# 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. = $$\sqrt{5}\over \sqrt{29}$$