Hey there, future engineers! Welcome to the exciting world of Three-Dimensional Geometry! You've probably dealt with lines on a flat piece of paper or a screen (that's 2D geometry). But now, we're going to add a whole new dimension – depth! And trust me, things get a lot more interesting when you step into 3D.

Today, we're going to dive into a super cool concept that's unique to 3D space: Skew Lines. We'll also figure out how to find the shortest distance between these special lines and even talk about the equation of that shortest distance line segment. So, grab your imaginary 3D glasses, and let's get started!

### What are Lines, Anyway? A Quick Recap!

Before we jump to skew lines, let's quickly remember what we know about lines in general, especially when we consider them in 3D space.

1. Intersecting Lines: These are lines that meet at a single, common point. Think of two roads crossing each other in a city. The distance between them at that intersection point is, well, zero!

2. Parallel Lines: These are lines that run in the same direction and never meet, no matter how far you extend them. Think of railway tracks or the opposite edges of a ruler. In 3D, parallel lines also never meet, and the distance between them is constant everywhere.

Now, for something that only happens in 3D...

### Introducing Skew Lines: The 3D Superstars!

Imagine you have two lines. Are they parallel? No. Do they intersect? No. What? How is that even possible? If they're not parallel, they *must* intersect, right? Well, that's true in 2D (on a flat surface). But in 3D, things are different!

Definition Time!

Two lines in three-dimensional space are called Skew Lines if they are neither parallel nor intersecting.

Let's break that down with an analogy:





* Think of it like this: Imagine two aeroplanes flying in the sky.
* If they're flying side-by-side in the same direction, they're like parallel lines. They won't crash into each other.
* If they're flying towards each other and meet at the same point at the same time, that's an intersection (and a very bad day!).
* But what if one plane is flying north at an altitude of 10,000 feet, and another plane is flying east at an altitude of 20,000 feet? Their paths might seem to "cross" when you look at a map (a 2D projection), but they are at completely different heights. They are not parallel, and they will never meet. These flight paths are perfect examples of skew lines!

Another great real-world example: Look at the edges of a room.
* The line where the ceiling meets the front wall, and the line where the floor meets the back wall. Are they parallel? No, they are usually perpendicular in projection. Do they intersect? No, one is up high, the other is down low. These are skew lines!

The key takeaway for skew lines is that they do not lie in the same plane. If two lines are in the same plane, they *must* either be parallel or intersect. But skew lines are like two completely separate entities in 3D space, each in its own "world" (or rather, its own plane, but not a common plane for both).

### The Challenge: Shortest Distance Between Skew Lines

Okay, so we have these two lines floating around in 3D, not touching, not running parallel. How do we measure the "distance" between them?

For parallel lines, it's easy: pick any point on one line and find its perpendicular distance to the other line. That's the distance.
For intersecting lines, the distance is 0.

But for skew lines, since they don't intersect, the distance is definitely not zero. And since they're not parallel, there isn't a *constant* distance everywhere. If you pick a random point on one line and find its distance to the other line, and then pick another random point, you'll likely get a different distance!

So, we're looking for something very specific: the shortest possible distance.

Key Concept:

The shortest distance between two skew lines is the length of the unique line segment that is perpendicular to BOTH lines.

Imagine our two aeroplane paths again. The shortest "bridge" or "connection" between them would be a line segment that drops straight down from one path to the other, at a 90-degree angle to *both* paths. This is a very special line segment!

#### Visualizing the Shortest Distance:

Hold two pencils in the air.
1. Make them parallel. The shortest distance is easy to see.
2. Make them intersect. The shortest distance is zero.
3. Now, hold them so they are not parallel, and definitely not touching. Try to imagine a tiny rubber band stretched between them that is as short as possible. That rubber band will naturally pull itself to be perpendicular to *both* pencils. That's your shortest distance segment!

This shortest distance segment is also sometimes called the line of shortest distance.

### How Do We Find This Magical Shortest Distance and Its Equation? (The Fundamental Idea)

You might be thinking, "This sounds complicated!" And yes, it involves a bit of vector math, but the underlying idea is quite intuitive.

1. Representing Lines in 3D: In 3D geometry, we usually represent lines using vector equations or Cartesian equations. For JEE, vector equations are super handy! A line passing through a point `a` and parallel to a vector `b` can be written as `r = a + λb`, where `λ` is a scalar parameter.

2. The Direction of Shortest Distance: Remember, the shortest distance segment is perpendicular to *both* skew lines. If we know the direction vectors of the two skew lines (let's call them `b₁` and `b₂`), then the direction of the shortest distance line segment must be perpendicular to both `b₁` and `b₂`. What mathematical operation gives us a vector perpendicular to two given vectors? You guessed it – the cross product! So, the direction of the shortest distance will be proportional to `b₁ × b₂`.

3. Finding the Magnitude (The Distance): Once we have the direction, we can use a clever trick involving projection. Imagine one skew line lying on a plane. The other skew line will be "above" or "below" this plane. The shortest distance is essentially the perpendicular distance between these two lines. We can pick a point on each line (say `A₁` on the first line and `A₂` on the second line) and consider the vector connecting them (`A₂ - A₁`). If we project this connecting vector onto the direction of the shortest distance (`b₁ × b₂`), the length of that projection will give us the shortest distance!

Think of it this way: You have a stick (`A₂ - A₁`) connecting two points, one on each skew line. You know the direction of the shortest path (`b₁ × b₂`). If you shine a light from the direction of the shortest path, the shadow of your stick on that direction would be the shortest distance! This is the essence of a scalar projection.

4. Finding the Equation (The Line Segment): This is a bit more involved but conceptually still based on perpendicularity. The shortest distance segment connects a point `P` on the first line to a point `Q` on the second line, such that `PQ` is perpendicular to both lines. We use the line equations (with parameters like `λ` and `μ`) to express `P` and `Q` in terms of these parameters. Then, we use the dot product (because perpendicular vectors have a dot product of zero) to set up equations to find `λ` and `μ`. Once we have `λ` and `μ`, we can find the exact coordinates of `P` and `Q`, and thus the equation of the line passing through `P` and `Q`.

### Why is this important for JEE?

The concept of skew lines and the shortest distance between them is a classic 3D geometry problem often tested in JEE. It combines your understanding of:
* Vector representation of lines.
* Cross product (for finding perpendicular directions).
* Dot product (for checking perpendicularity).
* Scalar projection (for finding distance).
* Solving systems of equations.

So, mastering this topic is not just about memorizing a formula; it's about understanding the geometric intuition behind the vector operations. This fundamental understanding will be your biggest asset when tackling complex problems.

In the upcoming sections, we'll dive deeper into the actual formulas and step-by-step calculations, but for now, I hope you have a clear picture of what skew lines are and why finding the shortest distance between them is a unique and important problem in 3D geometry! Keep visualizing, and you'll master this in no time!

Welcome to this deep dive into one of the most intriguing concepts in three-dimensional geometry: Skew Lines and the methods to find the shortest distance between them. This topic is a cornerstone for JEE Advanced, requiring a strong conceptual understanding and precise application of vector algebra.

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### Understanding Skew Lines: The 3D Conundrum

In two-dimensional geometry, two distinct lines can either intersect at a single point or be parallel. There are no other possibilities. However, once we step into the three-dimensional world, a new scenario emerges: lines that are neither intersecting nor parallel. These are what we call Skew Lines.

Imagine two airplanes flying in the sky. If their paths cross, they intersect. If they fly in the same direction, always maintaining the same separation, they are parallel. But what if one airplane is flying north at an altitude of 10,000 feet, and another is flying east at 15,000 feet? Their paths, if extended indefinitely, will never meet, nor are they parallel. Such flight paths represent skew lines.

Definition: Two lines in space are said to be skew if they are neither parallel nor intersecting. This implies that they lie in different planes. If two lines are parallel or intersect, they must lie in the same plane (they are coplanar). Skew lines are non-coplanar.

Let's represent two lines $L_1$ and $L_2$ in vector form:
Line $L_1$: $vec{r} = vec{a_1} + lambda vec{b_1}$
Line $L_2$: $vec{r} = vec{a_2} + mu vec{b_2}$

Here:
* $vec{a_1}$ and $vec{a_2}$ are the position vectors of points on lines $L_1$ and $L_2$ respectively.
* $vec{b_1}$ and $vec{b_2}$ are the direction vectors of lines $L_1$ and $L_2$ respectively.
* $lambda$ and $mu$ are scalar parameters.

For $L_1$ and $L_2$ to be skew lines, two conditions must be met:
1. Non-Parallel: Their direction vectors $vec{b_1}$ and $vec{b_2}$ are not collinear. That is, $vec{b_1}
e kvec{b_2}$ for any scalar $k$. This also means $vec{b_1} imes vec{b_2}
e vec{0}$.
2. Non-Intersecting: There are no values of $lambda$ and $mu$ for which $vec{a_1} + lambda vec{b_1} = vec{a_2} + mu vec{b_2}$. Mathematically, this can be checked by evaluating the scalar triple product: $(vec{a_2} - vec{a_1}) cdot (vec{b_1} imes vec{b_2})
e 0$. If this scalar triple product is zero, the lines are either intersecting or parallel (coplanar). Since we've already established non-parallelism, a zero scalar triple product would mean they intersect. For skew lines, it must be non-zero.

---

### Shortest Distance Between Skew Lines

The shortest distance between two skew lines is the length of the unique common perpendicular segment connecting them. This common perpendicular segment is perpendicular to both lines.

Intuition: Imagine holding two pencils in space such that they are skew. The shortest distance between them would be if you could place a tiny third pencil that touches both, and is perpendicular to both. That tiny pencil's length is the shortest distance.

Let $P_1$ be a point on $L_1$ with position vector $vec{p_1} = vec{a_1} + lambda vec{b_1}$ and $P_2$ be a point on $L_2$ with position vector $vec{p_2} = vec{a_2} + mu vec{b_2}$.
The vector joining $P_1$ and $P_2$ is $vec{P_1P_2} = vec{p_2} - vec{p_1} = (vec{a_2} - vec{a_1}) + mu vec{b_2} - lambda vec{b_1}$.

For $vec{P_1P_2}$ to be the shortest distance vector, it must be perpendicular to both $vec{b_1}$ and $vec{b_2}$.
This means $vec{P_1P_2} cdot vec{b_1} = 0$ and $vec{P_1P_2} cdot vec{b_2} = 0$.

The direction of the common perpendicular to both lines $L_1$ and $L_2$ is given by the cross product of their direction vectors, i.e., $vec{n} = vec{b_1} imes vec{b_2}$. The unit vector in this direction is $hat{n} = frac{vec{b_1} imes vec{b_2}}{|vec{b_1} imes vec{b_2}|}$.

The shortest distance, $d$, is the projection of the vector connecting any point on $L_1$ (say $vec{a_1}$) to any point on $L_2$ (say $vec{a_2}$) onto the direction of the common perpendicular.
So, $d = | ext{projection of } (vec{a_2} - vec{a_1}) ext{ onto } hat{n}|$.

#### Derivation of Shortest Distance Formula (Vector Form):

Let $P_1$ be a point on line $L_1$ with position vector $vec{a_1}$, and $P_2$ be a point on line $L_2$ with position vector $vec{a_2}$.
The vector connecting these two points is $vec{P_1P_2} = vec{a_2} - vec{a_1}$.

The direction of the common perpendicular line segment between $L_1$ and $L_2$ is given by $vec{b_1} imes vec{b_2}$.
Let $vec{n} = vec{b_1} imes vec{b_2}$.

The shortest distance $d$ is the magnitude of the projection of the vector $vec{P_1P_2}$ onto the direction $vec{n}$.
$d = |frac{(vec{a_2} - vec{a_1}) cdot vec{n}}{|vec{n}|}|$
Substitute $vec{n} = vec{b_1} imes vec{b_2}$:

Shortest Distance (Vector Form):

$d = frac{|(vec{a_2} - vec{a_1}) cdot (vec{b_1} imes vec{b_2})|}{|vec{b_1} imes vec{b_2}|}$

This formula involves the scalar triple product in the numerator, which confirms the coplanarity condition: if the scalar triple product is zero, the lines are coplanar (either intersecting or parallel). For parallel lines, $vec{b_1} imes vec{b_2} = vec{0}$, making the denominator zero. So, this formula is exclusively for skew lines.

#### Shortest Distance Formula (Cartesian Form):

Let the two lines be:
$L_1: frac{x-x_1}{l_1} = frac{y-y_1}{m_1} = frac{z-z_1}{n_1}$
$L_2: frac{x-x_2}{l_2} = frac{y-y_2}{m_2} = frac{z-z_2}{n_2}$

Here, $vec{a_1} = x_1hat{i} + y_1hat{j} + z_1hat{k}$ and $vec{b_1} = l_1hat{i} + m_1hat{j} + n_1hat{k}$.
And $vec{a_2} = x_2hat{i} + y_2hat{j} + z_2hat{k}$ and $vec{b_2} = l_2hat{i} + m_2hat{j} + n_2hat{k}$.

Then, $vec{a_2} - vec{a_1} = (x_2-x_1)hat{i} + (y_2-y_1)hat{j} + (z_2-z_1)hat{k}$.
And $vec{b_1} imes vec{b_2} = egin{vmatrix} hat{i} & hat{j} & hat{k} \ l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 end{vmatrix}$.

The scalar triple product $(vec{a_2} - vec{a_1}) cdot (vec{b_1} imes vec{b_2})$ can be written as a determinant:

Shortest Distance (Cartesian Form):

$d = frac{left| egin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 end{vmatrix}
ight|}{sqrt{(m_1n_2 - m_2n_1)^2 + (n_1l_2 - n_2l_1)^2 + (l_1m_2 - l_2m_1)^2}}$

This formula looks more daunting but is essentially the Cartesian representation of the vector formula. The denominator is $|vec{b_1} imes vec{b_2}|$.

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### Equation of the Line of Shortest Distance

Finding the equation of the line segment that represents the shortest distance requires finding the specific points $P_1$ and $P_2$ on each line.

Let $P_1$ be a point on $L_1$ and $P_2$ be a point on $L_2$ such that the vector $vec{P_1P_2}$ is perpendicular to both $vec{b_1}$ and $vec{b_2}$.
$vec{P_1} = vec{a_1} + lambda vec{b_1}$
$vec{P_2} = vec{a_2} + mu vec{b_2}$

The vector $vec{P_1P_2} = (vec{a_2} - vec{a_1}) + mu vec{b_2} - lambda vec{b_1}$.

For $vec{P_1P_2}$ to be the common perpendicular:
1. $vec{P_1P_2} cdot vec{b_1} = 0 implies [(vec{a_2} - vec{a_1}) + mu vec{b_2} - lambda vec{b_1}] cdot vec{b_1} = 0$
This expands to $(vec{a_2} - vec{a_1}) cdot vec{b_1} + mu (vec{b_2} cdot vec{b_1}) - lambda (vec{b_1} cdot vec{b_1}) = 0$ (Equation 1)
2. $vec{P_1P_2} cdot vec{b_2} = 0 implies [(vec{a_2} - vec{a_1}) + mu vec{b_2} - lambda vec{b_1}] cdot vec{b_2} = 0$
This expands to $(vec{a_2} - vec{a_1}) cdot vec{b_2} + mu (vec{b_2} cdot vec{b_2}) - lambda (vec{b_1} cdot vec{b_2}) = 0$ (Equation 2)

These two equations form a system of linear equations in terms of $lambda$ and $mu$. Solve for $lambda$ and $mu$.
Once $lambda$ and $mu$ are found, substitute them back into the equations for $vec{P_1}$ and $vec{P_2}$ to get the specific points $P_1$ and $P_2$.

The equation of the line of shortest distance can then be found using the two points $P_1$ and $P_2$:
$vec{r} = vec{P_1} + t(vec{P_2} - vec{P_1})$

Alternatively, the direction of the line of shortest distance is $vec{b_1} imes vec{b_2}$. If you know one point (say $P_1$), the equation is $vec{r} = vec{P_1} + t (vec{b_1} imes vec{b_2})$. This is less specific for the segment, but gives the line containing the segment.

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### JEE Focus: Shortest Distance Between Parallel Lines

While the primary focus is on skew lines, it's crucial to distinguish this from parallel lines. If two lines are parallel, their direction vectors are proportional ($vec{b_1} = kvec{b_2}$). In this case, $vec{b_1} imes vec{b_2} = vec{0}$, and the skew line formula becomes undefined.

For parallel lines, $L_1: vec{r} = vec{a_1} + lambda vec{b}$ and $L_2: vec{r} = vec{a_2} + mu vec{b}$.
The shortest distance $d$ is given by:
$d = frac{|(vec{a_2} - vec{a_1}) imes vec{b}|}{|vec{b}|}$

JEE Tip: Always check if the lines are parallel first by examining their direction vectors. If they are, use the parallel lines formula. If not, proceed to check the scalar triple product. If it's zero, they intersect (distance is 0). If non-zero, they are skew, and then use the skew lines formula.

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### Examples

Let's solidify our understanding with some examples.

#### Example 1: Finding Shortest Distance (Vector Form)

Find the shortest distance between the lines:
$L_1: vec{r} = (hat{i} + 2hat{j} + hat{k}) + lambda(hat{i} - hat{j} + hat{k})$
$L_2: vec{r} = (2hat{i} - hat{j} - hat{k}) + mu(2hat{i} + hat{j} + 2hat{k})$

Step 1: Identify $vec{a_1}, vec{a_2}, vec{b_1}, vec{b_2}$.
$vec{a_1} = hat{i} + 2hat{j} + hat{k}$
$vec{a_2} = 2hat{i} - hat{j} - hat{k}$
$vec{b_1} = hat{i} - hat{j} + hat{k}$
$vec{b_2} = 2hat{i} + hat{j} + 2hat{k}$

Step 2: Check for parallelism.
$vec{b_1}$ and $vec{b_2}$ are not proportional, e.g., $1/2
e -1/1$. So, the lines are not parallel.

Step 3: Calculate $vec{a_2} - vec{a_1}$.
$vec{a_2} - vec{a_1} = (2hat{i} - hat{j} - hat{k}) - (hat{i} + 2hat{j} + hat{k})$
$vec{a_2} - vec{a_1} = hat{i} - 3hat{j} - 2hat{k}$

Step 4: Calculate $vec{b_1} imes vec{b_2}$.
$vec{b_1} imes vec{b_2} = egin{vmatrix} hat{i} & hat{j} & hat{k} \ 1 & -1 & 1 \ 2 & 1 & 2 end{vmatrix}$
$= hat{i}(-1 cdot 2 - 1 cdot 1) - hat{j}(1 cdot 2 - 1 cdot 2) + hat{k}(1 cdot 1 - (-1) cdot 2)$
$= hat{i}(-2 - 1) - hat{j}(2 - 2) + hat{k}(1 + 2)$
$= -3hat{i} + 0hat{j} + 3hat{k} = -3hat{i} + 3hat{k}$

Step 5: Calculate $|vec{b_1} imes vec{b_2}|$.
$|vec{b_1} imes vec{b_2}| = sqrt{(-3)^2 + 3^2} = sqrt{9 + 9} = sqrt{18} = 3sqrt{2}$

Step 6: Calculate the scalar triple product $(vec{a_2} - vec{a_1}) cdot (vec{b_1} imes vec{b_2})$.
$(hat{i} - 3hat{j} - 2hat{k}) cdot (-3hat{i} + 3hat{k})$
$= (1)(-3) + (-3)(0) + (-2)(3)$
$= -3 + 0 - 6 = -9$
Since this is non-zero, the lines are skew.

Step 7: Apply the shortest distance formula.
$d = frac{|(vec{a_2} - vec{a_1}) cdot (vec{b_1} imes vec{b_2})|}{|vec{b_1} imes vec{b_2}|} = frac{|-9|}{3sqrt{2}} = frac{9}{3sqrt{2}} = frac{3}{sqrt{2}} = frac{3sqrt{2}}{2}$ units.

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#### Example 2: Finding the Equation of the Line of Shortest Distance (Cartesian Form)

Find the shortest distance and the equation of the line of shortest distance between the lines:
$L_1: frac{x-3}{1} = frac{y-5}{-2} = frac{z-7}{1}$
$L_2: frac{x+1}{7} = frac{y+1}{-6} = frac{z+1}{1}$

Step 1: Convert to Vector Form (or identify components).
$vec{a_1} = 3hat{i} + 5hat{j} + 7hat{k}$
$vec{b_1} = hat{i} - 2hat{j} + hat{k}$
$vec{a_2} = -hat{i} - hat{j} - hat{k}$
$vec{b_2} = 7hat{i} - 6hat{j} + hat{k}$

Step 2: Check for parallelism.
$vec{b_1}$ and $vec{b_2}$ are not proportional ($1/7
e -2/-6$). Not parallel.

Step 3: Calculate $vec{a_2} - vec{a_1}$.
$vec{a_2} - vec{a_1} = (-1-3)hat{i} + (-1-5)hat{j} + (-1-7)hat{k} = -4hat{i} - 6hat{j} - 8hat{k}$

Step 4: Calculate $vec{b_1} imes vec{b_2}$.
$vec{b_1} imes vec{b_2} = egin{vmatrix} hat{i} & hat{j} & hat{k} \ 1 & -2 & 1 \ 7 & -6 & 1 end{vmatrix}$
$= hat{i}(-2 - (-6)) - hat{j}(1 - 7) + hat{k}(-6 - (-14))$
$= hat{i}(-2+6) - hat{j}(1-7) + hat{k}(-6+14)$
$= 4hat{i} + 6hat{j} + 8hat{k}$

Step 5: Calculate $|vec{b_1} imes vec{b_2}|$.
$|vec{b_1} imes vec{b_2}| = sqrt{4^2 + 6^2 + 8^2} = sqrt{16 + 36 + 64} = sqrt{116}$

Step 6: Calculate $(vec{a_2} - vec{a_1}) cdot (vec{b_1} imes vec{b_2})$.
$(-4hat{i} - 6hat{j} - 8hat{k}) cdot (4hat{i} + 6hat{j} + 8hat{k})$
$= (-4)(4) + (-6)(6) + (-8)(8)$
$= -16 - 36 - 64 = -116$
Since this is non-zero, the lines are skew.

Step 7: Shortest Distance.
$d = frac{|-116|}{sqrt{116}} = frac{116}{sqrt{116}} = sqrt{116} = 2sqrt{29}$ units.

Step 8: Find the equation of the line of shortest distance.
Let $P_1(3+lambda, 5-2lambda, 7+lambda)$ be a point on $L_1$.
Let $P_2(-1+7mu, -1-6mu, -1+mu)$ be a point on $L_2$.
The vector $vec{P_1P_2} = (-1+7mu - (3+lambda))hat{i} + (-1-6mu - (5-2lambda))hat{j} + (-1+mu - (7+lambda))hat{k}$
$vec{P_1P_2} = (-4+7mu-lambda)hat{i} + (-6-6mu+2lambda)hat{j} + (-8+mu-lambda)hat{k}$

$vec{P_1P_2}$ must be perpendicular to $vec{b_1}$ and $vec{b_2}$.
We know $vec{P_1P_2}$ is in the direction of $vec{b_1} imes vec{b_2} = 4hat{i} + 6hat{j} + 8hat{k}$.
So, $vec{P_1P_2}$ must be proportional to $4hat{i} + 6hat{j} + 8hat{k}$. Let $vec{P_1P_2} = k(4hat{i} + 6hat{j} + 8hat{k})$.

Equating components:
$-4+7mu-lambda = 4k$ (i)
$-6-6mu+2lambda = 6k$ (ii)
$-8+mu-lambda = 8k$ (iii)

Divide (ii) by 2 and (iii) by 2:
$-3-3mu+lambda = 3k$ (ii')
$-4+0.5mu-0.5lambda = 4k$ (iii') - This is becoming complicated due to 'k'.

A more direct way to solve for $lambda$ and $mu$ is to use the dot product conditions:
1. $vec{P_1P_2} cdot vec{b_1} = 0$:
$(-4+7mu-lambda)(1) + (-6-6mu+2lambda)(-2) + (-8+mu-lambda)(1) = 0$
$-4+7mu-lambda + 12+12mu-4lambda - 8+mu-lambda = 0$
$0 + 20mu - 6lambda = 0 implies 10mu - 3lambda = 0$ (Eq A)

2. $vec{P_1P_2} cdot vec{b_2} = 0$:
$(-4+7mu-lambda)(7) + (-6-6mu+2lambda)(-6) + (-8+mu-lambda)(1) = 0$
$-28+49mu-7lambda + 36+36mu-12lambda - 8+mu-lambda = 0$
$0 + 86mu - 20lambda = 0 implies 43mu - 10lambda = 0$ (Eq B)

From (Eq A): $lambda = frac{10}{3}mu$.
Substitute into (Eq B): $43mu - 10(frac{10}{3}mu) = 0$
$43mu - frac{100}{3}mu = 0$
$frac{129mu - 100mu}{3} = 0 implies frac{29mu}{3} = 0 implies mu = 0$.

If $mu = 0$, then $lambda = 0$.

Substitute $lambda=0$ into $P_1$:
$P_1 = (3+0, 5-2(0), 7+0) = (3, 5, 7)$
Substitute $mu=0$ into $P_2$:
$P_2 = (-1+7(0), -1-6(0), -1+0) = (-1, -1, -1)$

The points on the lines giving the shortest distance are $P_1(3, 5, 7)$ and $P_2(-1, -1, -1)$.
The shortest distance is indeed the distance between these two points:
$d = sqrt{(-1-3)^2 + (-1-5)^2 + (-1-7)^2} = sqrt{(-4)^2 + (-6)^2 + (-8)^2}$
$d = sqrt{16 + 36 + 64} = sqrt{116} = 2sqrt{29}$ (matches our earlier calculation).

The equation of the line of shortest distance (passing through $P_1$ and $P_2$) is:
$vec{r} = vec{P_1} + t(vec{P_2} - vec{P_1})$
$vec{P_2} - vec{P_1} = (-4hat{i} - 6hat{j} - 8hat{k})$
$vec{r} = (3hat{i} + 5hat{j} + 7hat{k}) + t(-4hat{i} - 6hat{j} - 8hat{k})$
Or in Cartesian form:
$frac{x-3}{-4} = frac{y-5}{-6} = frac{z-7}{-8}$
This can be simplified by dividing the denominators by -2:

Equation of Line of Shortest Distance:

$frac{x-3}{2} = frac{y-5}{3} = frac{z-7}{4}$

This in-depth analysis of skew lines and shortest distance calculation, including the derivation and examples, provides a robust foundation for tackling JEE problems. Remember the key: identify the type of lines first, then apply the appropriate formula and methodology.

Mastering the concepts and formulas for skew lines and the shortest distance between them is crucial for JEE Main. Given the complexity of the formulas, mnemonics and shortcuts can significantly aid in quick recall during exams.

### Mnemonics & Shortcuts for Skew Lines and Shortest Distance

#### 1. Defining Skew Lines
Concept: Skew lines are non-parallel, non-intersecting lines in three-dimensional space. They cannot lie on the same plane.
Shortcut Mnemonic: "No Meet, No Greet, Just 3D Feat!"
* No Meet: They do not intersect.
* No Greet: They are not parallel (thus, their direction vectors are not proportional).
* 3D Feat: They exist specifically in 3D space, unlike lines in a plane.

#### 2. Shortest Distance Formula (Vector Form)
For two skew lines given by:
Line 1: $vec{r} = vec{a}_1 + lambda vec{b}_1$
Line 2: $vec{r} = vec{a}_2 + mu vec{b}_2$

The shortest distance, $d$, is given by:
$d = frac{|(vec{a}_2 - vec{a}_1) cdot (vec{b}_1 imes vec{b}_2)|}{|vec{b}_1 imes vec{b}_2|}$

Shortcut Mnemonic: "Diff-Dot-Cross, over Mag-Cross."
* Diff: Refers to the position vector difference $(vec{a}_2 - vec{a}_1)$. This is the vector connecting a point on Line 1 to a point on Line 2.
* Dot: Represents the dot product operator $(cdot)$.
* Cross: Refers to the cross product of the direction vectors $(vec{b}_1 imes vec{b}_2)$. This vector is perpendicular to both lines.
* Mag-Cross: Represents the magnitude of the cross product of the direction vectors, $|vec{b}_1 imes vec{b}_2|$.
This mnemonic helps you recall the entire structure of the numerator (scalar triple product) and the denominator (magnitude of the cross product).

#### 3. Shortest Distance Formula (Cartesian Form)
For two skew lines given by:
Line 1: $frac{x-x_1}{l_1} = frac{y-y_1}{m_1} = frac{z-z_1}{n_1}$
Line 2: $frac{x-x_2}{l_2} = frac{y-y_2}{m_2} = frac{z-z_2}{n_2}$

The shortest distance, $d$, is given by:
$d = frac{egin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 end{vmatrix}}{sqrt{(m_1n_2-m_2n_1)^2 + (n_1l_2-n_2l_1)^2 + (l_1m_2-l_2m_1)^2}}$

Shortcut Mnemonic: "Delta XYZ Determinant on top, Mag of Cross-Components below."
* Delta XYZ Determinant on top:
* The top row consists of the differences in the coordinates of the fixed points $(x_2-x_1, y_2-y_1, z_2-z_1)$.
* The second and third rows are the direction ratios of the two lines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.
* This directly represents the scalar triple product in Cartesian form.
* Mag of Cross-Components below:
* The denominator is the magnitude of the vector obtained by crossing the direction vectors $(vec{b}_1 imes vec{b}_2)$.
* The terms $(m_1n_2-m_2n_1)$, $(n_1l_2-n_2l_1)$, $(l_1m_2-l_2m_1)$ are precisely the components of $vec{b}_1 imes vec{b}_2$.
* This helps connect the Cartesian formula back to its vector counterpart, ensuring consistency.

#### 4. Equation of the Line of Shortest Distance
Finding the equation of the line segment that represents the shortest distance requires finding the specific points (say P and Q) on each skew line that are closest to each other.

Conceptual Shortcut: "Perpendicularity for Parameters P and Q."
1. Assume a general point P on Line 1 as $vec{p} = vec{a}_1 + lambda vec{b}_1$.
2. Assume a general point Q on Line 2 as $vec{q} = vec{a}_2 + mu vec{b}_2$.
3. Form the vector $vec{PQ} = vec{q} - vec{p} = (vec{a}_2 - vec{a}_1) + mu vec{b}_2 - lambda vec{b}_1$.
4. Key Condition (Perpendicularity): The vector $vec{PQ}$ must be perpendicular to both direction vectors $vec{b}_1$ and $vec{b}_2$.
* This means, $vec{PQ} cdot vec{b}_1 = 0$
* And, $vec{PQ} cdot vec{b}_2 = 0$
5. Solving these two scalar product equations will give you the unique values of $lambda$ and $mu$.
6. Substitute $lambda$ and $mu$ back into the expressions for $vec{p}$ and $vec{q}$ to get the specific coordinates of P and Q.
7. The line of shortest distance can then be found by using P (or Q) as a point and $vec{PQ}$ (or $vec{b}_1 imes vec{b}_2$) as its direction vector.

Quick Tips for Skew Lines and Shortest Distance

Mastering skew lines and the shortest distance between them is crucial for both JEE Main and board exams. Here are some quick tips to help you ace this topic efficiently:

1. Understanding Skew Lines

• Definition: Two lines are said to be skew if they are neither parallel nor intersecting. They exist in different planes.


• Quick Check (JEE Tip):
  
  
  
  • First, check if they are parallel: Their direction vectors ($vec{b}_1$ and $vec{b}_2$) must be non-collinear (i.e., $vec{b}_1
    eq kvec{b}_2$ for any scalar $k$).
  
  
  • Second, check if they intersect: If they don't, and they are not parallel, they are skew.
  
  
  • A quick way to check if they intersect for lines $vec{r} = vec{a}_1 + lambdavec{b}_1$ and $vec{r} = vec{a}_2 + muvec{b}_2$: Calculate $(vec{a}_2 - vec{a}_1) cdot (vec{b}_1 imes vec{b}_2)$. If this is zero, they are coplanar. If they are also not parallel, they must intersect. If it's non-zero, they are skew.
  
  
  
  

2. Shortest Distance (SD) between Skew Lines

Let the two skew lines be:

• Line 1: $vec{r}_1 = vec{a}_1 + lambda vec{b}_1$


• Line 2: $vec{r}_2 = vec{a}_2 + mu vec{b}_2$

Here, $vec{a}_1, vec{a}_2$ are position vectors of points on the lines, and $vec{b}_1, vec{b}_2$ are their direction vectors.

Formula (Vector Form - Most Important for JEE):

The shortest distance (d) between them is given by:

• Tip: The numerator is the scalar triple product $[(vec{a}_2 - vec{a}_1), vec{b}_1, vec{b}_2]$.


• Calculation Sequence:
  
  
  
  1. Calculate $(vec{a}_2 - vec{a}_1)$.
  
  
  2. Calculate $(vec{b}_1 imes vec{b}_2)$.
  
  
  3. Calculate the dot product of the results from (1) and (2).
  
  
  4. Calculate the magnitude of $(vec{b}_1 imes vec{b}_2)$.
  
  
  5. Divide the absolute value of (3) by (4).
  
  
  
  

3. Equation of the Shortest Distance Line

This is a more involved process but critical for complete understanding in JEE:

1. Direction Vector: The shortest distance line is perpendicular to both skew lines. Its direction vector $vec{n}$ is given by $vec{n} = vec{b}_1 imes vec{b}_2$.


2. Points on Lines: Let $P_1(vec{p}_1)$ be a point on Line 1 and $P_2(vec{p}_2)$ be a point on Line 2.
   
   
   
   • $vec{p}_1 = vec{a}_1 + lambda_0 vec{b}_1$
   
   
   • $vec{p}_2 = vec{a}_2 + mu_0 vec{b}_2$
   
   
   
   The vector $vec{P_1P_2} = vec{p}_2 - vec{p}_1$ represents the shortest distance.
   


3. Perpendicularity Conditions: $vec{P_1P_2}$ must be perpendicular to both $vec{b}_1$ and $vec{b}_2$.
   
   
   
   • $(vec{p}_2 - vec{p}_1) cdot vec{b}_1 = 0$
   
   
   • $(vec{p}_2 - vec{p}_1) cdot vec{b}_2 = 0$
   
   
   
   


4. Solve for Parameters: Use the two dot product equations to solve for $lambda_0$ and $mu_0$.


5. Find Points: Substitute $lambda_0$ and $mu_0$ back into the expressions for $vec{p}_1$ and $vec{p}_2$ to get the two points.


6. Equation of Line: Use one of these points (e.g., $vec{p}_1$) and the direction vector $vec{n} = vec{b}_1 imes vec{b}_2$ to write the equation of the shortest distance line: $vec{R} = vec{p}_1 + k(vec{b}_1 imes vec{b}_2)$.





4. CBSE vs. JEE Focus





• CBSE Boards: Primarily focuses on applying the shortest distance formula (vector and Cartesian forms). Derivations of the formula might be asked. Finding the equation of the shortest distance line is less common but can appear as a higher-order question.


• JEE Main: Expect direct application of the formula, but also questions requiring you to find the coordinates of the points defining the shortest distance, or the equation of the shortest distance line itself. Conceptual understanding of skew lines and coplanarity is also tested.






Keep practicing these concepts and formulas. You've got this!

Intuitive Understanding: Skew Lines and Shortest Distance

Understanding skew lines and the shortest distance between them is a fundamental concept in three-dimensional geometry. Unlike two-dimensional geometry where non-parallel lines always intersect, in 3D space, lines can be non-parallel *and* not intersect. These are what we call skew lines.

What are Skew Lines?

Imagine two roads. One road goes over a bridge, and another road goes under it. They are not parallel, but they also never meet or cross each other. This is the perfect analogy for skew lines.

In simple terms:

• Two lines are skew if they are not parallel.


• AND they do not intersect each other.


• This can only happen in three-dimensional space. In 2D, if lines aren't parallel, they must intersect.

Think of the edges of a cube: two opposite edges that are not in the same plane and are not parallel to each other are skew lines. For example, a bottom-front edge and a top-back edge.

Why "Shortest Distance"?

Since skew lines never intersect, there will always be a distance between them. However, this distance isn't constant. If you pick random points on each line and measure the distance between them, you'll get varying results. We are interested in the minimum possible distance between any pair of points, one on each line. This minimum distance is unique and geometrically significant.

Visualizing the Shortest Distance

The shortest distance between two skew lines can be intuitively understood as the length of a line segment that:

1. Connects a point on the first line to a point on the second line.


2. Is perpendicular to both the first skew line and the second skew line simultaneously.

Imagine a measuring stick that can perfectly bridge the gap between our two "roads" (skew lines) such that it stands upright (perpendicular) to both road surfaces at the points where it touches them. The length of this specific measuring stick is the shortest distance. This common perpendicular is unique.

JEE Insight: While the intuitive understanding is crucial for both CBSE and JEE, JEE questions often test your ability to apply vector or Cartesian formulas derived from this geometric understanding to find this shortest distance and the coordinates of the points where it connects the two lines.

The Line of Shortest Distance

The line segment that represents the shortest distance between the two skew lines is often called the line of shortest distance. Its equation defines this unique segment. Understanding this segment's existence and its perpendicularity to both skew lines is the key intuitive takeaway before diving into the mathematical formulas.

In essence, skew lines are truly 3D phenomena where lines bypass each other, and the shortest distance is a special, unique "bridge" that connects them at a perfect right angle to both.

Real-World Applications of Skew Lines and Shortest Distance

The concept of skew lines, which are non-parallel and non-intersecting lines in three-dimensional space, along with the calculation of the shortest distance between them, holds significant practical importance across various engineering, scientific, and technological domains. While direct application problems involving real-world scenarios might not be common in JEE Main, understanding these applications can deepen your conceptual understanding and highlight the utility of this mathematical tool.

Here are some key real-world applications:

• 
  Air Traffic Control and Aerospace Engineering: One of the most critical applications is in air traffic management. Flight paths of aircraft often represent skew lines in 3D space. Air traffic controllers use sophisticated systems that continuously calculate the shortest distance between these paths to ensure aircraft maintain safe separation distances, preventing mid-air collisions. This is crucial for both scheduled flights and during emergency landing procedures.
  


• 
  Robotics and Automation: In complex industrial robots or automated systems with multiple moving arms, understanding skew lines is vital for collision avoidance. The paths of robotic manipulators can be modeled as lines in space. Calculating the shortest distance between these paths helps in programming the robots to operate efficiently without self-intersecting or colliding with other machinery, optimizing movement and ensuring safety.
  


• 
  Civil Engineering and Architecture: When designing complex structures such as bridges, overpasses, or multi-level interchanges, different structural elements (like beams, cables, or utility pipes) might cross each other without actually intersecting. Engineers must ensure adequate clearance between these elements. Calculating the shortest distance between their projected lines helps in optimizing space, preventing structural interference, and ensuring construction feasibility. Similarly, in urban planning, designing complex road networks or subway tunnels involves ensuring that different routes maintain minimum separation.
  


• 
  Computer Graphics and Game Development: In 3D modeling and game engines, collision detection algorithms frequently employ concepts related to the shortest distance between objects. For simplifying calculations, objects or their bounding boxes might be approximated as line segments. Determining the closest points between these "skew lines" helps in detecting potential collisions between characters, vehicles, or environmental elements, contributing to realistic physics and interactions.
  


• 
  Telecommunications and Cable Routing: When routing fiber optic cables, power lines, or network infrastructure in densely populated areas or complex buildings, multiple cables often run in close proximity but on different planes. Ensuring minimum separation to prevent interference or for maintenance access can involve modeling these routes as skew lines and calculating the shortest distances.
  

In all these scenarios, the mathematical framework of vector and Cartesian equations of lines, coupled with the formula for the shortest distance between skew lines, provides the essential tools to analyze, predict, and optimize real-world spatial arrangements. Understanding these applications reinforces the practical value of three-dimensional geometry.

Understanding abstract mathematical concepts like skew lines and the shortest distance between them can be significantly simplified through relatable analogies. These real-world examples help build intuition and solidify comprehension, especially for 3D geometry.

Common Analogies for Skew Lines and Shortest Distance

1. Analogy for Skew Lines: Roads/Highways or Edges of a Room

• 
  Roads/Highways: Imagine a complex highway intersection. Consider a flyover (elevated road) that passes *over* another road at ground level. If the flyover is not parallel to the ground-level road and they clearly do not meet (one is above the other), they represent skew lines. They exist in different planes, are not parallel, and will never intersect.
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  

  Concept	Analogy Element
  Skew Lines	A flyover and a ground-level road that are not parallel and do not intersect.

  
  


• 
  Edges of a Room: Think about a rectangular room.
  
  
  
  • Pick one edge of the floor (e.g., where the wall meets the floor at the front left).
  
  
  • Now, consider an edge of the ceiling that is not parallel to the first edge and is not directly above it (e.g., the back right edge of the ceiling).
  
  
  
  These two edges are skew lines. They are in different planes (floor plane vs. ceiling plane), they are not parallel to each other, and they will never intersect, even if extended infinitely.
  

Key takeaway: The defining characteristics of skew lines are that they are non-parallel and non-intersecting, a condition only possible in three or more dimensions. In 2D, lines are either parallel or intersecting.

2. Analogy for Shortest Distance Between Skew Lines: The "Connecting Bridge"

Let's extend the room analogy to understand the shortest distance:

• 
  The "Connecting Bridge" in a Room: Using the same two skew edges (front left floor edge and back right ceiling edge), imagine you want to build the shortest possible "bridge" or support beam that connects these two edges.
  
  
  
  • This shortest connection would not be just any line segment between them.
  
  
  • Instead, it would be a unique line segment that is perpendicular to *both* skew lines simultaneously.
  
  
  
  

  Picture a specialized vertical pole that connects the floor edge to the ceiling edge. This pole would be the shortest distance, and critically, it would be perpendicular to the direction of both the floor edge and the ceiling edge at the points where it connects.

  
  
  
  
  
  
  
  
  
  
  
  
  
  
  

  Concept	Analogy Element
  Shortest Distance	The shortest possible 'bridge' or 'support beam' connecting the two skew edges, which is perpendicular to both edges.

  
  


• 
  Crane Dropping a Beam: Consider the highway analogy again. If a crane needed to drop the shortest possible beam from the flyover onto the ground-level road, that beam would need to be perfectly oriented such that it is perpendicular to the direction of travel on *both* the flyover and the ground road at the points of contact. The length of this beam represents the shortest distance.
  

For JEE/CBSE: While analogies aid understanding, remember that in exams, you'll need to apply the vector or Cartesian formulas. However, visualizing with these analogies can help you interpret the results and avoid common mistakes, especially regarding the perpendicularity condition for the shortest distance.

These analogies highlight that the shortest distance between skew lines is a unique segment that "bridges" them at a specific orientation, unlike intersecting lines (distance is zero) or parallel lines (distance is the perpendicular distance between them at any point).

To effectively grasp the concept of skew lines and the shortest distance between them, a strong foundation in several key mathematical areas is essential. This section outlines the prerequisite topics you should be comfortable with before delving into this advanced part of 3D Geometry.

Why are these important? The methods for finding the shortest distance between skew lines heavily rely on vector operations and their geometric interpretations. A clear understanding of these foundational concepts will make the derivation and application of formulas much more intuitive and less prone to errors.

1. Vector Algebra Fundamentals

Vector algebra forms the bedrock of 3D geometry, especially for topics involving lines and their relationships.

• 
  Position Vectors: Understanding how to represent a point in 3D space using a position vector (e.g., $vec{r} = xhat{i} + yhat{j} + zhat{k}$).
  


• 
  Vector Operations: Proficiency in vector addition, subtraction, and scalar multiplication.
  


• 
  Dot Product (Scalar Product):
  
  
  
  • Definition: $vec{a} cdot vec{b} = |vec{a}| |vec{b}| cos heta$, and in component form $a_1b_1 + a_2b_2 + a_3b_3$.
  
  
  • Geometric Significance: Used to find the angle between two vectors and to check for perpendicularity ($vec{a} cdot vec{b} = 0$ if $vec{a} perp vec{b}$). This is crucial for understanding the common perpendicular between skew lines.
  
  
  
  


• 
  Cross Product (Vector Product):
  
  
  
  • Definition: $vec{a} imes vec{b} = |vec{a}| |vec{b}| sin heta hat{n}$, where $hat{n}$ is a unit vector perpendicular to both $vec{a}$ and $vec{b}$. Its magnitude gives the area of the parallelogram formed by $vec{a}$ and $vec{b}$.
  
  
  • Geometric Significance: Generates a vector perpendicular to the plane containing $vec{a}$ and $vec{b}$. This is indispensable for finding the direction of the common perpendicular to two skew lines.
  
  
  • Computation: Ability to compute the cross product using determinants.
  
  
  
  


• 
  Scalar Triple Product (Box Product):
  
  
  
  • Definition: $[vec{a} vec{b} vec{c}] = vec{a} cdot (vec{b} imes vec{c})$.
  
  
  • Geometric Significance: Represents the volume of the parallelepiped formed by the three vectors. For skew lines, it helps in calculating the shortest distance using the volume of a parallelepiped formed by the vector connecting two points on the lines and their direction vectors.
  
  
  • Computation: Ability to compute using a 3x3 determinant.
  
  
  
  

2. Lines in 3D Space

Understanding the basic representation and properties of lines is fundamental.

• 
  Vector Equation of a Line: Knowing how to represent a line passing through a point $vec{a}$ and parallel to a direction vector $vec{b}$ as $vec{r} = vec{a} + lambda vec{b}$.
  


• 
  Cartesian Equation of a Line: Conversion between vector and Cartesian forms: $frac{x-x_1}{a} = frac{y-y_1}{b} = frac{z-z_1}{c}$.
  


• 
  Direction Ratios (DRs) and Direction Cosines (DCs): Understanding their meaning, relationship, and how they define the direction of a line.
  


• 
  Conditions for Parallel and Perpendicular Lines:
  
  
  
  • Parallel: Direction vectors are proportional ($vec{b}_1 = k vec{b}_2$).
  
  
  • Perpendicular: Dot product of direction vectors is zero ($vec{b}_1 cdot vec{b}_2 = 0$).
  
  
  
  

3. Determinants

The ability to accurately and quickly calculate determinants is a practical skill required for vector cross products and scalar triple products, which are extensively used in shortest distance problems.

• 2x2 Determinants: For cross product components.


• 3x3 Determinants: For the full cross product and scalar triple product.

JEE vs. CBSE Relevance: While these foundational concepts are covered in both CBSE board exams and JEE, the JEE often demands a more profound conceptual understanding and faster application, especially in complex problem-solving scenarios involving these vector operations. Ensure you can not only define but also geometrically interpret and apply each concept efficiently.

Related Topics for Skew Lines and Shortest Distance

Understanding skew lines and calculating the shortest distance between them requires a strong foundation in several core concepts of Three-Dimensional Geometry and Vector Algebra. These related topics are not just prerequisites but often appear in conjunction with or as part of the solution process for problems involving skew lines. Mastery of these areas will significantly enhance your ability to tackle complex problems.

Here are the key related topics:

• Vector Algebra (Dot Product and Cross Product):
  
  
  
  • Relevance: Vector algebra is the backbone of 3D geometry. The direction vectors of lines are central to determining their orientation.
    
    
    
    • The cross product of the direction vectors of two skew lines gives a vector perpendicular to both lines, which is crucial for finding the direction of the shortest distance. Its magnitude is also used in the shortest distance formula.
    
    
    • The dot product is used to check for perpendicularity between vectors and is implicitly involved in calculating scalar projections, which are part of the shortest distance formula.
    
    
    
    
  
  
  • JEE Focus: Expect direct applications of vector operations in almost every 3D geometry problem, especially those involving distances and angles.
  
  
  
  


• Equation of a Line in 3D Space:
  
  
  
  • Relevance: This is fundamental. You must be proficient in representing a line in both its vector form ($vec{r} = vec{a} + lambdavec{b}$) and Cartesian form ($frac{x-x_1}{l} = frac{y-y_1}{m} = frac{z-z_1}{n}$). Skew line problems will always provide lines in one of these forms. Identifying the position vector ($vec{a}$) and direction vector ($vec{b}$) for each line is the first step.
  
  
  • CBSE & JEE Focus: Both syllabi extensively cover line equations. JEE problems often require converting between forms or extracting specific information from them.
  
  
  
  


• Scalar Triple Product (STP) and Coplanarity of Lines:
  
  
  
  • Relevance: The Scalar Triple Product (STP), $(vec{a}_2 - vec{a}_1) cdot (vec{b}_1 imes vec{b}_2)$, is a powerful tool to determine if two lines are coplanar or not.
    
    
    
    • If the STP is zero, the lines are coplanar (meaning they are either intersecting or parallel).
    
    
    • If the STP is non-zero, the lines are non-coplanar, which is the definition of skew lines.
    
    
    
    The numerator of the shortest distance formula for skew lines is essentially the magnitude of this Scalar Triple Product.
  
  
  • JEE Focus: This is a common method for initial classification of lines before proceeding with distance calculations.
  
  
  
  


• Distance between Parallel Lines:
  
  
  
  • Relevance: While distinct from skew lines, the concept of finding the shortest distance between two parallel lines is a simpler case of distance calculation in 3D geometry. Understanding this simpler case provides context and helps appreciate the additional complexity involved with skew lines. The approach for parallel lines often involves picking a point on one line and finding its distance from the other line.
  
  
  • CBSE & JEE Focus: Both cover this topic, often before or alongside skew lines, offering a comparative understanding of 3D distance calculations.
  
  
  
  


• Equation of a Plane in 3D Space:
  
  
  
  • Relevance: Although the shortest distance between skew lines is typically found using vector methods involving the cross product, the concept of planes is subtly related. The shortest distance between two skew lines can be seen as the perpendicular distance between two parallel planes, where one plane contains one line and is parallel to the other line. This conceptual understanding can be helpful, though not always directly used in calculation.
  
  
  • JEE Focus: While not central to the direct shortest distance formula for skew lines, a solid understanding of plane equations is essential for overall 3D geometry problems, and some alternative derivations might involve planes.
  
  
  
  

By strengthening your understanding of these interconnected topics, you will build a robust framework to confidently approach problems involving skew lines and the shortest distance between them, a frequently tested concept in the JEE Main examination.

Key Takeaways: Skew Lines and Shortest Distance

Understanding skew lines and calculating the shortest distance between them is a crucial concept in Three-Dimensional Geometry, particularly for the JEE Main and Advanced examinations. Mastery of both vector and Cartesian approaches is essential.

• 
  Definition of Skew Lines:
  
  
  
  • Two lines in 3D space are called skew lines if they are neither parallel nor intersecting.
  
  
  • This implies that skew lines are also non-coplanar. They lie in different planes.
  
  
  
  


• 
  Condition for Skew Lines:
  
  
  
  • Given two lines:
    
    
    
    • L₁: &vec;r = &vec;a₁ + λ&vec;b₁
    
    
    • L₂: &vec;r = &vec;a₂ + μ&vec;b₂
    
    
    
    
  
  
  • They are skew if:
    
    
    
    1. &vec;b₁ is not parallel to &vec;b₂ (i.e., &vec;b₁ ≠ k&vec;b₂ for any scalar k).
    
    
    2. They do not intersect, meaning the shortest distance between them is non-zero. The condition for non-intersection is   (&vec;a₂ - &vec;a₁) ċ (&vec;b₁ × &vec;b₂) ≠ 0.
    
    
    
    
  
  
  
  


• 
  Shortest Distance Formula (Vector Form):
  
  
  
  • For two skew lines L₁: &vec;r = &vec;a₁ + λ&vec;b₁ and L₂: &vec;r = &vec;a₂ + μ&vec;b₂, the shortest distance (d) between them is given by:
    
    
    d = |(&vec;a₂ - &vec;a₁) ċ (&vec;b₁ × &vec;b₂)| / |&vec;b₁ × &vec;b₂|
    
  
  
  • Key insight: The direction of the shortest distance vector is perpendicular to both &vec;b₁ and &vec;b₂, which is given by &vec;b₁ × &vec;b₂.
  
  
  
  


• 
  Shortest Distance Formula (Cartesian Form):
  
  
  
  • If the lines are given as:
    
    
    
    • L₁: (x - x₁)/l₁ = (y - y₁)/m₁ = (z - z₁)/n₁
    
    
    • L₂: (x - x₂)/l₂ = (y - y₂)/m₂ = (z - z₂)/n₂
    
    
    
    
  
  
  • The shortest distance (d) can be found using the determinant:
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    

    	Numerator
    d =	x₂ - x₁	y₂ - y₁	z₂ - z₁
    	l₁	m₁	n₁
    	l₂	m₂	n₂
    	÷ √[(m₁n₂ - m₂n₁)² + (n₁l₂ - n₂l₁)² + (l₁m₂ - l₂m₁)²]

    
    
    
    
  
  
  
  


• 
  Equation of the Line of Shortest Distance:
  
  
  
  • This line is perpendicular to both skew lines. Its direction vector is parallel to &vec;b₁ × &vec;b₂.
  
  
  • To find its equation, you need a point on it. This point can be found by determining the points P on L₁ and Q on L₂ that are closest to each other.
  
  
  • Let P(&vec;a₁ + λ&vec;b₁) and Q(&vec;a₂ + μ&vec;b₂) be the points on L₁ and L₂ respectively, where the shortest distance exists.
  
  
  • The vector &vec;PQ = (&vec;a₂ + μ&vec;b₂) - (&vec;a₁ + λ&vec;b₁) must be perpendicular to both &vec;b₁ and &vec;b₂.
    
    
    
    • (&vec;PQ) ċ &vec;b₁ = 0
    
    
    • (&vec;PQ) ċ &vec;b₂ = 0
    
    
    
    
  
  
  • Solving these two equations for λ and μ allows you to find the coordinates of P and Q. The line passing through P and Q is the line of shortest distance.
  
  
  
  


• 
  Special Case: Intersecting Lines (Distance = 0):
  
  
  
  • If (&vec;a₂ - &vec;a₁) ċ (&vec;b₁ × &vec;b₂) = 0, the lines are either parallel or intersecting.
  
  
  • If they are not parallel, then they must intersect, and the shortest distance between them is zero.
  
  
  
  


• 
  JEE Focus:
  
  
  
  • This topic is frequently tested in both JEE Main and Advanced.
  
  
  • Be proficient in both vector and Cartesian approaches, as problems can be presented in either form.
  
  
  • Finding the equation of the line of shortest distance is a more complex and common problem in advanced exams compared to just calculating the distance.
  
  
  • Practice problems involving finding the coordinates of the feet of the common perpendicular.
  
  
  
  

Solving problems involving skew lines and the shortest distance between them requires a systematic approach, often leveraging vector algebra. This section outlines the practical steps to tackle such problems for JEE Main and board exams.

1. Identifying Skew Lines

First, confirm that the given lines are indeed skew. Skew lines are non-parallel and non-intersecting. To check this:

• Parallelism Check: Two lines $vec{r_1} = vec{a_1} + lambda vec{b_1}$ and $vec{r_2} = vec{a_2} + mu vec{b_2}$ are parallel if $vec{b_1}$ is parallel to $vec{b_2}$ (i.e., $vec{b_1} = k vec{b_2}$ for some scalar k). If they are parallel, they are not skew; they are either coincident or parallel and distinct.


• Intersection Check (if not parallel): If lines are not parallel, they either intersect or are skew. For intersection, there must exist $lambda$ and $mu$ such that $vec{a_1} + lambda vec{b_1} = vec{a_2} + mu vec{b_2}$. If no such $lambda, mu$ exist, the lines are skew.

2. Calculating the Shortest Distance (SD) between Skew Lines

This is a core concept for both CBSE and JEE. Always prefer the vector form for its elegance.

Method 1: Vector Form

Given two skew lines:

• Line L1: $vec{r} = vec{a_1} + lambda vec{b_1}$ (passes through point A with position vector $vec{a_1}$ and is parallel to vector $vec{b_1}$)


• Line L2: $vec{r} = vec{a_2} + mu vec{b_2}$ (passes through point B with position vector $vec{a_2}$ and is parallel to vector $vec{b_2}$)

The shortest distance (SD) between them is given by the formula:

$SD = left| frac{(vec{b_1} imes vec{b_2}) cdot (vec{a_2} - vec{a_1})}{|vec{b_1} imes vec{b_2}|}
ight|$

Problem-Solving Steps:

1. Identify Vectors: Extract $vec{a_1}, vec{b_1}, vec{a_2}, vec{b_2}$ from the given line equations. Remember $vec{a_1}$ and $vec{a_2}$ are position vectors of *any* point on the respective lines, and $vec{b_1}$ and $vec{b_2}$ are direction vectors.


2. Calculate $(vec{a_2} - vec{a_1})$: Find the vector connecting a point on L1 to a point on L2.


3. Calculate $(vec{b_1} imes vec{b_2})$: This vector is perpendicular to both lines, representing the direction of the shortest distance.


4. Calculate Dot Product: Find $(vec{b_1} imes vec{b_2}) cdot (vec{a_2} - vec{a_1})$. This gives the scalar triple product, which can also be calculated as a determinant (see Cartesian form below).


5. Calculate Magnitude: Find $|vec{b_1} imes vec{b_2}|$.


6. Apply Formula: Substitute the calculated values into the SD formula. Take the absolute value to ensure distance is positive.

Method 2: Cartesian Form (Determinant Approach)

If lines are given in Cartesian form:

• Line L1: $frac{x-x_1}{a_1} = frac{y-y_1}{b_1} = frac{z-z_1}{c_1}$


• Line L2: $frac{x-x_2}{a_2} = frac{y-y_2}{b_2} = frac{z-z_2}{c_2}$

The shortest distance (SD) is given by:

$SD = left| frac{egin{vmatrix} (x_2-x_1) & (y_2-y_1) & (z_2-z_1) \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 end{vmatrix}}{sqrt{(b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2 + (a_1b_2-a_2b_1)^2}}
ight|$

This is essentially the same as the vector formula, where the numerator is the scalar triple product $(vec{a_2} - vec{a_1}) cdot (vec{b_1} imes vec{b_2})$ and the denominator is $|vec{b_1} imes vec{b_2}|$. Choose the form you are more comfortable with.

3. Finding the Equation of the Line of Shortest Distance

This is generally a more complex problem, frequently appearing in JEE. The line of shortest distance (LSD) is perpendicular to both skew lines.

Problem-Solving Steps:

1. Represent General Points: Let P be a general point on L1 and Q be a general point on L2.
   
   
   
   • P = $vec{a_1} + lambda vec{b_1}$ (in terms of $lambda$)
   
   
   • Q = $vec{a_2} + mu vec{b_2}$ (in terms of $mu$)
   
   
   
   


2. Form Vector PQ: Calculate the vector $vec{PQ} = Q - P = (vec{a_2} - vec{a_1}) + mu vec{b_2} - lambda vec{b_1}$.


3. Apply Perpendicularity Conditions: The line of shortest distance, represented by $vec{PQ}$, is perpendicular to both L1 ($vec{b_1}$) and L2 ($vec{b_2}$).
   
   
   
   • $vec{PQ} cdot vec{b_1} = 0$
   
   
   • $vec{PQ} cdot vec{b_2} = 0$
   
   
   
   


4. Solve for $lambda$ and $mu$: These two dot product equations will give you a system of two linear equations in $lambda$ and $mu$. Solve them to find their unique values.


5. Find Points P and Q: Substitute the values of $lambda$ and $mu$ back into the expressions for P and Q to get the specific points on each line that define the shortest distance.


6. Form the Equation: The line of shortest distance passes through P and Q. Its equation can be written as:
   
   
   
   • Vector Form: $vec{R} = P + t (vec{Q} - vec{P})$ or $vec{R} = P + t (vec{PQ})$
   
   
   • Cartesian Form: Use the coordinates of P (or Q) and the direction ratios of $vec{PQ}$.
   
   
   
   

JEE Specific Tip:

For JEE, ensure proficiency in handling cross products and dot products quickly and accurately. The algebraic computations can be extensive, so careful calculation is key. Also, remember that the direction vector of the line of shortest distance is proportional to $(vec{b_1} imes vec{b_2})$, though finding the actual points P and Q is usually necessary to define its equation.

Mastering these steps will equip you to solve a wide range of problems on skew lines and their shortest distance.

CBSE Focus Areas: Skew Lines and Shortest Distance

For CBSE Board Examinations, understanding skew lines and the method to calculate the shortest distance between them, along with the equation of the line representing this distance, is a critical concept, often featuring in higher-mark questions (4 or 6 marks).

1. Understanding Skew Lines (CBSE Perspective)

• Definition: Two lines in space are called skew lines if they are neither parallel nor intersecting. This means they lie in different planes.


• Distinction:
  
  
  
  • Important: Unlike parallel or intersecting lines (which are coplanar), skew lines are non-coplanar.
  
  
  • If lines are parallel, their direction vectors are proportional. If they intersect, there's a common point. Skew lines satisfy neither condition.
  
  
  
  

2. Shortest Distance Between Two Skew Lines

The shortest distance between two skew lines is the length of the unique common perpendicular between them.

a) Vector Form:

Consider two skew lines given by:

• $L_1: vec{r} = vec{a_1} + lambda vec{b_1}$


• $L_2: vec{r} = vec{a_2} + mu vec{b_2}$

Where $vec{a_1}, vec{a_2}$ are position vectors of points on $L_1, L_2$ respectively, and $vec{b_1}, vec{b_2}$ are their direction vectors.

The shortest distance (d) between them is given by:

$$d = left| frac{(vec{b_1} imes vec{b_2}) cdot (vec{a_2} - vec{a_1})}{|vec{b_1} imes vec{b_2}|}
ight|$$

CBSE Application Focus:

• Students must be able to correctly identify $vec{a_1}, vec{a_2}, vec{b_1}, vec{b_2}$ from the given line equations.


• Accuracy in calculating the cross product $vec{b_1} imes vec{b_2}$ and the dot product is crucial.


• Remember that the magnitude of $vec{b_1} imes vec{b_2}$ should not be zero for skew lines (as it implies they are parallel).

b) Cartesian Form:

Consider two skew lines given by:

• $L_1: frac{x-x_1}{l_1} = frac{y-y_1}{m_1} = frac{z-z_1}{n_1}$


• $L_2: frac{x-x_2}{l_2} = frac{y-y_2}{m_2} = frac{z-z_2}{n_2}$

The shortest distance (d) between them is given by the determinant form:

$$d = left| frac{egin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ l_1 & m_1 & n_1 \ l_2 & m_2 & n_2 end{vmatrix}}{sqrt{(m_1n_2 - m_2n_1)^2 + (n_1l_2 - n_2l_1)^2 + (l_1m_2 - l_2m_1)^2}}
ight|$$

CBSE Application Focus:

• This formula is derived from the vector form. The numerator is $(vec{a_2} - vec{a_1}) cdot (vec{b_1} imes vec{b_2})$ expressed as a scalar triple product, and the denominator is $|vec{b_1} imes vec{b_2}|$.


• Students should be proficient in calculating 3x3 determinants.


• Tip: For most CBSE problems, converting Cartesian to vector form, applying the vector formula, and then converting back (if needed) is often less prone to error than memorizing the lengthy Cartesian determinant formula directly.

3. Equation of the Line of Shortest Distance

While finding the shortest distance is more frequent, CBSE sometimes asks for the equation of the line of shortest distance. This line is perpendicular to both skew lines.

• Let P be a point on $L_1$ and Q be a point on $L_2$ such that PQ is the shortest distance.


• The direction vector of the line PQ will be parallel to $vec{b_1} imes vec{b_2}$.


• To find the equation, assume P and Q in terms of $lambda$ and $mu$ respectively. Use the condition that vector $vec{PQ}$ is perpendicular to both $vec{b_1}$ and $vec{b_2}$ (i.e., $vec{PQ} cdot vec{b_1} = 0$ and $vec{PQ} cdot vec{b_2} = 0$) to solve for $lambda$ and $mu$.


• Once P and Q are found, the equation of the line PQ can be determined using the two-point form.

Common CBSE Mistakes to Avoid:

• Confusing the formula for shortest distance between skew lines with that for parallel lines.


• Incorrectly identifying $vec{a_1}, vec{a_2}, vec{b_1}, vec{b_2}$ from the given equations.


• Errors in calculating cross products or dot products.


• Forgetting the absolute value in the final shortest distance calculation.

Mastering these formulas and their application is key to scoring well on this topic in CBSE exams. Practice solving problems involving both vector and Cartesian forms.

JEE Focus Areas: Skew Lines and Shortest Distance

Understanding skew lines and the shortest distance between them is a frequently tested concept in JEE Main and Advanced. It requires a strong grasp of vector algebra and 3D geometry principles. Mastering this topic can fetch you easy marks if you are familiar with the formulas and their applications.

1. What are Skew Lines?

• Two lines in three-dimensional space are called skew lines if they are neither parallel nor intersecting.


• They lie in different planes. For two lines to be skew, they cannot be coplanar.


• JEE Tip: Always first check if lines are parallel or intersecting before concluding they are skew.

2. Condition for Skew Lines

Given two lines:

• Line 1: r = a₁ + λb₁


• Line 2: r = a₂ + μb₂

The lines are skew if and only if:

(a₂ - a₁) · (b₁ x b₂) ≠ 0

This is the scalar triple product of the vectors (a₂ - a₁), b₁, and b₂. If it is zero, the lines are coplanar (either parallel or intersecting).

3. Shortest Distance Between Skew Lines

The shortest distance (d) between two skew lines r = a₁ + λb₁ and r = a₂ + μb₂ is given by:

Vector Form:

d = | (a₂ - a₁) · (b₁ x b₂) | / | b₁ x b₂ |

Cartesian Form:

Given lines:

L₁: (x - x₁)/a₁ = (y - y₁)/b₁ = (z - z₁)/c₁

L₂: (x - x₂)/a₂ = (y - y₂)/b₂ = (z - z₂)/c₂

The shortest distance is:

Determinant Term
(x₂-x₁)   (y₂-y₁)   (z₂-z₁) a₁   b₁   c₁ a₂   b₂   c₂

d = | (x₂-x₁)   (y₂-y₁)   (z₂-z₁) | / √[(b₁c₂-b₂c₁)² + (c₁a₂-c₂a₁)² + (a₁b₂-a₂b₁)²]

• JEE Tip: The vector form is generally more efficient and less prone to calculation errors. Recognize that the denominator is the magnitude of (b₁ x b₂).


• Important: If lines are parallel (i.e., b₁ and b₂ are parallel vectors, so b₁ x b₂ = 0), this formula is not applicable. For parallel lines, d = | (a₂ - a₁) x b | / | b | where b is the common direction vector.

4. Equation of the Line of Shortest Distance

This is a more advanced concept often tested in JEE Advanced or tougher JEE Main problems.

• The line of shortest distance is perpendicular to both direction vectors b₁ and b₂.


• Its direction vector will be parallel to N = b₁ x b₂.


• Let P be a point on L₁ and Q be a point on L₂ such that PQ is the shortest distance segment.
  
  
  
  • Vector PQ must be parallel to N.
  
  
  • This means PQ · b₁ = 0 and PQ · b₂ = 0.
  
  
  
  


• To find the equation of the line of shortest distance:
  
  
  
  1. Assume points P and Q on the lines L₁ and L₂ respectively, using parameters λ and μ.
     
     
     
     • P = a₁ + λb₁
     
     
     • Q = a₂ + μb₂
     
     
     
     
  
  
  2. Form the vector PQ = Q - P.
  
  
  3. Use the conditions PQ · b₁ = 0 and PQ · b₂ = 0 to set up a system of two linear equations in λ and μ.
  
  
  4. Solve for λ and μ. Substitute these values back into the expressions for P and Q to get the coordinates of the feet of the perpendicular.
  
  
  5. The equation of the line of shortest distance can then be formed using points P and Q, or P (or Q) and the direction vector N = b₁ x b₂.
  
  
  
  

5. JEE Main & Advanced Strategy

• Accuracy in Vector Operations: Scalar triple product and cross product calculations must be flawless. A single sign error can lead to an incorrect answer.


• Formula Recall: Memorize the shortest distance formula in vector form.


• Distinguish Cases: Always first check if the lines are parallel, intersecting, or skew. The formulas differ.


• Conceptual Depth: For finding the equation of the line of shortest distance, understand that it's perpendicular to both original lines. This is a critical insight for constructing the solution.

Stay focused and practice diligently! These concepts are highly rewarding with consistent effort.