Consider the lines L1 and L2 given by L1: (x-1)/2 = (y-3)/1 = (z-2)/2, L2: (x-2)/1 = (y-2)/2 = (z-3)

Q: Wait, why do we use two different parameters, $\lambda$ and $\mu$? Can't we just use one?

That is a fantastic question! If we used the same parameter for both lines, we would be forcing the points $P$ and $Q$ to be at the same 'relative' position along their respective lines, which is not necessarily true. By using $\lambda$ and $\mu$, we allow $P$ and $Q$ to be any independent points on their lines. We then let the algebra determine the specific values of $\lambda$ and $\mu$ that satisfy the condition of being connected by the transversal $L_3$.

Q: Why can't we just use the shortest distance formula between two skew lines?

The shortest distance formula is specifically for the unique line segment that is perpendicular to both skew lines. In this problem, $L_3$ is a specific transversal with a given direction $(1, -1, -2)$. It is highly unlikely that this specific line is the one representing the shortest distance. We must solve for the intersection points $P$ and $Q$ directly using the given direction ratios.

Q: How do we know the direction ratios of $PQ$ are proportional to $(1, -1, -2)$?

Because the problem states that the line $L_3$ intersects $L_1$ at $P$ and $L_2$ at $Q$. This means the line segment $PQ$ is a segment of the line $L_3$. Since $L_3$ has the direction ratios $(1, -1, -2)$, any segment lying on $L_3$ must also have direction ratios proportional to $(1, -1, -2)$. It is a fundamental property of lines in 3D space!

Question

Consider the lines $L_1$ and $L_2$ given by $L_1: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2}$, $L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}$. A line $L_3$ having direction ratios $1, -1, -2$ intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively. Then the length of line segment $PQ$ is

Options:

(A) $2\sqrt{6}$

(B) $3\sqrt{2}$

(C) $4\sqrt{3}$

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast 3D coordinate system. You have two lines, $L_1$ and $L_2$, floating in space like two non-intersecting, non-parallel paths—these are what we call skew lines.

They are like two airplanes flying at different altitudes, never crossing, yet perfectly defined by their equations. Now, a third line, $L_3$, acts as a bridge, a transversal that connects these two paths. Our mission is to find the length of this bridge, the segment $PQ$.

### The GPS of 3D Lines: Parametrization

To find the length of $PQ$, we first need to know exactly where $P$ and $Q$ are. In 3D geometry, we use parameters to act as a GPS for any point on a line.

For $L_1$, we set its equation equal to a scalar $\lambda$. This allows us to write the coordinates of any point $P$ as $P = (2\lambda+1, \lambda+3, 2\lambda+2)$.

Similarly, for $L_2$, we use a different parameter, $\mu$, because $Q$ is independent of $P$. This gives us $Q = (\mu+2, 2\mu+2, 3\mu+3)$. Now, we have the coordinates of our two mystery points, locked in the language of algebra.

### The Vector Bridge

The segment $PQ$ is a vector connecting these two points. Its direction ratios are simply the differences in their coordinates:

$$DR_x = (2\lambda+1) - (\mu+2) = 2\lambda - \mu - 1$$
$$DR_y = (\lambda+3) - (2\mu+2) = \lambda - 2\mu + 1$$
$$DR_z = (2\lambda+2) - (3\mu+3) = 2\lambda - 3\mu - 1$$

This vector represents the bridge itself.

### The Proportionality Constraint: The Secret Sauce

Here is the moment of truth. We know that $PQ$ lies on $L_3$, and $L_3$ has a fixed direction $(1, -1, -2)$. In the world of vectors, if two lines are parallel, their direction ratios must be proportional.

This gives us the beautiful equality:

$$\frac{2\lambda - \mu - 1}{1} = \frac{\lambda - 2\mu + 1}{-1} = \frac{2\lambda - 3\mu - 1}{-2}$$

This is the key that unlocks the door. By solving the first two parts, we find that $\lambda = \mu$. Substituting this back into the first and third parts, we discover that $\lambda = 3$ and $\mu = 3$.

### The Final Triumph

With $\lambda = 3$ and $\mu = 3$, we can finally pinpoint $P$ and $Q$. Substituting these values back into our parametric equations, we get $P = (7, 6, 8)$ and $Q = (5, 8, 12)$.

The distance between these two points is the final step of our journey. Using the 3D distance formula:

$$PQ = \sqrt{(7-5)^2 + (6-8)^2 + (8-12)^2}$$

This simplifies to:

$$PQ = \sqrt{2^2 + (-2)^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24}$$

Simplifying this, we arrive at our final answer: **$2\sqrt{6}$**. You have successfully navigated the 3D space, bridged the skew lines, and calculated the exact distance. This is the elegance of JEE Advanced mathematics—taking a complex spatial problem and reducing it to a series of logical, beautiful steps.

Select Answer:

Visualized Solution

Visualizing the 3D Lines

Parametric Form of Point $P$

Parametric Form of Point $Q$

Direction Ratios of Segment $P Q$

Applying Proportionality Condition

Solving for $λ$ and $μ$ (Part 1)

Finding the Value of $λ$

Exact Coordinates of $P$ and $Q$

Distance Formula Setup

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The GPS of 3D Lines

The Vector Bridge

The Proportionality Constraint

The Final Triumph

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