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