$ \end{tabular}$$