Shortest Distance between Two Lines - Practice Questions & MCQ

There are three possible types of relations that two different lines can have in a three-dimensional space. They can be

1. Parallel lines: when their direction vectors are parallel and the two lines never meet.

2. Intersecting lines: when their direction vectors are not parallel and the two lines intersect.

3. Skew lines: When two lines neither parallel nor intersecting at a point.  

For example, consider a cuboid 

Edges AB and CD  are parallel. Edges AB and BC intersect at a single point B. Edges AB and EH are skew, since they are not parallel and never meet. 

For skew lines, the line of the shortest distance will be perpendicular to both the lines.

So, the shortest distance between edges AB and EH is |AE|. 

Distance between two skew lines

If L₁ and L₂  are two skew lines, then there is one and only one line perpendicular to each of lines L1 and L2 which is known as the line of shortest distance.

Vector form

$\mathrm{L}_1: \quad \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}}$
$\mathrm{L}_2: \quad \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\mu \overrightarrow{\mathbf{b}^{\prime}}$
Let $S$ be any point on the line $L_1$ with position vector $\overrightarrow{\mathbf{r}}_0$ and $T$ on $L_2$ with position vector $\overrightarrow{\mathbf{r}}_0$. Then the magnitude of the shortest distance vector will be equal to that of the projection of $S T$ along the direction of the line of shortest distance.

${ }_{\text {If }} \overrightarrow{P Q}$ is the shortest distance vector between $\mathrm{L}_1$ and $\mathrm{L}_2$, then it being perpendicular to both $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{b}^{\prime}}$, therefore, the unit vector $\hat{\mathbf{n}}$ along $\overrightarrow{P Q}$ would be
$$
\hat{\mathbf{n}}=\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathbf{b}^{\prime}}}{\left|\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}^{\prime}}\right|}
$$

Then,
$$
\overrightarrow{P Q}=d \hat{\mathbf{n}}
$$
where "d" is the magnitude of the shortest distance vector. Let $\theta$ be the angle between $\overrightarrow{S T}$ and $\overrightarrow{P Q}$.

Then  

$$
\begin{aligned}
\mathrm{PQ} & =\mathrm{ST}|\cos \theta| \\
\cos \theta & =\left|\frac{\overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{ST}}}{|\overrightarrow{\mathrm{PQ}}||\overrightarrow{\mathrm{ST}}|}\right| \\
& =\left|\frac{d \hat{n} \cdot\left(\overrightarrow{r_0^{\prime}}-\vec{r}_0\right)}{d \mathrm{ST}}\right|\left(\text { since } \overrightarrow{\mathrm{ST}}=\vec{r}_0^{\prime}-\vec{r}_0\right) \\
& =\left|\frac{\left(\vec{b} \times \overrightarrow{b^{\prime}}\right) \cdot\left(\overrightarrow{r_0^{\prime}}-\vec{r}_0\right)}{\mathrm{ST}\left|\vec{b} \times \overrightarrow{b^{\prime}}\right|}\right|
\end{aligned}
$$

Hence, the required shortest distance is
or
$$
\begin{aligned}
d & =\mathrm{PQ}=\mathrm{ST}|\cos \theta| \\
\mathbf{d} & =\left|\frac{\left(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}^{\prime}}\right) \cdot\left(\overrightarrow{\mathbf{r}^{\prime}}-\overrightarrow{\mathbf{r}}_0\right)}{\left|\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}^{\prime}}\right|}\right|
\end{aligned}
$$

For Intersecting lines

Their shortest distance should be 0

Vector form

$$
\left(\vec{b} \times \overrightarrow{b_1}\right) \cdot\left(\overrightarrow{\mathbf{r}_0^{\prime}}-\overrightarrow{\mathbf{r}}_0\right)=0
$$

Distance between parallel lines
Let two lines $L_1$ and $L_2$ be parallel. Let the equation of lines be given by
$\mathrm{L}_1: \quad \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0+\lambda \overrightarrow{\mathbf{b}}$
$\mathrm{L}_2: \quad \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{r}}_0^{\prime}+\mu \overrightarrow{\mathbf{b}}$
where, $\overrightarrow{\mathbf{r}}_0$ is the position vector of a point $S$ on $\mathrm{L}_1$ and $\overrightarrow{\mathbf{r}}_0$ is the position vector of a point $T$ on $\mathrm{L}_2$.
Let $\theta$ be the angle between the vectors ST and $\overrightarrow{\mathbf{b}}$.

Then,
$$
\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathrm{ST}}=(|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathrm{ST}}| \sin \theta)) \hat{\mathbf{n}}
$$
where $\hat{\mathbf{n}}$ is the unit vector perpendicular to the plane of the lines $L_1$ and $\underline{L_2}$.
but
$$
\begin{aligned}
& \overrightarrow{S T}=\overrightarrow{\mathbf{r}}_0-\overrightarrow{\mathbf{r}}_0 \\
& \Rightarrow \quad \overrightarrow{\mathbf{b}} \times\left(\overrightarrow{\mathbf{r}}_0-\overrightarrow{\mathbf{r}}_0\right)=(|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathrm{PT}}|) \hat{\mathbf{n}} \quad \text { (since } \mathrm{PT}=\mathrm{ST} \sin \theta \text { ) } \\
& \text { i.e. } \left.\quad\left|\overrightarrow{\mathbf{b}} \times\left(\overrightarrow{\mathbf{r}}_0-\overrightarrow{\mathbf{r}}_0\right)\right|=(|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathrm{PT}}|) \cdot 1 \quad \text { (as }|\hat{\mathbf{n}}|=1\right)
\end{aligned}
$$

Hence, the distance between the given parallel lines is
$$
\mathbf{d}=|\overrightarrow{\mathrm{PT}}|=\left|\frac{\overrightarrow{\mathbf{b}} \times\left(\overrightarrow{\mathbf{r}}_0^{\prime}-\overrightarrow{\mathbf{r}}_0\right)}{|\overrightarrow{\mathbf{b}}|}\right|
$$