If the shortest distance between the lines (x-λ)/2 = (y-4)/3 = (z-3)/4 and (x-2)/4 = (y-4)/6 = (z-7)

Q: Wait, why can't I just use the shortest distance formula for skew lines? It should work for parallel lines too, right?

That is a fantastic question! While the skew line formula $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ is mathematically robust, it will lead you to a division by zero in the case of parallel lines because the cross product of parallel vectors is the zero vector. Parallel lines are a special, simpler case of skew lines. By recognizing the parallelism early, you save yourself from a mathematical dead end and use a much more elegant, direct approach.

Q: How do I know for sure if the lines are parallel? Is there a quick test?

Absolutely! The direction ratios of a line are the components of its direction vector. If you have two lines with direction vectors $\vec{b}_1$ and $\vec{b}_2$, they are parallel if and only if $\vec{b}_1 = k\vec{b}_2$ for some scalar $k$. In this problem, the ratios $(2, 3, 4)$ and $(4, 6, 8)$ clearly satisfy this, as $(4, 6, 8) = 2(2, 3, 4)$. Always check this ratio proportionality first; it is the key that unlocks the door to the simplest solution.

Q: I ended up with two values for $\lambda$. Did I make a mistake, or are both correct?

You did not make a mistake! Mathematically, both values are correct. The distance between two parallel lines is fixed, but the position of the lines depends on $\lambda$. There are two distinct positions for the first line that result in the same perpendicular distance from the second line—one on either side of the second line. Since this is a multiple-choice question, we simply check which of your two solutions appears in the options provided.

Question

If the shortest distance between the lines $\frac{x-\lambda}{2} = \frac{y-4}{3} = \frac{z-3}{4}$ and $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda$ is :

Options:

(A) -1

(B) $-\frac{13}{25}$

(C) $\frac{13}{25}$

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. Before you, two infinite lines stretch out into the darkness. At first glance, they look like any other pair of lines in space—perhaps they intersect, perhaps they are skew, or perhaps they are parallel.

In the world of JEE Advanced, the first step is always to be a detective. We do not rush into calculations; we observe. We look at the denominators of the line equations:

$$\frac{x-\lambda}{2} = \frac{y-4}{3} = \frac{z-3}{4} \quad 	ext{and} \quad \frac{x-2}{4} = \frac{y-4}{6} = \frac{z-7}{8}$$

The direction ratios are $(2, 3, 4)$ and $(4, 6, 8)$. Do you see the hidden symmetry? The second set is exactly twice the first.

This is our 'Aha!' moment. The lines are parallel. This realization is not just a shortcut; it is a fundamental shift in our perspective. We are no longer dealing with the complexity of skew lines; we are dealing with the elegant, constant distance between two parallel paths.

### Defining the Bridge

To find the distance between these parallel lines, we need to build a bridge. We identify a point on the first line, $A_1 = (\lambda, 4, 3)$, and a point on the second line, $A_2 = (2, 4, 7)$. These points are our anchors.

We then define the vector connecting them:
$$\vec{a}_2 - \vec{a}_1 = (2 - \lambda)\hat{i} + (4 - 4)\hat{j} + (7 - 3)\hat{k} = (2 - \lambda)\hat{i} + 0\hat{j} + 4\hat{k}$$

This vector is the displacement between our two anchors. Now, we have our common direction vector, $\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

The distance between parallel lines is the height of a parallelogram formed by the connecting vector and the direction vector. The formula is:

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

This is where the physics of the problem meets the algebra.

### The Algebraic Dance

Now, we perform the cross product. We set up the determinant:

$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2-\lambda & 0 & 4 \ 2 & 3 & 4 \end{vmatrix}$$

Expanding this, we get $\hat{i}(0 - 12) - \hat{j}(4(2-\lambda) - 8) + \hat{k}(3(2-\lambda) - 0)$. Simplifying this, we arrive at $-12\hat{i} + 4\lambda\hat{j} + (6 - 3\lambda)\hat{k}$.

This vector represents the area of the parallelogram in a sense, and its magnitude is the numerator of our distance formula. The magnitude is:

$$\sqrt{(-12)^2 + (4\lambda)^2 + (6-3\lambda)^2}$$

Expanding this, we get $\sqrt{144 + 16\lambda^2 + 36 - 36\lambda + 9\lambda^2}$, which simplifies to $\sqrt{25\lambda^2 - 36\lambda + 180}$.

### The Final Resolution

We know the distance $d$ is $\frac{13}{\sqrt{29}}$. The magnitude of our direction vector $\vec{b}$ is $\sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$.

Equating our expression to the given distance, we have:

$$\frac{\sqrt{25\lambda^2 - 36\lambda + 180}}{\sqrt{29}} = \frac{13}{\sqrt{29}}$$

The $\sqrt{29}$ cancels out, leaving us with $\sqrt{25\lambda^2 - 36\lambda + 180} = 13$. Squaring both sides gives:

$$25\lambda^2 - 36\lambda + 180 = 169 \quad \Rightarrow \quad 25\lambda^2 - 36\lambda + 11 = 0$$

Factoring this quadratic, we find $(25\lambda - 11)(\lambda - 1) = 0$. Thus, $\lambda = 1$ or $\lambda = \frac{11}{25}$.

Looking at our options, **$\lambda = 1$** is the clear winner. You have navigated the 3D space, identified the symmetry, and solved the equation with precision. This is the essence of JEE Advanced—not just solving, but understanding the geometry behind the numbers.

Animated Solution for Mathematics - Three Dimensional Geometry: If the shortest distance between the lines $\frac{x - λ}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{29}$ , then a value of $λ$ is :

Select Answer:

Visualized Solution

Direction Ratios

Points and Direction Vector

Connecting Vector $a_{2} - a_{1}$

Distance Formula for Parallel Lines

Cross Product Setup

Expanding the Determinant

Magnitude of the Cross Product

Magnitude of Vector $b$

Equating to Given Distance

Forming the Quadratic Equation

Solving for $λ$

Final Answer

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

Defining the Bridge

The Algebraic Dance

The Final Resolution

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