Shortest Distance Between Two Lines

Hello JEE aspirants! The "Shortest Distance Between Two Lines" is a crucial concept in 3D geometry. You'll often encounter problems based on this in JEE Main, and mastering it will significantly boost your score. It's all about finding the minimum distance between lines in 3D space, a skill that finds applications in various fields like physics and engineering.

Conceptual Explanation and Intuition

Imagine two lines floating in 3D space. They can be parallel, intersecting, or skew (non-parallel and non-intersecting). The shortest distance is the length of the line segment that is perpendicular to both lines. For intersecting lines, the shortest distance is, of course, zero.

The key idea is to understand the vector representation of lines. Each line can be defined by a point on the line and a direction vector. The shortest distance calculations use vector algebra to find the perpendicular distance.

1. Shortest Distance Between Skew Lines

Skew lines are lines that are neither parallel nor intersecting. Think of them as two lines that pass over and under each other in 3D space without ever meeting.

SD = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}

Here, the lines are given in vector form as: $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ , where:

$\vec{a_1}$ and $\vec{a_2}$ are position vectors of points on the two lines.
$\vec{b_1}$ and $\vec{b_2}$ are the direction vectors of the two lines.
$\lambda$ and $\mu$ are scalar parameters.

Derivation/Explanation:

The vector $\vec{a_2} - \vec{a_1}$ connects a point on line 1 to a point on line 2. We want the component of this vector that is perpendicular to *both* lines. The cross product $\vec{b_1} \times \vec{b_2}$ gives us a vector that *is* perpendicular to both lines.

The dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ gives us the component of $\vec{a_2} - \vec{a_1}$ that lies in the direction of $\vec{b_1} \times \vec{b_2}$ . The absolute value ensures we have a positive distance. We then divide by the magnitude of $\vec{b_1} \times \vec{b_2}$ to normalize and get the shortest distance.

Example: Consider the lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda (\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu (2\hat{i} + \hat{j} + \hat{k})$ . Here, $\vec{a_1} = \hat{i} + \hat{j}$ , $\vec{b_1} = \hat{i} - \hat{j} + \hat{k}$ , $\vec{a_2} = 2\hat{i} + \hat{j} - \hat{k}$ , and $\vec{b_2} = 2\hat{i} + \hat{j} + \hat{k}$ . Calculate $\vec{a_2} - \vec{a_1}$ , $\vec{b_1} \times \vec{b_2}$ , and then apply the formula.

2. Shortest Distance Between Parallel Lines

When lines are parallel, their direction vectors are proportional. Finding the shortest distance is a slightly different calculation.

SD = \frac{|\vec{b} \times (\vec{a_2}-\vec{a_1})|}{|\vec{b}|}

Here, the lines are given as $\vec{r} = \vec{a_1} + \lambda \vec{b}$ and $\vec{r} = \vec{a_2} + \mu \vec{b}$ , where:

$\vec{a_1}$ and $\vec{a_2}$ are position vectors of points on the two lines.
$\vec{b}$ is the direction vector (same for both lines).

Derivation/Explanation:

The vector $\vec{a_2} - \vec{a_1}$ connects a point on line 1 to a point on line 2. The cross product $\vec{b} \times (\vec{a_2} - \vec{a_1})$ gives a vector whose magnitude represents the area of the parallelogram formed by $\vec{b}$ and $\vec{a_2} - \vec{a_1}$ . Dividing this area by the base ( $|\vec{b}|$ ) gives the height of the parallelogram, which is the shortest distance between the parallel lines.

Example: Consider the parallel lines $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (\hat{i} + \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} + 3\hat{j} + 4\hat{k}) + \mu (\hat{i} + \hat{j} + \hat{k})$ . Here, $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$ , $\vec{a_2} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ , and $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ . Compute $\vec{a_2} - \vec{a_1}$ , $\vec{b} \times (\vec{a_2} - \vec{a_1})$ , and then apply the formula.

3. Condition for Lines to Intersect

Two lines intersect if the shortest distance between them is zero.

SD = 0

For skew lines, this means: $|(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})| = 0$ . For parallel lines, the formula $SD = \frac{|\vec{b} \times (\vec{a_2}-\vec{a_1})|}{|\vec{b}|}$ isn't helpful (it will either be 0 or non-zero, trivially determining if they are the *same* line). Instead, check if $(\vec{a_2} - \vec{a_1}) = k\vec{b}$ , where $k$ is some scalar. If this condition holds, the point $\vec{a_2}$ lies on the line $\vec{r} = \vec{a_1} + \lambda \vec{b}$ .

Example: If $(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0$ , the lines intersect (or are coplanar). This means the vector $\vec{a_2} - \vec{a_1}$ is perpendicular to the normal vector of the plane containing $\vec{b_1}$ and $\vec{b_2}$ , implying the lines lie in the same plane and intersect.

Tip: Always check if the lines are parallel first. If they are, use the parallel lines formula; otherwise, use the skew lines formula. Calculating

\vec{b_1} \times \vec{b_2}

early on helps in both cases!

Common Mistake: Forgetting the absolute value in the shortest distance formula. Distance is always non-negative! Also, make sure you are using the correct formula (skew vs. parallel). A quick check: if

\vec{b_1}

and

\vec{b_2}

are proportional, the lines are parallel.

JEE Trick: Sometimes, JEE questions will give you the lines in Cartesian form. Convert them to vector form

\vec{r} = \vec{a} + \lambda \vec{b}

to easily apply the formulas. Remember the conversion:

\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}

translates to

\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}

and

\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}

Master these formulas and practice applying them to various problems. Good luck with your JEE preparation!