Hello future engineers! Welcome to the exciting world of Three-Dimensional Geometry. Today, we're going to unravel the mystery of how to describe a straight line in space. You've been dealing with lines on a 2D plane (like your notebook paper) since childhood, right? Think of equations like `y = mx + c` or `Ax + By = C`. But what happens when our line isn't confined to a flat surface? What if it's floating freely in the vastness of 3D space, like a laser beam cutting through a room? That's what we're here to figure out!

### Understanding the Basics: What defines a Line in 3D?

Before we jump into equations, let's intuitively understand what makes a line unique in 3D space. Imagine you're trying to describe a straight path.

1. Two Points are Enough: If you tell me two distinct points through which a line passes, say Point A and Point B, then there's only one unique straight line that can connect them. Think of stretching a string between two fixed nails. That string forms a unique straight line.
2. One Point and a Direction: Alternatively, if you tell me one specific point where the line starts (or passes through) and the exact direction it's heading in, again, there's only one unique straight line possible. Imagine a flashlight: the light beam starts at the bulb (a point) and goes in a specific direction. That beam is a unique straight line.

These two fundamental ideas will be the bedrock of forming our equations. We'll predominantly use the concept of a position vector and a direction vector because vectors are incredibly powerful tools for handling 3D geometry!

### The Vector Equation of a Line: Point-Direction Form

Let's start with the second idea: defining a line using a point it passes through and its direction.

Imagine a point in space, let's call it A, with a position vector a relative to the origin. This vector a simply tells us "where" point A is in space.

Now, imagine the line needs to go in a specific direction. We can represent this direction using another vector, let's call it b. This vector b is parallel to our desired line. It doesn't tell us where the line starts, just its orientation. Think of b as an arrow showing the line's path.

Now, let's take any arbitrary point P that lies on this line. Let its position vector be r. Our goal is to find a relationship between r, a, and b.

Look at the vector connecting point A to point P, which is AP. Since P lies on the line passing through A and parallel to b, the vector AP must be parallel to the direction vector b.

Mathematically, if two vectors are parallel, one is a scalar multiple of the other. So, we can write:

$vec{AP} = lambda vec{b}$

Here, $lambda$ (lambda) is a scalar parameter. It's just a number that can be positive, negative, or zero.
* If $lambda = 0$, then $vec{AP} = vec{0}$, meaning P is the same as A.
* If $lambda > 0$, P is on one side of A.
* If $lambda < 0$, P is on the other side of A.
As $lambda$ varies from $-infty$ to $+infty$, the point P "travels" along the entire line!

Now, using the triangle law of vector addition, we know that $vec{AP} = vec{r} - vec{a}$.
Substituting this into our equation:

$vec{r} - vec{a} = lambda vec{b}$

Rearranging this, we get the Vector Equation of a Line:

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

Term	Meaning
$vec{r}$	The position vector of any arbitrary point (x, y, z) on the line. This is what makes it a general equation for *all* points on the line.
$vec{a}$	The position vector of a known point ($x_1, y_1, z_1$) through which the line passes.
$lambda$	A scalar parameter (a real number) that varies to trace all points on the line.
$vec{b}$	A vector parallel to the line, indicating its direction. Its components are the direction ratios of the line.

JEE Focus: This vector form is incredibly versatile and often preferred in competitive exams as it's concise and works directly with vector algebra.

### The Cartesian Equation of a Line: Point-Direction Form

While the vector form is elegant, sometimes we need to work with coordinates (x, y, z). Let's convert our vector equation into its Cartesian equivalent.

Let:
* The known point A have coordinates $(x_1, y_1, z_1)$, so its position vector is $vec{a} = x_1hat{i} + y_1hat{j} + z_1hat{k}$.
* The arbitrary point P on the line have coordinates $(x, y, z)$, so its position vector is $vec{r} = xhat{i} + yhat{j} + zhat{k}$.
* The direction vector b have components $(a, b, c)$. These components are called the direction ratios of the line. So, $vec{b} = ahat{i} + bhat{j} + chat{k}$. (Don't confuse these 'a, b, c' with the position vector $vec{a}$ components; they represent direction ratios here).

Substitute these into the vector equation $vec{r} = vec{a} + lambda vec{b}$:

$xhat{i} + yhat{j} + zhat{k} = (x_1hat{i} + y_1hat{j} + z_1hat{k}) + lambda (ahat{i} + bhat{j} + chat{k})$

$xhat{i} + yhat{j} + zhat{k} = (x_1 + lambda a)hat{i} + (y_1 + lambda b)hat{j} + (z_1 + lambda c)hat{k}$

For two vectors to be equal, their corresponding components must be equal:
1. $x = x_1 + lambda a$
2. $y = y_1 + lambda b$
3. $z = z_1 + lambda c$

From each of these equations, we can express $lambda$:
1. $lambda = frac{x - x_1}{a}$
2. $lambda = frac{y - y_1}{b}$
3. $lambda = frac{z - z_1}{c}$

Since $lambda$ is the same for all three, we can equate these expressions to get the Cartesian Equation of a Line:

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

Here:
* $(x_1, y_1, z_1)$ are the coordinates of the known point the line passes through.
* $(a, b, c)$ are the direction ratios of the line (components of the direction vector $vec{b}$).

Important Note: If any of the direction ratios ($a, b,$ or $c$) are zero, say $a=0$, then the line is perpendicular to the X-axis (or parallel to the YZ-plane). In this case, the equation would be written as $x = x_1$ and $frac{y - y_1}{b} = frac{z - z_1}{c}$. You cannot divide by zero! This means the line runs parallel to the coordinate plane whose normal contains the zero direction ratio.

CBSE vs JEE Focus: Both forms are equally important for CBSE. For JEE, understanding the relationship between direction ratios and direction cosines is key. Direction cosines are normalized direction ratios and provide the direction vector with unit magnitude. We'll explore that further in 'Direction Cosines and Ratios'.

#### Example 1: Point-Direction Form
Find the vector and Cartesian equations of the line passing through the point (1, 2, 3) and parallel to the vector $3hat{i} + 2hat{j} - 2hat{k}$.

Step-by-step Solution:
1. Identify the known point and direction vector:
* Position vector of the known point, $vec{a} = hat{i} + 2hat{j} + 3hat{k}$ (from point (1, 2, 3)).
* Direction vector, $vec{b} = 3hat{i} + 2hat{j} - 2hat{k}$.
* From $vec{b}$, we get direction ratios $a=3, b=2, c=-2$.

2. Write the Vector Equation:
* Using $vec{r} = vec{a} + lambda vec{b}$:
* $vec{r} = (hat{i} + 2hat{j} + 3hat{k}) + lambda (3hat{i} + 2hat{j} - 2hat{k})$

3. Write the Cartesian Equation:
* Using $frac{x - x_1}{a} = frac{y - y_1}{b} = frac{z - z_1}{c}$:
* Here $(x_1, y_1, z_1) = (1, 2, 3)$ and $(a, b, c) = (3, 2, -2)$.
* $frac{x - 1}{3} = frac{y - 2}{2} = frac{z - 3}{-2}$

### The Equation of a Line Passing Through Two Given Points

Now let's tackle our first idea: defining a line using two points it passes through.
Suppose the line passes through two distinct points A and B, with position vectors $vec{a}$ and $vec{b}$ respectively.

Can we convert this scenario back to the "point-direction" form? Absolutely!
1. Choose a point: We can pick either A (with position vector $vec{a}$) or B (with position vector $vec{b}$) as our known point. Let's choose A.
2. Find a direction vector: If the line passes through A and B, then the vector $vec{AB}$ is parallel to the line itself!
* We know $vec{AB} = vec{b} - vec{a}$.
* So, our direction vector $vec{b}_{direction}$ is $(vec{b} - vec{a})$.

Now, substitute these into the general vector form $vec{r} = vec{a}_{point} + lambda vec{b}_{direction}$:

$vec{r} = vec{a} + lambda (vec{b} - vec{a})$

This can also be written as:

$vec{r} = (1 - lambda)vec{a} + lambdavec{b}$

This is the Vector Equation of a Line passing through two given points.

### Cartesian Equation of a Line Passing Through Two Given Points

Let the two given points be $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$.
The direction ratios of the line passing through these two points can be found from the components of the vector $vec{AB}$:

$vec{AB} = (x_2 - x_1)hat{i} + (y_2 - y_1)hat{j} + (z_2 - z_1)hat{k}$

So, the direction ratios are $a = (x_2 - x_1)$, $b = (y_2 - y_1)$, and $c = (z_2 - z_1)$.

Now, use the Cartesian form for a line through one point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$:

$frac{x - x_1}{x_2 - x_1} = frac{y - y_1}{y_2 - y_1} = frac{z - z_1}{z_2 - z_1}$

This is the Cartesian Equation of a Line passing through two given points.

#### Example 2: Two-Point Form
Find the vector and Cartesian equations of the line passing through the points (1, -1, 0) and (2, 3, 4).

Step-by-step Solution:
1. Identify the two points:
* Let $A = (1, -1, 0)$, so $vec{a} = hat{i} - hat{j} + 0hat{k}$.
* Let $B = (2, 3, 4)$, so $vec{b} = 2hat{i} + 3hat{j} + 4hat{k}$.

2. Find the direction vector:
* $vec{b} - vec{a} = (2hat{i} + 3hat{j} + 4hat{k}) - (hat{i} - hat{j} + 0hat{k})$
* $vec{b} - vec{a} = (2-1)hat{i} + (3-(-1))hat{j} + (4-0)hat{k}$
* $vec{b} - vec{a} = hat{i} + 4hat{j} + 4hat{k}$
* The direction ratios are $(1, 4, 4)$.

3. Write the Vector Equation:
* Using $vec{r} = vec{a} + lambda (vec{b} - vec{a})$:
* $vec{r} = (hat{i} - hat{j}) + lambda (hat{i} + 4hat{j} + 4hat{k})$

4. Write the Cartesian Equation:
* Using $frac{x - x_1}{x_2 - x_1} = frac{y - y_1}{y_2 - y_1} = frac{z - z_1}{z_2 - z_1}$:
* Here $(x_1, y_1, z_1) = (1, -1, 0)$ and $(x_2, y_2, z_2) = (2, 3, 4)$.
* $frac{x - 1}{2 - 1} = frac{y - (-1)}{3 - (-1)} = frac{z - 0}{4 - 0}$
* $frac{x - 1}{1} = frac{y + 1}{4} = frac{z}{4}$

### Key Takeaways for Fundamentals

* A line in 3D space is uniquely determined by either two distinct points or one point and a direction.
* Vector equations are concise and powerful, especially for advanced problems.
* Cartesian equations are useful for plotting and when dealing with coordinate geometry problems.
* The parameter $lambda$ is crucial! It effectively "sweeps" across all points on the line.
* Always be careful when a direction ratio is zero in the Cartesian form, as it implies the line is parallel to a coordinate plane.

Understanding these fundamental forms is essential before we move on to more complex concepts like angles between lines, shortest distance between lines (especially skew lines), and planes. Keep practicing these basic forms, and you'll build a strong foundation for 3D geometry!

Welcome to this deep dive into understanding and forming the equations of a line in three-dimensional space! Unlike lines in two dimensions, which can be fully described by a simple equation like $y = mx + c$, lines in 3D space require a bit more information and a richer mathematical framework. For JEE, a solid grasp of these concepts is absolutely fundamental, as lines form the building blocks for understanding planes, distances, and various geometric configurations.

At its core, to uniquely define a line in 3D space, we need two fundamental pieces of information:
1. A point that the line passes through: This fixes its position in space.
2. The direction in which the line extends: This defines its orientation.

Let's explore how these two pieces of information are woven into both vector and Cartesian forms of a line's equation.

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I. Equation of a Line in Vector Form

The vector form is often the most intuitive way to understand the equation of a line because it directly incorporates a point and a direction vector.

Case 1: Line Passing Through a Given Point and Parallel to a Given Vector

Imagine you're standing at a specific point in space, say point A, and you're told to walk in a particular direction, say along vector $vec{b}$. If you keep walking in that direction, you'll trace a straight line.

* Let $A$ be a fixed point in space with position vector $vec{a}$ relative to the origin O.
* Let $vec{b}$ be a vector that determines the direction of the line.
* Let $P$ be any arbitrary point on the line, with position vector $vec{r}$.

Component	Description
$vec{r}$	Position vector of any general point P(x, y, z) on the line. This is the variable part of the equation.
$vec{a}$	Position vector of a specific known point A($x_1, y_1, z_1$) through which the line passes.
$vec{b}$	Direction vector of the line. This vector is parallel to the line. Its components are proportional to the direction ratios.
$lambda$	A scalar parameter. As $lambda$ varies, $vec{r}$ traces out different points on the line. It can be any real number ($lambda in mathbb{R}$).

Derivation:
Since point P lies on the line passing through A and is parallel to $vec{b}$, the vector $vec{AP}$ must be parallel to $vec{b}$.
This means $vec{AP}$ can be expressed as a scalar multiple of $vec{b}$.
So, $vec{AP} = lambda vec{b}$, where $lambda$ is a scalar.
We know that $vec{AP} = vec{r} - vec{a}$.
Therefore, $vec{r} - vec{a} = lambda vec{b}$.
Rearranging this, we get the vector equation of the line:

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

This equation is paramount for describing a line in 3D space.

Example 1:
Find the vector equation of the line passing through the point $(1, 2, -3)$ and parallel to the vector $3hat{i} - 2hat{j} + 5hat{k}$.

Solution:
Here, the fixed point A has position vector $vec{a} = hat{i} + 2hat{j} - 3hat{k}$.
The direction vector is $vec{b} = 3hat{i} - 2hat{j} + 5hat{k}$.
Using the formula $vec{r} = vec{a} + lambda vec{b}$, we get:

$vec{r} = (hat{i} + 2hat{j} - 3hat{k}) + lambda (3hat{i} - 2hat{j} + 5hat{k})$

This is the required vector equation of the line.

Case 2: Line Passing Through Two Given Points

What if you're given two points on the line instead of a point and a direction vector? No problem! We can easily deduce the direction vector.

* Let $A$ be a fixed point with position vector $vec{a}$.
* Let $B$ be another fixed point with position vector $vec{b}$.
* Let $P$ be any arbitrary point on the line, with position vector $vec{r}$.

Derivation:
The line passes through points A and B. We can use one of these points, say A($vec{a}$), as our fixed point.
The direction of the line is given by the vector connecting A to B, i.e., $vec{AB}$.
So, our direction vector $vec{b}_{direction} = vec{AB} = vec{b} - vec{a}$.
Now, substituting this into the general form $vec{r} = vec{a}_{fixed} + lambda vec{b}_{direction}$, we get:

$vec{r} = vec{a} + lambda (vec{b} - vec{a})$

This equation represents the line passing through two distinct points A and B.

Example 2:
Find the vector equation of the line passing through the points $A(1, 2, -1)$ and $B(2, -1, 3)$.

Solution:
Here, the position vectors of the two points are:
$vec{a} = hat{i} + 2hat{j} - hat{k}$
$vec{b} = 2hat{i} - hat{j} + 3hat{k}$

The direction vector of the line is $vec{b} - vec{a}$:
$vec{b} - vec{a} = (2hat{i} - hat{j} + 3hat{k}) - (hat{i} + 2hat{j} - hat{k})$
$vec{b} - vec{a} = (2-1)hat{i} + (-1-2)hat{j} + (3-(-1))hat{k}$
$vec{b} - vec{a} = hat{i} - 3hat{j} + 4hat{k}$

Using the formula $vec{r} = vec{a} + lambda (vec{b} - vec{a})$, we get:

$vec{r} = (hat{i} + 2hat{j} - hat{k}) + lambda (hat{i} - 3hat{j} + 4hat{k})$

This is the required vector equation of the line.

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II. Equation of a Line in Cartesian Form

While the vector form is elegant, the Cartesian form is often more practical for computations involving coordinates. It's simply an expansion of the vector form using components.

Recall that:
* $vec{r} = xhat{i} + yhat{j} + zhat{k}$ (position vector of any point P(x,y,z) on the line)
* $vec{a} = x_1hat{i} + y_1hat{j} + z_1hat{k}$ (position vector of the known point A($x_1, y_1, z_1$))
* $vec{b} = ahat{i} + bhat{j} + chat{k}$ (direction vector, where $a, b, c$ are the direction ratios of the line)

Case 1: Line Passing Through a Given Point $(x_1, y_1, z_1)$ and Having Direction Ratios $(a, b, c)$

Derivation from Vector Form:
Start with the vector form: $vec{r} = vec{a} + lambda vec{b}$
Substitute the component forms:
$xhat{i} + yhat{j} + zhat{k} = (x_1hat{i} + y_1hat{j} + z_1hat{k}) + lambda (ahat{i} + bhat{j} + chat{k})$
Group terms by $hat{i}, hat{j}, hat{k}$:
$xhat{i} + yhat{j} + zhat{k} = (x_1 + lambda a)hat{i} + (y_1 + lambda b)hat{j} + (z_1 + lambda c)hat{k}$

Equating the coefficients of $hat{i}, hat{j}, hat{k}$:
$x = x_1 + lambda a implies lambda = frac{x - x_1}{a}$
$y = y_1 + lambda b implies lambda = frac{y - y_1}{b}$
$z = z_1 + lambda c implies lambda = frac{z - z_1}{c}$

Since $lambda$ is the same for all three equations, we can equate them to get the Cartesian equation:

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

Here, $(x_1, y_1, z_1)$ is a point on the line, and $(a, b, c)$ are the direction ratios (DRs) of the line. The direction ratios are any set of numbers proportional to the components of the direction vector. If $(l, m, n)$ are direction cosines (DCs) of the line, then $a, b, c$ can be taken as $l, m, n$ respectively. Remember, $l^2+m^2+n^2=1$, while $a^2+b^2+c^2$ can be any positive number. Direction cosines are essentially normalized direction ratios.

JEE Focus: Parametric Form
From the steps above, we obtained:
$x = x_1 + lambda a$
$y = y_1 + lambda b$
$z = z_1 + lambda c$
This is known as the parametric form of the line. Any point on the line can be represented as $(x_1 + lambda a, y_1 + lambda b, z_1 + lambda c)$ for some real value of $lambda$. This form is extremely useful in JEE problems, especially when you need to find the coordinates of a specific point on the line (e.g., foot of perpendicular, intersection with a plane).

Example 3:
Find the Cartesian equation of the line passing through $(1, 2, -3)$ and having direction ratios $(3, -2, 5)$.

Solution:
Given point $(x_1, y_1, z_1) = (1, 2, -3)$.
Given direction ratios $(a, b, c) = (3, -2, 5)$.
Using the formula $frac{x - x_1}{a} = frac{y - y_1}{b} = frac{z - z_1}{c}$:

$frac{x - 1}{3} = frac{y - 2}{-2} = frac{z - (-3)}{5}$

$frac{x - 1}{3} = frac{y - 2}{-2} = frac{z + 3}{5}$

This is the required Cartesian equation of the line.

Case 2: Line Passing Through Two Given Points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$

Derivation from Vector Form:
We established the vector form as $vec{r} = vec{a} + lambda (vec{b} - vec{a})$.
Let $vec{a} = x_1hat{i} + y_1hat{j} + z_1hat{k}$ and $vec{b} = x_2hat{i} + y_2hat{j} + z_2hat{k}$.
Then the direction vector components are $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$. These become our direction ratios $(a, b, c)$.
So, $a = x_2 - x_1$, $b = y_2 - y_1$, $c = z_2 - z_1$.
Substituting these into the Cartesian form from Case 1:

$frac{x - x_1}{x_2 - x_1} = frac{y - y_1}{y_2 - y_1} = frac{z - z_1}{z_2 - z_1}$

This is the Cartesian equation of a line passing through two specified points.

Example 4:
Find the Cartesian equation of the line passing through the points $A(1, 2, -1)$ and $B(2, -1, 3)$.

Solution:
Given points: $(x_1, y_1, z_1) = (1, 2, -1)$ and $(x_2, y_2, z_2) = (2, -1, 3)$.
The direction ratios are:
$a = x_2 - x_1 = 2 - 1 = 1$
$b = y_2 - y_1 = -1 - 2 = -3$
$c = z_2 - z_1 = 3 - (-1) = 4$

Using the formula $frac{x - x_1}{x_2 - x_1} = frac{y - y_1}{y_2 - y_1} = frac{z - z_1}{z_2 - z_1}$:

$frac{x - 1}{1} = frac{y - 2}{-3} = frac{z - (-1)}{4}$

$frac{x - 1}{1} = frac{y - 2}{-3} = frac{z + 1}{4}$

This is the required Cartesian equation of the line.

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III. Special Cases and JEE Focus: Handling Zero Direction Ratios

What happens if one or more of the direction ratios $(a, b, c)$ are zero?
Consider the equation $frac{x - x_1}{a} = frac{y - y_1}{b} = frac{z - z_1}{c}$.
If, for example, $a = 0$, then the form $frac{x - x_1}{0}$ seems problematic. However, in the context of lines, it implies that the numerator must also be zero for the proportion to hold.
Key Interpretation: If a direction ratio is zero, it means the line is perpendicular to the corresponding axis (or parallel to the plane formed by the other two axes).
For example, if $a=0$, it means the direction vector $vec{b}$ has no $hat{i}$ component, i.e., $vec{b} = bhat{j} + chat{k}$. This line is parallel to the YZ-plane.
In the parametric form, $x = x_1 + lambda a$, if $a=0$, then $x = x_1$. This means all points on the line have the same x-coordinate $x_1$. The line lies entirely in the plane $x = x_1$.

So, if $a=0$, the Cartesian equation is written as:

$frac{x - x_1}{0} = frac{y - y_1}{b} = frac{z - z_1}{c}$, which implies $x = x_1$ and $frac{y - y_1}{b} = frac{z - z_1}{c}$

This represents the intersection of the plane $x=x_1$ and the plane defined by $frac{y - y_1}{b} = frac{z - z_1}{c}$.

Example with Zero Direction Ratio:
Find the Cartesian equation of a line passing through $(2, -1, 4)$ and parallel to the Z-axis.

Solution:
A line parallel to the Z-axis has direction ratios $(0, 0, 1)$ (or any non-zero multiple, e.g., $(0,0,k)$ where $k
eq 0$).
Point $(x_1, y_1, z_1) = (2, -1, 4)$.
Direction ratios $(a, b, c) = (0, 0, 1)$.
Using the form $frac{x - x_1}{a} = frac{y - y_1}{b} = frac{z - z_1}{c}$:
$frac{x - 2}{0} = frac{y - (-1)}{0} = frac{z - 4}{1}$
This implies:
$x - 2 = 0 implies x = 2$
$y - (-1) = 0 implies y = -1$
And $frac{z - 4}{1}$ remains as is.
So the equation of the line is defined by the two equations:

$x = 2, y = -1$

This makes perfect sense: it's a line where the x-coordinate is always 2 and the y-coordinate is always -1, allowing only the z-coordinate to vary, thus making it parallel to the Z-axis. This is an example of the non-symmetric form of a line, where it's expressed as the intersection of two planes ($x=2$ and $y=-1$).

JEE Advanced Perspective: Non-symmetric Form
A line can also be given as the intersection of two planes:
$A_1x + B_1y + C_1z + D_1 = 0$
$A_2x + B_2y + C_2z + D_2 = 0$
To convert this to the symmetric Cartesian form, you need to:
1. Find a point on the line (e.g., set one coordinate to zero and solve for the other two).
2. Find the direction vector: The direction vector of the line is perpendicular to the normal vectors of both planes. So, the direction vector $vec{b}$ can be found by taking the cross product of the normal vectors of the two planes: $vec{b} = vec{n_1} imes vec{n_2}$, where $vec{n_1} = A_1hat{i} + B_1hat{j} + C_1hat{k}$ and $vec{n_2} = A_2hat{i} + B_2hat{j} + C_2hat{k}$.

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Conclusion

Understanding the equation of a line in space is fundamental to 3D Geometry. Whether in vector form ($vec{r} = vec{a} + lambda vec{b}$) or Cartesian form ($frac{x - x_1}{a} = frac{y - y_1}{b} = frac{z - z_1}{c}$), the core idea remains the same: a line is defined by a point it passes through and its direction. Master these forms and their derivations, and you'll be well-equipped to tackle more complex problems involving lines, planes, and distances in 3D space, which are frequently tested in JEE. The parametric form is particularly crucial for finding coordinates of specific points on the line, a common technique in problem-solving. Always pay attention to special cases like zero direction ratios, as they often simplify to specific geometrical interpretations.

Ready to conquer "Equation of a line in space"? Here are some quick mnemonics and shortcuts to help you remember the crucial forms for your JEE and board exams.

🚀 Mnemonics & Shortcuts for Equation of a Line in Space 🚀

Remembering the different forms of a line's equation can be tricky. These mnemonics will simplify recall, especially under exam pressure.

• 
  General Tip for JEE & CBSE: Always understand the fundamental idea: A line needs a point it passes through and a direction vector. All forms are variations of this core concept.
  

1. Line through a Point and Parallel to a Vector

This is the most fundamental form.

• 
  Vector Form:
  
  $vec{r} = vec{a} + lambdavec{b}$
  
  
  
  • Mnemonic: "Reach Anywhere by a Lambda-jump in B-direction."
    
    
    
    • $vec{r}$: position vector of any point on the line.
    
    
    • $vec{a}$: position vector of the fixed point the line passes through.
    
    
    • $lambda$: scalar parameter (any real number).
    
    
    • $vec{b}$: direction vector of the line (parallel vector).
    
    
    
    
  
  
  • Shortcut Focus: Think of $vec{a}$ as your starting point, and $lambdavec{b}$ as how far you travel along the line's direction.
  
  
  
  


• 
  Cartesian Form:
  
  $frac{x - x_1}{a} = frac{y - y_1}{b} = frac{z - z_1}{c}$
  
  
  
  • Mnemonic: "XYZ minus Point, All Over Direction."
    
    
    
    • Numerator: $(x-x_1)$, $(y-y_1)$, $(z-z_1)$ – always subtract the coordinates of the known point.
    
    
    • Denominator: $a, b, c$ – these are the direction ratios of the line (components of the direction vector $vec{b}$).
    
    
    
    
  
  
  • Shortcut Focus: Keep the "minus point" in the numerator and "direction ratios" in the denominator clearly distinct. All three fractions must be equal.
  
  
  
  

2. Line Passing Through Two Points

Here, the direction vector is derived from the two given points.

• 
  Vector Form:
  
  $vec{r} = vec{a} + lambda(vec{b} - vec{a})$
  
  
  
  • Mnemonic: "Result starts at 'A', then takes a 'Lambda-Jump' along the path from 'A' to 'B' (B-A)."
    
    
    
    • $vec{a}$: position vector of the first known point.
    
    
    • $vec{b}$: position vector of the second known point.
    
    
    • $(vec{b} - vec{a})$: This term represents the direction vector of the line, which is the vector connecting the two points.
    
    
    
    
  
  
  • Shortcut Focus: The direction vector is simply the difference between the position vectors of the two points. You can also use $vec{r} = vec{b} + lambda(vec{a} - vec{b})$.
  
  
  
  


• 
  Cartesian Form:
  
  $frac{x - x_1}{x_2 - x_1} = frac{y - y_1}{y_2 - y_1} = frac{z - z_1}{z_2 - z_1}$
  
  
  
  • Mnemonic: "XYZ minus First Point, All Over Second Point minus First Point."
    
    
    
    • Numerator: $(x-x_1)$, $(y-y_1)$, $(z-z_1)$ – always subtract the coordinates of the *first* known point.
    
    
    • Denominator: $(x_2-x_1)$, $(y_2-y_1)$, $(z_2-z_1)$ – these are the direction ratios, found by subtracting the coordinates of the first point from the second point.
    
    
    
    
  
  
  • Shortcut Focus: This is a direct extension of the previous Cartesian form, where $(a,b,c)$ are replaced by $(x_2-x_1)$, $(y_2-y_1)$, $(z_2-z_1)$. Ensure consistency: if you use $(x-x_1)$ in the numerator, use $(x_2-x_1)$ in the denominator, not $(x_1-x_2)$.
  
  
  
  

---

Keep practicing these forms, and they will become second nature!

Here are some quick tips to master the 'Equation of a Line in Space' for your JEE Main and board exams:

• 
  Understand the Core Requirement: To uniquely define a line in 3D space, you always need two fundamental pieces of information:
  
  
  
  • A point that lies on the line.
  
  
  • A direction vector that is parallel to the line.
  
  
  
  


• 
  Vector Form: $vec{r} = vec{a} + lambda vec{b}$
  
  
  
  • Identify $vec{a}$ (Point Vector): This is the position vector of a known point on the line. For example, if a point is $(x_1, y_1, z_1)$, then $vec{a} = x_1hat{i} + y_1hat{j} + z_1hat{k}$.
  
  
  • Identify $vec{b}$ (Direction Vector): This vector is parallel to the line and defines its orientation in space.
  
  
  • Parameter $lambda$: This is a scalar parameter. Each value of $lambda$ corresponds to a unique point on the line. This is extremely useful for representing any generic point on the line.
  
  
  
  


• 
  Cartesian Form: $frac{x-x_1}{a} = frac{y-y_1}{b} = frac{z-z_1}{c}$
  
  
  
  • Point $(x_1, y_1, z_1)$ (on the line): You can directly read this point from the numerators (pay close attention to the signs!).
  
  
  • Direction Ratios $(a, b, c)$: These are the components of the direction vector $vec{b} = ahat{i} + bhat{j} + chat{k}$, found in the denominators.
  
  
  • Important Caution: If any direction ratio is zero (e.g., $a=0$), it means the line is parallel to the plane defined by the other two axes (e.g., YZ-plane if $a=0$). The Cartesian form is then written as $x-x_1=0$ (or $x=x_1$) and $frac{y-y_1}{b} = frac{z-z_1}{c}$. Do NOT write a zero in the denominator like $frac{x-x_1}{0}$.
  
  
  
  


• 
  Line Passing Through Two Points $vec{a}$ and $vec{d}$ (or $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$):
  
  
  
  • Vector Form: $vec{r} = vec{a} + lambda (vec{d} - vec{a})$
  
  
  • Cartesian Form: $frac{x-x_1}{x_2-x_1} = frac{y-y_1}{y_2-y_1} = frac{z-z_1}{z_2-z_1}$
  
  
  • Tip: In this case, the direction vector is simply the vector connecting the two given points, i.e., $(vec{d}-vec{a})$ or $(x_2-x_1)hat{i} + (y_2-y_1)hat{j} + (z_2-z_1)hat{k}$.
  
  
  
  


• 
  JEE Specific Strategy: Parameterization is Key!
  
  
  
  • For any problem involving finding a specific point on a line (e.g., intersection with a plane, foot of perpendicular from an external point, image of a point), always represent a generic point on the line using the parameter $lambda$.
  
  
  • If $frac{x-x_1}{a} = frac{y-y_1}{b} = frac{z-z_1}{c} = lambda$, then a generic point on the line is $(x_1+lambda a, y_1+lambda b, z_1+lambda c)$. This is your go-to representation for all calculations.
  
  
  
  


• 
  Relationship between Direction Ratios and Direction Cosines:
  
  
  
  • Direction Ratios $(a, b, c)$ can be any set of numbers proportional to the actual direction cosines.
  
  
  • Direction Cosines $(l, m, n)$ are unique for a given direction and satisfy the condition $l^2+m^2+n^2=1$.
  
  
  • $l = frac{a}{sqrt{a^2+b^2+c^2}}$, $m = frac{b}{sqrt{a^2+b^2+c^2}}$, $n = frac{c}{sqrt{a^2+b^2+c^2}}$.
  
  
  • Tip: When calculating angles between lines, using the dot product of their direction vectors (or direction cosines) is the most straightforward method.
  
  
  
  


• 
  CBSE vs. JEE Focus:
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  

  Aspect	CBSE Boards	JEE Main
  Formulas	Direct application and inter-conversion.	Application, understanding underlying concepts.
  Problem Solving	Focus on writing equations, finding basic properties (e.g., angle between lines).	Extensive use of parameterization for complex problems (e.g., finding foot of perpendicular, image, intersection points, shortest distance between skew lines).
  Key Skill	Ability to recall and apply formulas correctly.	Mastery of parameterization to represent any point on the line and solve further conditions.

  
  

Keep practicing problems involving both forms and their applications to build confidence!

While the study of lines in 3D space might seem abstract, its principles are fundamental to understanding and modeling the physical world around us. From designing complex machinery to simulating virtual environments, the ability to represent and manipulate lines in three dimensions is crucial. Understanding the equation of a line in space, both in vector and Cartesian form, provides a powerful mathematical tool for a wide array of practical applications.

Here are some key real-world applications:

• Computer Graphics and Animation:
  
  In 3D computer graphics, every object is composed of vertices connected by edges (lines). The equation of a line is used extensively for:
  
  
  
  • Rendering: Determining the path of light rays (ray tracing) from a light source to a surface and then to the camera.
  
  
  • Collision Detection: Checking if a moving object's path (a line segment) intersects with another object.
  
  
  • Camera Movement: Defining the line of sight for a virtual camera as it moves through a scene.
  
  
  • Animation: Guiding the movement of characters or objects along specific linear paths.
  
  
  
  


• Robotics and Automation:
  
  Robots often need to perform tasks that involve linear movements. The equation of a line is vital for:
  
  
  
  • Path Planning: Defining the precise straight-line trajectory for a robot arm's end-effector (gripper, tool) to move from one point to another without collision.
  
  
  • Assembly Lines: Guiding automated machinery to move components along a straight path for assembly.
  
  
  • Navigation: Autonomous vehicles use linear path segments to navigate their environment.
  
  
  
  


• Aerospace and Satellite Technology:
  
  Although celestial bodies and projectiles often follow curved trajectories, straight-line approximations are essential for short durations or specific calculations:
  
  
  
  • Flight Paths: In air traffic control, flight paths are often approximated by straight lines between waypoints for planning and communication.
  
  
  • Satellite Tracking: Predicting the line of sight between a ground station and a satellite, or determining potential collision courses for space debris.
  
  
  • Missile Trajectories: Initial ballistic paths or terminal guidance often involve linear estimations.
  
  
  
  


• Civil Engineering and Architecture:
  
  The design and construction of structures heavily rely on 3D geometry:
  
  
  
  • Structural Design: Representing beams, columns, and trusses as lines in 3D space to analyze stress, strain, and stability.
  
  
  • Surveying: Establishing baselines and alignments for roads, bridges, and tunnels using coordinates and linear equations.
  
  
  • Building Information Modeling (BIM): All structural elements are defined by their geometric properties, often including linear equations for edges and axes.
  
  
  
  


• Physics and Engineering Mechanics:
  
  In many physical scenarios, the motion or arrangement of objects can be simplified to lines:
  
  
  
  • Projectile Motion: While overall path is parabolic, the instantaneous velocity vector is a line tangent to the path. In a vacuum, and without gravity, motion is purely linear.
  
  
  • Force Vectors: Representing forces acting on objects as vectors along specific lines in space.
  
  
  • Light Propagation: In homogeneous media, light travels in straight lines, which can be modeled using the equation of a line.
  
  
  
  

The ability to formulate and interpret the equation of a line in space is not just a theoretical exercise for competitive exams like JEE; it's a foundational skill for innovation in diverse scientific and engineering fields. Master these concepts not just for scores, but for their widespread utility in the real world!

Understanding abstract mathematical concepts like the 'Equation of a line in space' can be significantly enhanced by drawing parallels with simpler, more familiar ideas. Analogies help bridge the gap between known concepts and new, complex ones, making them more intuitive and memorable for exams like JEE Main.

---

Common Analogies for Lines in 3D Space

• 
  Line in 2D vs. Line in 3D:
  
  
  
  • 2D Analogy: Think of a line on a piece of paper. To define it, you typically need a point on the line and its slope (which tells you its direction).
  
  
  • 3D Concept: Similarly, in 3D space, to define a line, you need a point that lies on it and a direction vector. The direction vector serves the same purpose as the slope, indicating the orientation of the line in space. Just as a 2D line extends infinitely in two directions, a 3D line also extends infinitely.
  
  
  
  


• 
  Position Vector as a GPS Location:
  
  
  
  • Analogy: A position vector (a or r) for a point in 3D space is like a GPS coordinate or a unique address. It tells you exactly where that point is relative to a fixed origin (your "home base" or "start point").
  
  
  • 3D Concept: When we write the equation of a line as r = a + λb, a is the position vector of a known point on the line – it's your starting GPS coordinate from which you begin tracing the line.
  
  
  
  


• 
  Direction Vector as a Compass Bearing or Trajectory:
  
  
  
  • Analogy: The direction vector (b) is like the compass bearing or the direction an arrow is shot. It tells you 'which way' the line is going from that starting point. If you were walking, it's the specific path you're instructed to follow (e.g., "walk 2 units east, 3 units north, and 1 unit up").
  
  
  • 3D Concept: In r = a + λb, the vector b dictates the line's orientation. Any scalar multiple of b (e.g., 2b, -0.5b) would still represent the same line's direction, just like walking in the exact opposite direction still defines the same line path.
  
  
  
  


• 
  Parametric Form as a Journey:
  
  
  
  • Analogy: The parametric form of a line, r = a + λb, can be thought of as a continuous journey. Imagine you start at point a (your initial location). The variable λ (lambda) is like 'time' or a 'scalar multiplier' for your movement. The vector b is your 'velocity' or 'step'.
  
  
  • 3D Concept: As λ changes, your position vector r traces out all points on the line.
    
    
    
    • If λ = 0, you are at a.
    
    
    • If λ = 1, you move one 'step' in direction b from a.
    
    
    • If λ = -1, you move one 'step' in the opposite direction from a.
    
    
    
    Every value of λ gives you a different point on the line, just as every moment in time defines your position on a journey. This is particularly useful for JEE problems involving dynamic points or conditions on the line.
  
  
  
  


• 
  Skew Lines as Intersecting Flyovers/Underpasses:
  
  
  
  • Analogy: Skew lines are a uniquely 3D concept. Imagine two roads at different altitudes – one road is an overpass and the other is an underpass. They are not parallel (they are headed in different directions), and they definitely don't intersect (because they are at different heights). Yet, in 2D projection, they might appear to cross.
  
  
  • 3D Concept: This perfectly illustrates skew lines. They are lines in space that are neither parallel nor intersecting. This concept is crucial for JEE, especially for finding the shortest distance between two skew lines.
  
  
  
  

By relating these 3D geometric concepts to everyday experiences, you can build a stronger intuitive foundation, which is invaluable for solving complex problems efficiently in competitive exams.

To effectively grasp the concept of the "Equation of a line in space," a solid understanding of several fundamental topics is essential. These prerequisites form the building blocks upon which 3D geometry, particularly lines, is constructed. Mastering them will significantly ease the learning curve for this crucial JEE and CBSE topic.

🎯 Why are prerequisites important for this topic?

The equation of a line in space fundamentally relies on vector representations and directional properties. Without a strong foundation in these, understanding how these equations are derived and applied becomes challenging.

• 
  1. Vectors (Fundamental):
  

  A thorough understanding of vectors is the absolute cornerstone for 3D Geometry. The entire topic of lines in space is often first introduced and most intuitively understood using vector concepts.

  
  
  
  
  • 
    Definition of a Vector: Understanding what a vector represents (magnitude and direction) versus a scalar.
    
  
  
  • 
    Position Vector: Knowing how to represent a point in space as a position vector from the origin (e.g., $$
    
    r
    
    =
    x
    
    i
    
    +
    y
    
    j
    
    +
    z
    
    k
    
    for point (x,y,z)). This is crucial for defining 'a point on the line'.
    
  
  
  • 
    Direction Vector: Understanding that a vector can represent a direction in space, irrespective of its starting point. This is the most critical vector concept for lines, as a line's direction is defined by a parallel vector.
    
  
  
  • 
    Vector Operations: Proficiency in:
    
    
    
    • Scalar Multiplication: How multiplying a vector by a scalar changes its magnitude but not its direction (unless negative). Essential for parametric forms of a line.
    
    
    • Vector Addition/Subtraction: Used to combine position vectors and direction vectors to define points on the line.
    
    
    
    
  
  
  • 
    Magnitude of a Vector: Calculating $$
    |
    
    a
    
    |
    =
    
    
    
    x
    2
    
    +
    
    y
    2
    
    2
    
    +
    
    z
    2
    
    
    
    . This is needed for finding unit vectors and direction cosines.
    
  
  
  • 
    Unit Vector: A vector with magnitude 1, often used to solely represent direction.
    
  
  
  
  

  JEE & CBSE Relevance: Vectors form the backbone of all 3D geometry topics, including lines, planes, and more complex shortest distance problems. Strong vector skills are non-negotiable.

  
  


• 
  2. Direction Ratios (DRs) and Direction Cosines (DCs):
  

  These are the numerical representations of a vector's direction, fundamental for both vector and Cartesian forms of a line.

  
  
  
  
  • 
    Definition:
    
    
    
    • Direction Ratios (DRs): Any set of three numbers (a, b, c) that are proportional to the direction cosines of a line. They essentially give the components of a vector parallel to the line.
    
    
    • Direction Cosines (DCs): The cosines of the angles (α, β, γ) made by a line with the positive x, y, and z axes respectively (l = cosα, m = cosβ, n = cosγ). Remember l² + m² + n² = 1.
    
    
    
    
  
  
  • 
    Relationship between DRs and DCs: Knowing how to convert DRs to DCs and vice-versa. If (a, b, c) are DRs, then $$
    l
    =
    
    a
    
    
    
    a
    2
    
    +
    
    b
    2
    
    +
    
    c
    2
    
    
    
    
    ,
    m
    =
    
    b
    
    
    
    a
    2
    
    +
    
    b
    2
    
    +
    
    c
    2
    
    
    
    
    ,
    n
    =
    
    c
    
    
    
    a
    2
    
    +
    
    b
    2
    
    +
    
    c
    2
    
    
    
    
    .
    
  
  
  • 
    Finding DRs from Two Points: Given two points $$
    P
    (
    
    x
    1
    
    ,
    
    y
    1
    
    ,
    
    z
    1
    
    )
    and $$
    Q
    (
    
    x
    2
    
    ,
    
    y
    2
    
    ,
    
    z
    2
    
    )
    , the DRs of the line PQ are $$
    (
    
    x
    2
    
    −
    
    x
    1
    
    ,
    
    y
    2
    
    −
    
    y
    1
    
    ,
    
    z
    2
    
    −
    
    z
    1
    
    )
    . This directly translates to the two-point form of a line.
    
  
  
  
  

  JEE & CBSE Relevance: DRs and DCs are critical for both the vector and Cartesian forms of the line equation. They are frequently tested in various problem types.

  
  


• 
  3. Basic 3D Coordinate Geometry:
  

  A foundational understanding of the Cartesian coordinate system in three dimensions is necessary to visualize and interpret the equations.

  
  
  
  
  • 
    Points in Space: Understanding how (x, y, z) coordinates define a unique position.
    
  
  
  • 
    Axes and Planes: Familiarity with the x, y, z axes and the xy, yz, zx planes.
    
  
  
  
  

  JEE & CBSE Relevance: While seemingly basic, spatial visualization is key to conceptualizing lines and their relationships with other geometric entities in space.

  
  

💡 Quick Tip: If any of these concepts feel shaky, revisit your notes or textbook sections on Vectors and 3D Geometry basics before diving deep into the Equation of a Line. A strong foundation here will make your learning journey smoother and more successful!

Understanding the "Equation of a Line in Space" is foundational in 3D Geometry. Several other topics are intrinsically linked to this concept, often serving as prerequisites or direct applications. A strong grasp of these related areas is crucial for excelling in JEE Main and Board exams.

Related Topics

• Vector Algebra:
  
  
  
  • Relevance: The vector form of a line equation, $vec{r} = vec{a} + lambdavec{b}$, directly uses vector concepts. A thorough understanding of vector addition, scalar multiplication, dot product, and cross product is indispensable.
  
  
  • Application: These operations are vital for finding angles between lines, determining perpendicularity, calculating shortest distances, and solving various geometrical problems involving lines.
  
  
  
  


• Direction Ratios and Direction Cosines:
  
  
  
  • Relevance: These define the direction of a line in Cartesian coordinates. They are directly used to form the Cartesian equation of a line ($frac{x-x_1}{a} = frac{y-y_1}{b} = frac{z-z_1}{c}$).
  
  
  • Application: Understanding their relationship ($l^2+m^2+n^2=1$ for direction cosines) is key for finding angles between lines and for many distance-related problems.
  
  
  
  


• Equation of a Plane:
  
  
  
  • Relevance: Lines and planes are often studied together. A line can be the intersection of two planes, or a plane can be defined containing a given line and a point.
  
  
  • Application: Problems involving the intersection of a line and a plane, distance from a point to a plane, or finding a plane parallel/perpendicular to a line, all require knowledge of both line and plane equations.
  
  
  
  


• Shortest Distance Between Two Lines (Skew Lines):
  
  
  
  • Relevance: This is a direct and significant application of the equation of a line in space. The formula for the shortest distance between two skew lines relies heavily on their vector equations and the scalar triple product.
  
  
  • JEE Focus: This is a highly tested concept in JEE Main, requiring a deep understanding of line direction vectors and position vectors.
  
  
  
  


• Intersection of Lines and Line with a Plane:
  
  
  
  • Relevance: Finding the point of intersection between two lines (if they are coplanar and non-parallel) or between a line and a plane are common problem types.
  
  
  • Application: This involves solving a system of linear equations derived from the parametric forms of the line and/or plane equations.
  
  
  
  


• Projection of a Point/Line onto a Plane/Line:
  
  
  
  • Relevance: This concept involves finding the foot of the perpendicular from a point to a line or a plane, or the projection of a line segment onto another line or plane.
  
  
  • Application: It utilizes the direction vectors of lines and normal vectors of planes, requiring proficiency in dot products and vector projection formulas.
  
  
  
  

Mastering these interconnected topics will provide a holistic understanding of 3D Geometry, allowing you to tackle complex problems with confidence. Regularly practicing problems that integrate these concepts is highly recommended for both CBSE and JEE preparation.

Understanding the equation of a line in space is fundamental, but exams often set up traps to test your conceptual clarity and attention to detail. Being aware of these common pitfalls can significantly improve your accuracy and score.

• 
  Trap 1: Incorrectly Identifying Point and Direction Ratios from Cartesian Form
  
  
  
  • 
    The Mistake: Students frequently misinterpret the coordinates of the point `(x₁, y₁, z₁)` and the direction ratios `(a, b, c)` from the Cartesian equation `(x - x₁)/a = (y - y₁)/b = (z - z₁)/c`.
    
    
    Example: For `(x + 2)/3 = (y - 1)/4 = z/5`, many incorrectly identify the point as `(2, -1, 0)`.
    
    
    Another common error is with forms like `(2 - x)/1 = (y - 3)/2 = (z + 1)/4`, where the direction ratio for `x` is taken as `1`.
    
  
  
  • 
    The Fix: Always ensure the numerators are strictly in the format `(x - x₁)`, `(y - y₁)`, and `(z - z₁)`.
    
    
    
    • If you see `(x + 2)`, it means `(x - (-2))`, so `x₁ = -2`.
    
    
    • If you see `(2 - x)/1`, rewrite it as `(x - 2)/(-1)`. The direction ratio for `x` is then `-1`, not `1`.
    
    
    
    Tip: Always standardize the Cartesian form to `(x - x₁)/a = (y - y₁)/b = (z - z₁)/c` before extracting the point and direction ratios.
    
  
  
  
  



• 
  Trap 2: Confusing Position Vector 'a' and Direction Vector 'b' in Vector Form
  
  
  
  • 
    The Mistake: In the vector equation of a line, `r = a + λb`, students sometimes mix up the roles of vector `a` and vector `b`.
    
  
  
  • 
    The Fix:
    
    
    
    • `a` is the position vector of a specific point on the line (e.g., `x₁i + y₁j + z₁k`). It tells you where the line is located in space.
    
    
    • `b` is the vector parallel to the line (e.g., `ai + bj + ck`). It tells you the direction of the line.
    
    
    
    These two vectors are fundamentally different and serve distinct purposes. Understanding this distinction is crucial for problems involving parallelism, perpendicularity, and angles.
    
  
  
  
  



• 
  Trap 3: Improper Handling of Zero Direction Ratios in Cartesian Form
  
  
  
  • 
    The Mistake: When one or more direction ratios are zero, students may incorrectly write '0' in the denominator, which is mathematically undefined, or misinterpret the resulting equation.
    
    
    Example: A line through `(1, 2, 3)` parallel to `2i + 0j + 4k`. Incorrectly written as `(x-1)/2 = (y-2)/0 = (z-3)/4`.
    
  
  
  • 
    The Fix: If a direction ratio is zero, it implies that the corresponding coordinate is constant. Never write zero in the denominator.
    
    
    
    • If the direction vector is `(a, 0, c)`, the Cartesian equation will be `(x - x₁)/a = (z - z₁)/c` and `y = y₁`.
    
    
    • If the direction vector is `(0, b, 0)`, the equation is `x = x₁` and `z = z₁`, with `y` being variable. This represents a line parallel to the Y-axis.
    
    
    
    JEE & CBSE Alert: This is a common point of confusion. Always state the constant coordinate separately.
    
  
  
  
  



• 
  Trap 4: Algebraic Errors during Conversion or Parameterization
  
  
  
  • 
    The Mistake: Simple sign errors, misplacing components, or incorrect algebraic manipulations are frequent sources of mistakes when converting between vector and Cartesian forms, or when expressing a general point on a line using the parameter `λ`.
    
  
  
  • 
    The Fix:
    
    
    
    • When converting `r = a + λb` to Cartesian, remember `x = x₁ + λa`, `y = y₁ + λb`, `z = z₁ + λc`.
    
    
    • When converting `(x - x₁)/a = (y - y₁)/b = (z - z₁)/c = λ` to vector form, ensure that `a = x₁i + y₁j + z₁k` (the point) and `b = ai + bj + ck` (the direction) are correctly identified.
    
    
    
    Always double-check your signs and coefficients, especially when dealing with negative values or fractions.
    
  
  
  
  



• 
  Trap 5: Assuming a Line Passes Through the Origin by Default
  
  
  
  • 
    The Mistake: Unless explicitly stated or derived, students sometimes implicitly assume that if no specific point is mentioned, the line passes through the origin. This is often incorrect in 3D geometry.
    
  
  
  • 
    The Fix: A line always requires a point to fix its position. If a line passes through the origin, then the position vector `a` (or `(x₁, y₁, z₁)` in Cartesian form) will be `0i + 0j + 0k` or `(0, 0, 0)` respectively. Never assume the origin unless it's explicitly given or deducible from the problem statement.
    
  
  
  
  

By carefully avoiding these common traps, you can build a strong foundation for solving more complex problems involving lines in three dimensions.

Key Takeaways: Equation of a Line in Space

Understanding the equation of a line in three-dimensional space is fundamental for solving problems in 3D Geometry for both CBSE board exams and JEE Main. The core idea is that a line is uniquely determined if you know a point it passes through and its direction.

1. Vector Form of the Equation of a Line

The most general and often preferred form for JEE, as it's concise and algebraic manipulations are often simpler.

• 
  Line passing through a given point and parallel to a given vector:
  
  If a line passes through a point A with position vector $vec{a}$ and is parallel to a vector $vec{b}$, its equation is:
  
  
  

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

  
  where $vec{r}$ is the position vector of any arbitrary point P(x, y, z) on the line, and $lambda$ is a scalar parameter (any real number).
  


• 
  Line passing through two given points:
  
  If a line passes through two points A and B with position vectors $vec{a}$ and $vec{b}$ respectively, its equation is:
  
  
  

  $vec{r} = vec{a} + lambda (vec{b} - vec{a})$

  
  Here, the vector $(vec{b} - vec{a})$ gives the direction of the line.
  

2. Cartesian Form of the Equation of a Line

This form is often more intuitive for visualization and is frequently used in CBSE. It can be easily derived from the vector form.

• 
  Line passing through a given point $(x_1, y_1, z_1)$ and having direction ratios $(a, b, c)$:
  
  The equation of the line is:
  
  
  

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

  
  Here, $(a, b, c)$ are the direction ratios of the line, which are the components of the direction vector $vec{b} = ahat{i} + bhat{j} + chat{k}$.
  
  JEE Tip: Direction cosines $(l, m, n)$ can also be used, in which case $frac{x - x_1}{l} = frac{y - y_1}{m} = frac{z - z_1}{n}$. Remember that $(l, m, n)$ are directly proportional to $(a, b, c)$.
  


• 
  Line passing through two given points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:
  
  The equation of the line is:
  
  
  

  $frac{x - x_1}{x_2 - x_1} = frac{y - y_1}{y_2 - y_1} = frac{z - z_1}{z_2 - z_1}$

  
  Here, $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$ are the direction ratios of the line.
  

3. Crucial Points for JEE and CBSE

• 
  Interchangeability: You must be proficient in converting between vector and Cartesian forms, as problems may present information in one form and require answers in another.
  


• 
  Direction Ratios vs. Direction Cosines: While direction ratios $(a, b, c)$ can be any set of numbers proportional to the direction of the line, direction cosines $(l, m, n)$ are unique and satisfy $l^2 + m^2 + n^2 = 1$. For competitive exams, understanding their relationship is key. If $(a, b, c)$ are direction ratios, then $(l, m, n) = left( frac{a}{sqrt{a^2+b^2+c^2}}, frac{b}{sqrt{a^2+b^2+c^2}}, frac{c}{sqrt{a^2+b^2+c^2}}
  ight)$.
  


• 
  Finding a General Point on a Line: To find coordinates of any point on a line, equate the Cartesian form to $lambda$:
  
  $frac{x - x_1}{a} = frac{y - y_1}{b} = frac{z - z_1}{c} = lambda$
  
  This gives $x = x_1 + alambda$, $y = y_1 + blambda$, $z = z_1 + clambda$. This technique is invaluable for problems involving distances, intersections, and specific point properties.
  

Mastering these forms and their underlying principles will set a strong foundation for advanced topics in 3D geometry. Keep practicing conversion and application problems!

A systematic problem-solving approach is crucial for tackling questions on the equation of a line in 3D space, which is a frequently tested topic in both JEE Main and CBSE Board exams. Many problems involve interpreting given conditions to form the line equation, and then using this equation to find distances, angles, or points of intersection.

Key Information to Extract

Before attempting to solve, identify the following from the problem statement:

• A point on the line: (x₁, y₁, z₁) or its position vector $vec{a}$.


• Direction Ratios (DRs) of the line: (a, b, c) or a vector parallel to the line $vec{b}$.


• Two points on the line: If given, use them to find the DRs and then a point.

Common Problem Types & Approach

1. 
   Finding the Equation of a Line:
   
   
   
   • Given a point and parallel vector: Directly use the forms:
     
     
     
     • Vector form (JEE Focus): $vec{r} = vec{a} + lambda vec{b}$
     
     
     • Cartesian form (JEE & CBSE): $frac{x-x_1}{a} = frac{y-y_1}{b} = frac{z-z_1}{c}$
     
     
     
     
   
   
   • Given two points ($A(vec{a})$ and $B(vec{b})$):
     
     
     
     • The direction vector $vec{b}$ can be taken as $(vec{b} - vec{a})$.
     
     
     • Vector form: $vec{r} = vec{a} + lambda (vec{b} - vec{a})$
     
     
     • Cartesian form: $frac{x-x_1}{x_2-x_1} = frac{y-y_1}{y_2-y_1} = frac{z-z_1}{z_2-z_1}$
     
     
     
     
   
   
   
   


2. 
   Checking if a Point Lies on a Line:
   
   
   
   • Substitute the coordinates of the point into the Cartesian equation of the line. If all ratios are equal, the point lies on the line.
   
   
   • In vector form, if $vec{p}$ is the position vector of the point, then $vec{p} = vec{a} + lambda vec{b}$ for some scalar $lambda$. Equate components and check for consistency.
   
   
   
   


3. 
   Finding the Intersection of Two Lines:
   
   
   
   • Represent the general point on the first line using a parameter (e.g., $lambda_1$).
   
   
   • Represent the general point on the second line using another parameter (e.g., $lambda_2$).
   
   
   • Equate the corresponding coordinates (x, y, z). This gives three equations with two variables ($lambda_1, lambda_2$).
   
   
   • Solve any two equations for $lambda_1$ and $lambda_2$.
   
   
   • Substitute these values into the third equation. If it holds true, the lines intersect. The intersection point is found by substituting $lambda_1$ or $lambda_2$ back into its respective line equation.
   
   
   • JEE Tip: If the third equation does not hold, the lines are either parallel or skew.
   
   
   
   


4. 
   Angle Between Two Lines:
   
   
   
   • Identify the direction vectors ($vec{b_1}$ and $vec{b_2}$) of the two lines.
   
   
   • Use the dot product formula: $cos heta = frac{|vec{b_1} cdot vec{b_2}|}{|vec{b_1}| |vec{b_2}|}$. Remember to take the absolute value for the acute angle.
   
   
   
   


5. 
   Shortest Distance Between Skew Lines (JEE Important!):
   
   
   
   • Let the lines be $vec{r_1} = vec{a_1} + lambda vec{b_1}$ and $vec{r_2} = vec{a_2} + mu vec{b_2}$.
   
   
   • The shortest distance $d = frac{|(vec{a_2} - vec{a_1}) cdot (vec{b_1} imes vec{b_2})|}{|vec{b_1} imes vec{b_2}|}$.
   
   
   • Approach:
     
     
     
     1. Identify $vec{a_1}, vec{a_2}, vec{b_1}, vec{b_2}$.
     
     
     2. Calculate $(vec{a_2} - vec{a_1})$.
     
     
     3. Calculate the cross product $(vec{b_1} imes vec{b_2})$.
     
     
     4. Calculate the dot product of the results from steps 2 and 3.
     
     
     5. Calculate the magnitude of $(vec{b_1} imes vec{b_2})$.
     
     
     6. Substitute into the formula.
     
     
     
     
   
   
   
   

Example Problem & Solution Strategy

Problem: Find the shortest distance between the lines $frac{x+1}{7} = frac{y+1}{-6} = frac{z+1}{1}$ and $frac{x-3}{1} = frac{y-5}{-2} = frac{z-7}{1}$.

Strategy:

1. Convert to Vector Form:
   Line 1: $vec{a_1} = -hat{i} - hat{j} - hat{k}$, $vec{b_1} = 7hat{i} - 6hat{j} + hat{k}$
   Line 2: $vec{a_2} = 3hat{i} + 5hat{j} + 7hat{k}$, $vec{b_2} = hat{i} - 2hat{j} + hat{k}$
   


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


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


4. Calculate $(vec{a_2} - vec{a_1}) cdot (vec{b_1} imes vec{b_2})$:
   $(4hat{i} + 6hat{j} + 8hat{k}) cdot (-4hat{i} - 6hat{j} - 8hat{k})$
   $= (4)(-4) + (6)(-6) + (8)(-8) = -16 - 36 - 64 = -116$
   


5. Calculate $|vec{b_1} imes vec{b_2}|$:
   $|-4hat{i} - 6hat{j} - 8hat{k}| = sqrt{(-4)^2 + (-6)^2 + (-8)^2} = sqrt{16 + 36 + 64} = sqrt{116}$
   


6. Apply the Shortest Distance Formula:
   $d = frac{|-116|}{sqrt{116}} = frac{116}{sqrt{116}} = sqrt{116}$ units.
   

By following these structured steps, you can confidently approach and solve a wide range of problems involving lines in 3D space. Practice converting between vector and Cartesian forms as it's a common requirement.

For CBSE board examinations, the topic of "Equation of a Line in Space" is fundamental, with a strong emphasis on understanding both its vector and Cartesian forms, as well as their interconversion and direct application. Unlike JEE, which often delves into complex manipulations and multi-concept problems, CBSE primarily assesses your ability to recall formulas, apply them accurately, and solve standard problem types.

Key Forms and Concepts for CBSE

• Equation of a Line Passing Through a Given Point and Parallel to a Given Vector:
  
  
  
  • Vector Form: The equation of a line passing through a point with position vector a and parallel to a vector b is given by $mathbf{r} = mathbf{a} + lambda mathbf{b}$, where $lambda$ is a scalar parameter.
  
  
  • Cartesian Form: If the point is $P(x_1, y_1, z_1)$ and the direction ratios of the line are $a, b, c$, then the equation is $frac{x - x_1}{a} = frac{y - y_1}{b} = frac{z - z_1}{c}$.
    
    CBSE Focus: Be proficient in converting between these two forms and identifying the components (point and direction vector/ratios) from a given equation.
  
  
  
  


• Equation of a Line Passing Through Two Given Points:
  
  
  
  • Vector Form: If the line passes through points with position vectors a and b, its equation is $mathbf{r} = mathbf{a} + lambda (mathbf{b} - mathbf{a})$.
  
  
  • Cartesian Form: If the points are $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, the equation is $frac{x - x_1}{x_2 - x_1} = frac{y - y_1}{y_2 - y_1} = frac{z - z_1}{z_2 - z_1}$.
    
    CBSE Focus: Direct application of these formulas is frequently tested.
  
  
  
  

Standard Problems for CBSE

CBSE questions often revolve around these core problem types:

• Finding Direction Ratios and Direction Cosines: From the Cartesian form, the denominators are the direction ratios. Direction cosines are derived by dividing each direction ratio by the magnitude of the direction vector.


• Angle Between Two Lines:
  
  
  
  • If two lines have direction vectors b$_1$ and b$_2$, the cosine of the angle $ heta$ between them is given by $cos heta = left| frac{mathbf{b}_1 cdot mathbf{b}_2}{|mathbf{b}_1| |mathbf{b}_2|}
    ight|$.
  
  
  • In Cartesian form, with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$, $cos heta = left| frac{a_1a_2 + b_1b_2 + c_1c_2}{sqrt{a_1^2+b_1^2+c_1^2} sqrt{a_2^2+b_2^2+c_2^2}}
    ight|$.
  
  
  • Conditions: For perpendicular lines, $mathbf{b}_1 cdot mathbf{b}_2 = 0$. For parallel lines, $mathbf{b}_1 = k mathbf{b}_2$.
  
  
  
  


• Shortest Distance Between Two Skew Lines:
  
  
  
  • This is a very important topic for CBSE, often involving detailed calculation. If the lines are $mathbf{r}_1 = mathbf{a}_1 + lambda mathbf{b}_1$ and $mathbf{r}_2 = mathbf{a}_2 + mu mathbf{b}_2$, the shortest distance $d$ is given by $d = left| frac{(mathbf{b}_1 imes mathbf{b}_2) cdot (mathbf{a}_2 - mathbf{a}_1)}{|mathbf{b}_1 imes mathbf{b}_2|}
    ight|$.
  
  
  • CBSE Focus: Memorize the formula and practice the vector cross product, dot product, and magnitude calculations. Questions may also ask for the equation of the line of shortest distance, though less common.
  
  
  
  


• Shortest Distance Between Parallel Lines:
  
  
  
  • If two lines are parallel, $mathbf{b}_1 = mathbf{b}_2 = mathbf{b}$. The distance is $d = left| frac{mathbf{b} imes (mathbf{a}_2 - mathbf{a}_1)}{|mathbf{b}|}
    ight|$.
  
  
  
  

CBSE vs. JEE Approach

For CBSE, the emphasis is on direct application of formulas, step-by-step calculations, and clear presentation of solutions. While JEE might test your conceptual depth by embedding these concepts in more complex problems (e.g., finding the image of a point in a line, or properties involving multiple lines/planes), CBSE typically keeps it straightforward. Understanding the derivation of formulas, especially for shortest distance, enhances conceptual clarity, which is valued in CBSE.

Study Tip: Practice a wide variety of problems from your NCERT textbook and exemplars. Focus on correctly identifying the given information (point, direction vector/ratios) and applying the appropriate formula.

The concept of the equation of a line in three-dimensional space is fundamental for JEE Main and Advanced, forming the basis for understanding planes, shortest distances, and various geometric problems. A strong grasp of both vector and Cartesian forms is essential.

1. Equation of a Line Passing Through a Given Point and Parallel to a Given Vector

This is the most basic form and the foundation for many problems.

• 
  Vector Form:
  

  The equation of a line passing through a point with position vector $vec{a}$ and parallel to a vector $vec{b}$ is given by:

  
  

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

  
  

  Here, $vec{r}$ is the position vector of any point on the line, and $lambda$ is a scalar parameter (any real number).

  
  
  
  
  • $vec{a} = x_1hat{i} + y_1hat{j} + z_1hat{k}$ (Position vector of the given point $(x_1, y_1, z_1)$)
  
  
  • $vec{b} = ahat{i} + bhat{j} + chat{k}$ (Vector parallel to the line, where $a, b, c$ are its direction ratios)
  
  
  
  


• 
  Cartesian Form:
  

  If the line passes through $(x_1, y_1, z_1)$ and has direction ratios $a, b, c$, its equation is:

  
  

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

  
  

  Each ratio is equal to the parameter $lambda$ from the vector form. This form is particularly useful for finding coordinates of points on the line, e.g., $(x_1+lambda a, y_1+lambda b, z_1+lambda c)$.

  
  

2. Equation of a Line Passing Through Two Given Points

If a line passes through two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ with position vectors $vec{a}$ and $vec{b}$ respectively, its equation can be found by treating the vector $(vec{b} - vec{a})$ as the direction vector.

• 
  Vector Form:
  

  $vec{r} = vec{a} + lambda (vec{b} - vec{a})$

  
  


• 
  Cartesian Form:
  

  $frac{x-x_1}{x_2-x_1} = frac{y-y_1}{y_2-y_1} = frac{z-z_1}{z_2-z_1}$

  
  

  Here, $(x_2-x_1), (y_2-y_1), (z_2-z_1)$ act as the direction ratios of the line.

  
  

3. Direction Ratios (DRs) and Direction Cosines (DCs)

• Direction Ratios (DRs): Any set of three numbers $a, b, c$ that are proportional to the direction cosines of a line are called its DRs. These define the direction of the line.


• Direction Cosines (DCs): If a line makes angles $alpha, eta, gamma$ with the positive X, Y, Z axes respectively, then $cosalpha, coseta, cosgamma$ are its DCs, usually denoted as $l, m, n$. They satisfy $l^2 + m^2 + n^2 = 1$.


• Relationship: If $a, b, c$ are DRs, then DCs are $l = frac{a}{sqrt{a^2+b^2+c^2}}$, $m = frac{b}{sqrt{a^2+b^2+c^2}}$, $n = frac{c}{sqrt{a^2+b^2+c^2}}$.

JEE Focus Areas and Problem Types

Understanding the equation of a line is a prerequisite for these frequently tested concepts:

• Angle Between Two Lines: If two lines have direction vectors $vec{b_1}$ and $vec{b_2}$, the angle $ heta$ between them is given by $cos heta = frac{|vec{b_1} cdot vec{b_2}|}{|vec{b_1}| |vec{b_2}|}$. (In Cartesian: $cos heta = frac{|a_1a_2 + b_1b_2 + c_1c_2|}{sqrt{a_1^2+b_1^2+c_1^2}sqrt{a_2^2+b_2^2+c_2^2}}$).


• Condition for Perpendicular Lines: $vec{b_1} cdot vec{b_2} = 0 implies a_1a_2 + b_1b_2 + c_1c_2 = 0$.


• Condition for Parallel Lines: $vec{b_1} = k vec{b_2}$ for some scalar $k implies frac{a_1}{a_2} = frac{b_1}{b_2} = frac{c_1}{c_2}$.


• Intersection of Two Lines: Equate the general points of two lines (using different parameters, e.g., $lambda_1$ and $lambda_2$) and solve for the parameters. If a consistent solution exists, the lines intersect.


• Perpendicular Distance from a Point to a Line: This involves finding the foot of the perpendicular. Let the given point be $P(x_0, y_0, z_0)$ and the line be $vec{r} = vec{a} + lambda vec{b}$. Let $Q$ be the foot of the perpendicular on the line. Then $vec{PQ}$ must be perpendicular to $vec{b}$. This condition helps find $lambda$ for point $Q$. The distance is then $|vec{PQ}|$.


• Image of a Point with respect to a Line: Once the foot of the perpendicular $Q$ is found, the image $P'$ is such that $Q$ is the midpoint of $PP'$.


• Coplanarity of Two Lines: Two lines are coplanar if they are parallel or intersect. If non-parallel and non-intersecting, they are skew. (Formula for coplanarity involves a scalar triple product, linking to planes).

CBSE vs JEE: While CBSE focuses on direct application of formulas for line equations, JEE problems often combine these concepts with other topics like planes, vectors, and even properties of geometric figures, requiring a deeper conceptual understanding and problem-solving strategy.

Mastering the various forms of the line equation and their interconversions is crucial. Practice problems involving intersection, angles, and perpendicular distances extensively to solidify your understanding for JEE.