Mathematics · Ch 11 — Three-Dimensional Geometry

Distance Between Parallel Lines

11.5.2

Distance Between Parallel Lines

11.5.2 Distance Between Parallel Lines

When two lines in space are parallel, the shortest distance between them is the length of the perpendicular segment connecting a point on one line to the other. This distance is constant — it does not depend on which point you choose.

Why the perpendicular distance?

For parallel lines, a single perpendicular to one line is automatically perpendicular to the other, so the shortest distance is simply the perpendicular distance from any point on one line to the other line.

Formula for the distance between two parallel lines

Let the two parallel lines be given in vector form:

\vec{r} = \vec{a}_1 + \lambda \vec{b} \quad \text{and} \quad \vec{r} = \vec{a}_2 + \mu \vec{b}

Here $\vec{a}_1$ and $\vec{a}_2$ are position vectors of points on the two lines, and $\vec{b}$ is the common direction vector. The shortest distance $d$ is:

d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}

Derivation of the formula

›Proof

Let $P$ (position $\vec{a}_1$ ) be a point on the first line and $Q$ (position $\vec{a}_2$ ) on the second, so $\vec{PQ} = \vec{a}_2 - \vec{a}_1$ . The magnitude $|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|$ equals the area of the parallelogram formed by $\vec{a}_2 - \vec{a}_1$ and $\vec{b}$ . With $\vec{b}$ as base, the perpendicular distance from $Q$ to the line is the height of this parallelogram:

$d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$

Alternative form using unit vector

If $\hat{b} = \frac{\vec{b}}{|\vec{b}|}$ is the unit direction vector, the distance can also be written as:

d = |(\vec{a}_2 - \vec{a}_1) \times \hat{b}|

Important observation

The distance $d$ is independent of the choice of points $\vec{a}_1$ and $\vec{a}_2$ . Picking different points changes $\vec{a}_2 - \vec{a}_1$ , but its component perpendicular to $\vec{b}$ stays the same.

Note

This formula is valid only when the lines are parallel. For skew lines (non-parallel, non-intersecting), a different formula is used.

Worked example

Problem: Find the shortest distance between the parallel lines

\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 (2\hat{i} + 3\hat{j} + 4\hat{k})

Solution:

Here $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ , $\vec{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}$ , and $\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ .

\vec{a}_2 - \vec{a}_1 = \hat{i} + 2\hat{j} + 2\hat{k}

(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(8 - 6) - \hat{j}(4 - 4) + \hat{k}(3 - 4) = 2\hat{i} + 0\hat{j} - \hat{k}

So $|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{2^2 + 0^2 + (-1)^2} = \sqrt{5}$ and $|\vec{b}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$ .

Therefore:

d = \frac{\sqrt{5}}{\sqrt{29}} = \sqrt{\frac{5}{29}}

Key formulas recap …

Figure 11.7Two parallel lines l1 and l2 with points S (position vector a1) and T (position vector a2), the join ST making angle theta with l1, and the perpendicular TP giving the distance between the lines.

Fig. 11.7 — Two parallel lines l1 and l2 with points S (position vector a1) and T (position vector a2), the join ST making angle theta with l1, and the perpendicular TP giving the distance between the lines.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

Figure 11.7 is the key visual for understanding the shortest distance between two parallel lines in space. The diagram shows two horizontal, parallel lines: the lower line is labelled $l_1$ and the upper line is labelled $l_2$ . Both are drawn as indigo arrows pointing to the right, indicating that they extend infinitely in that direction.

On $l_1$ , a point $S$ is marked. Its position vector from the origin is $\vec{a}_1$ . On $l_2$ , a point $T$ is marked, with position vector $\vec{a}_2$ . The segment $ST$ is drawn connecting these two points, and at $S$ it makes an angle $\theta$ with $l_1$ . From $T$ , a perpendicular is dropped onto $l_1$ , meeting it at point $P$ . Right-angle marks are shown at both $T$ and $P$ , confirming that $TP$ is perpendicular to $l_1$ . The length $TP$ is the perpendicular distance between the two parallel lines — this is the shortest distance.

The physical idea is straightforward: for two parallel lines, the shortest distance between them is the length of the perpendicular segment from any point on one line to the other line. The segment $ST$ is not the shortest distance because it is slanted; only when $\theta = 90^\circ$ does $ST$ coincide with $TP$ . The figure makes clear that the perpendicular distance is unique and independent of which point you choose on $l_1$ or $l_2$ .

Important

For parallel lines, the shortest distance is always the perpendicular distance. The slanted segment $ST$ is longer than $TP$ unless $\theta = 90^\circ$ .

The textbook uses this figure to derive the formula for the shortest distance between two parallel lines. If the lines are given in vector form as

\vec{r} = \vec{a}_1 + \lambda \vec{b} \quad \text{and} \quad \vec{r} = \vec{a}_2 + \mu \vec{b},

where $\vec{b}$ is the common direction vector, then the shortest distance $d$ is

d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}.

Here, $\vec{a}_2 - \vec{a}_1$ is the vector from $S$ to $T$ (the slanted segment). The cross product $(\vec{a}_2 - \vec{a}_1) \times \vec{b}$ gives a vector whose magnitude equals the area of the parallelogram formed by $\vec{a}_2 - \vec{a}_1$ and $\vec{b}$ . Dividing by $|\vec{b}|$ gives the height of that parallelogram — which is exactly the perpendicular distance $TP$ .

d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}

In the figure, $|\vec{a}_2 - \vec{a}_1|$ is the length $ST$ , and $|\vec{b}|$ is the magnitude of the direction vector along the lines. The angle $\theta$ between $ST$ and $l_1$ satisfies $\sin\theta = \frac{TP}{ST}$ , which is exactly what the cross product formula captures: $|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = |\vec{a}_2 - \vec{a}_1|\,|\vec{b}|\,\sin\theta$ , so dividing by $|\vec{b}|$ yields $|\vec{a}_2 - \vec{a}_1|\sin\theta = TP$ . …

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