Name: Distance of a Point From a Plane
Uploaded: 2019-07-19
Duration: 9 min 54 s
Description: Three-dimensional space has an infinite number of points and planes in it! This video discusses the distance of a point from a plane, both in vector and cart...

Question

Find the shortest distance between the pair of lines

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

and

\vec{r} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k})

.

Accepted Answer

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

Given lines are $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - 3 \hat{j} + 2 \hat{k})$ and $\vec{r} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$

Comparing given lines with $\vec{r} = \vec{a_{1}} + λ \vec{b_{1}}$ and $\vec{r} = \vec{a_{2}} + μ \vec{b_{2}}$

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

$\vec{a_{2}} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}, \vec{b_{2}} = 2 \hat{i} + 3 \hat{j} + \hat{k}$

Now, $\vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & 2 \\ 2 & 3 & 1 \end{matrix}|$

$= \hat{i} (- 3 - 6) - \hat{j} (1 - 4) + \hat{k} (3 + 6)$

$= - 9 \hat{i} + 3 \hat{j} + 9 \hat{k}$

$|\overset{⇀}{b_{1}} \times \vec{b_{2}}| = \sqrt{{(- 9)}^{2} + {(3)}^{2} + {(9)}^{2}} = \sqrt{171}$

and $\vec{a_{2}} - \vec{a_{1}} = (4 - 1) \hat{i} + (5 - 2) \hat{j} + (6 - 3) \hat{k} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k}$

Shortest distance between the lines $= |\frac{(\vec{b_{1}} \times \vec{b_{2}}) \cdot (\vec{a_{2}} - \vec{a_{1}})}{|\vec{b_{1}} \times \vec{b_{2}}|}|$

$= |\frac{(3 \hat{i} + 3 \hat{j} + 3 \hat{k}) \cdot (- 9 \hat{i} + 3 \hat{j} + 9 \hat{k})}{\sqrt{171}}|$

$= |\frac{- 27 + 9 + 27}{3 \sqrt{19}}|$

$= \frac{9}{3 \sqrt{19}} = \frac{3}{\sqrt{19}}$

Therefore, the shortest distance between given pair of lines is $\frac{3}{\sqrt{19}}$ units.

Find the shortest distance between the pair of lines $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - 3 \hat{j} + 2 \hat{k})$ and $\vec{r} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$ .

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

Find the shortest distance between the pair of lines r→=(i^+2j^+3k^)+λ(i^-3j^+2k^) and r→=(4i^+5j^+6k^)+μ(2i^+3j^+k^).

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

Find the shortest distance between the pair of lines $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - 3 \hat{j} + 2 \hat{k})$ and $\vec{r} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$ .