Name: Distance Between Two Parallel Lines in 3D
Uploaded: 2019-07-19
Duration: 6 min 41 s
Description: What do you know about parallel lines in three-dimensional space? Do they stay at a constant distance apart? How do we find that distance? This video answers...

Question

Find the shortest distance between the lines

\vec{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{j} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k})

and

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

.

Accepted Answer

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

We have to find the shortest distance between the two lines.

Let $\vec{a} = 6 \hat{i} + 2 \hat{j} + 2 \hat{j}, \vec{b} = (\hat{i} - 2 \hat{j} + 2 \hat{k}), \vec{c} = - 4 \hat{i} - \hat{k}$ and $\vec{d} = 3 \hat{i} - 2 \hat{j} - 2 \hat{k}$ .

Shortest distance between the given lines is $\frac{∣ [(\vec{a} - \vec{c}) \vec{b} \vec{d}] |}{| \vec{b} \times \vec{d} |}$

$[(\vec{a} - \vec{c}) \vec{b} \vec{d}]$ $= | \begin{array}{ccc} 10 & 2 & 3 \\ 1 & - 2 & 2 \\ 3 & - 2 & - 2 \end{array} |$

$= 10 (4 + 4) - 2 (- 2 - 6) + 3 (- 2 + 6)$

$= 108$

$\vec{b} \times \vec{d} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 2 \\ 3 & - 2 & - 2 \end{array} |$

$= \hat{i} (4 + 4) - \hat{j} (- 2 - 6) + \hat{k} (- 2 + 6)$

$= 8 \hat{i} + 8 \hat{j} + 4 \hat{k}$

$| \vec{b} \times \vec{d} | = \sqrt{64 + 64 + 16} = 12$

$∴ shortest distance = \frac{108}{12} = 9$

Thus, the shortest distance between the given two lines is $9$ units.

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

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

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

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

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