Name: Shortest Distance Between Skew Lines
Uploaded: 2019-07-19
Duration: 5 min 52 s
Description: In this video, let us learn about how to find the shortest distance between two skewed or non-coplanar lines. Peruse this tutorial video now!

Question

Find the shortest distance between the following pair of lines

\vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2 t) \hat{k}

and

\vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}

.

Accepted Answer

The shortest distance between two vector lines $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}}$ and $\vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$ is given by $|\frac{(\vec{b_{1}} \times \vec{b_{2}}) . (\vec{a_{2}} - \vec{a_{1}})}{|\vec{b_{1}} \times \vec{b_{2}}|}|$

Given equations:

$\overset{⇀}{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2 t) \hat{k} . . . (i)$

$\vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k} . . . (ii)$

$\Rightarrow \vec{r} = \hat{i} - \hat{i} t + t \hat{j} - 2 \hat{j} + 3 \hat{k} - 2 t \hat{k}$

$\vec{r} = s \hat{i} + \hat{i} + 2 s \hat{j} - \hat{j} - 2 s \hat{k} - \hat{k}$

$\Rightarrow \vec{r} = \hat{i} - 2 \hat{j} + 3 \hat{k} + t (- \hat{i} + \hat{j} - 2 \hat{k})$

$\vec{r} = \hat{i} - \hat{j} - \hat{k} + s (\hat{i} + 2 \hat{j} - 2 \hat{k})$

Comparing with standard forms:

$\Rightarrow \vec{a_{1}} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ $\vec{a_{2}} = \hat{i} - \hat{j} - \hat{k}$

$\Rightarrow \vec{b_{1}} = - \hat{i} + \hat{j} - 2 \hat{k}$ $\vec{b_{2}} = \hat{i} + 2 \hat{j} - 2 \hat{k}$

$\Rightarrow (\vec{a_{2}} - \vec{a_{1}}) = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2 \hat{j} + 3 \hat{k})$

$= (\hat{i} - \hat{i}) + (- \hat{j} + 2 \hat{j}) + (- \hat{k} - 3 \hat{k}) = \hat{j} - 4 \hat{k}$

Now,

$(\vec{b_{1}} \times \vec{b_{2}}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & - 2 \\ 1 & 2 & - 2 \end{matrix}|$

$= [\hat{i} (- 2 - (- 4)) - \hat{j} (2 - (- 2)) + \hat{k} (- 2 - 1)]$

$= 2 \hat{i} - 4 \hat{j} - 3 \hat{k}$

Magnitude of

$(\vec{b_{1}} \times \vec{b_{2}}) = \sqrt{{(2)}^{2} + {(- 4)}^{2} + {(- 3)}^{2}}$

$= \sqrt{4 + 16 + 9} = \sqrt{29}$

Also, $\Rightarrow (\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}}) = (\hat{j} - 4 \hat{k}) (2 \hat{i} - 4 \hat{j} - 3 \hat{k})$

$= - 4 + 12 = 8$

The shortest distance is given by:

$\Rightarrow |\frac{(\vec{b_{1}} \times \vec{b_{2}}) . (\vec{a_{2}} - \vec{a_{1}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| = |\frac{8}{\sqrt{29}}| = \frac{8}{\sqrt{29}}$

Find the shortest distance between the following pair of lines $\vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2 t) \hat{k}$ and $\vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}$ .

Important Questions on Straight Lines

Learn and Prepare for any exam you want !

Find the shortest distance between the following pair of lines r→=1-ti^+t-2j^+3-2tk^ and r→=s+1i^+2s-1j^-2s+1k^.

Important Questions on Straight Lines

Learn and Prepare for any exam you want !

Find the shortest distance between the following pair of lines $\vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2 t) \hat{k}$ and $\vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}$ .