Shortest Distance Between Two Lines

In three-dimensional geometry, two lines may intersect, may be parallel, or may be skew. The concept of shortest distance helps in finding the perpendicular distance between such lines when they do not meet directly.

\[SD=\left|\frac{\left(a_{2}-a_{1}\right)\times b}{\left|b\right|}\right|\]

Vector Form:

\[\mathbf{d}=\left|\frac{(\overline{\mathbf{b}}_{1}\times\overline{\mathbf{b}}_{2}).(\overline{\mathbf{a}}_{2}-\overline{\mathbf{a}}_{1})}{\left|\overline{\mathbf{b}}_{1}\times\overline{\mathbf{b}}_{2}\right|}\right|\]

Cartesian Form:

\[\mathbf{d}=\left|\frac{ \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ \mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\ \mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \end{vmatrix}}{\sqrt{\left(\mathbf{a}_1\mathbf{b}_2-\mathbf{a}_2\mathbf{b}_1\right)^2+\left(\mathbf{a}_1\mathbf{c}_2-\mathbf{a}_2\mathbf{c}_1\right)^2+\left(\mathbf{b}_1\mathbf{c}_2-\mathbf{b}_2\mathbf{c}_1\right)^2}}\right|\]

Find the distance between \[\vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k})\] and \[\vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})\].

Solution:

Calculate \[\vec{a}_2 - \vec{a}_1\]:

Therefore

Find the distance between \[\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})\] and \[\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})\].

Solution:

Identify: \[\vec{a}_2 - \vec{a}_1 = 2\hat{i} + \hat{j} - \hat{k}\]

\[\vec{a}_2 - \vec{a}_1 = 2\hat{i} + \hat{j} - \hat{k}\]

\[\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}\].

Result:

or

• Intersecting lines: SD = 0

• Parallel lines: \[SD = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}\]

• Skew lines: \[SD = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}\]