Distance Between Two Parallel Lines in 3d

Here you will learn formula to find the distance between two parallel lines in 3d with example.

Let’s begin –

Distance Between Two Parallel Lines in 3d

Let \(l_1\) and \(l_2\) be two parallel lines having vector equations

\(l_1\) : \(\vec{r}\) = \(\vec{a_1}\) + \(\lambda\)\(\vec{b}\)

and \(l_2\) : \(\vec{r}\) = \(\vec{a_2}\) + \(\mu\)\(\vec{b}\).

The shortest distance (S.D.) between two the parallel lines \(\vec{r}\) = \(\vec{a_1}\) + \(\lambda\)\(\vec{b}\) and \(\vec{r}\) = \(\vec{a_2}\) + \(\mu\)\(\vec{b}\) is given by

d = |\((\vec{a_2} -\vec{ a_1}) \times \vec{b}\over |\vec{b}|\)|

Example : Find the shortest distance between the lines whose vector equations are

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

and \(\vec{r}\) = (\(2\hat{i} + 4\hat{j} + 5\hat{k}\)) + \(\mu\)(\(4\hat{i} + 6\hat{j} + 8\hat{k}\))

Solution : The vector equations of given lines are

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

and \(\vec{r}\) = (\(2\hat{i} + 4\hat{j} + 5\hat{k}\)) + 2\(\mu\)(\(2\hat{i} + 3\hat{j} + 4\hat{k}\))                         ………..(ii)

Equation (ii) can be re-written as

\(\vec{r}\) = (\(2\hat{i} + 4\hat{j} + 5\hat{k}\)) + \({\mu}’\)(\(2\hat{i} + 3\hat{j} + 4\hat{k}\))                             …………(iii)

where  \({\mu}’\) = 2\(\mu\)

These two lines passes through the points having position vectors \(\vec{a_1}\) = \(\hat{i} + 2\hat{j} + 3\hat{k}\) and \(a_2\) = \(2\hat{i} + 4\hat{j} + 5\hat{k}\) respectively and both are parallel to the vector \(\vec{b}\) = (\(2\hat{i} + 3\hat{j} + 4\hat{k}\)).

So, the shortest distance between them is given by

S.D. =  |\((\vec{a_2} -\vec{ a_1}) \times \vec{b}\over |\vec{b}|\)|                                  ………(iv)

Now, \((\vec{a_2} -\vec{ a_1}) \times \vec{b}\) = \(\hat{i} + 2\hat{j} + 2\hat{k}\)  \(\times\) (\(2\hat{i} + 3\hat{j} + 4\hat{k}\))

= \(\begin{vmatrix} \hat{i} & \hat{j} &  \hat{k} \\  1 & 2 & 2 \\  2 & 3 & 4 \end{vmatrix}\) = \(2\hat{i} – 0\hat{j} – k\hat{k}\)

\(\therefore\)  |\((\vec{a_2} -\vec{ a_1}) \times \vec{b}\)| = \(\sqrt{4 + 0 + 1}\) = \(\sqrt{5}\)

and |\(\vec{b}\)| = \(\sqrt{4 + 9 + 16}\) = \(\sqrt{29}\)

Substituting the values of |\((\vec{a_2} -\vec{ a_1}) \times \vec{b}\) | and |\(\vec{b}\)| in (iv), we obtain

S.D. = \(\sqrt{5}\over \sqrt{29}\)

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