If the shortest distance between the straight lines 3(x - 1) = 6(y - 2) = 2(z - 1) and 4(x - 2) = 2(

Q: Wait, why do we divide by the LCM in the first step? Can't I just use the coefficients as they are?

That is a great question! The standard symmetric form of a line, $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, requires the coefficients of $x, y,$ and $z$ to be exactly $1$. If they are not, the denominators do not correctly represent the direction vector of the line. Dividing by the LCM is the cleanest way to normalize these coefficients to $1$ without changing the line itself.

Q: How do I know if the lines are skew or intersecting? Does the formula change?

If the lines were intersecting, the shortest distance $d$ would be $0$. In that case, the numerator of our formula, $|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|$, would equal $0$. Since we are given a non-zero distance, we know for certain they are skew. The formula works for both cases, but it is specifically designed to handle the skew scenario.

Q: Why do we take the absolute value in the numerator? Does distance have to be positive?

Absolutely! Distance is a physical quantity and must always be non-negative. The dot product can result in a negative value depending on the order of vectors or the direction of the cross product. The absolute value ensures that our final distance calculation remains physically meaningful.

Question

If the shortest distance between the straight lines $3(x - 1) = 6(y - 2) = 2(z - 1)$ and $4(x - 2) = 2(y - \lambda) = (z - 3), \lambda \in \mathbb{R}$ is $\frac{1}{\sqrt{38}}$, then the integral value of $\lambda$ is equal to :

Options:

(A) 3

(B) 2

(C) 5

(D) -1

Sigma Percentile · Accepted Answer

### The Geometry of Skew Lines: A 3D Odyssey

Imagine you are standing in a vast, three-dimensional void. Two straight lines are flying through this space. They don't touch, and they aren't parallel. These are what we call **skew lines**.

They are like two ships passing in the night, separated by a specific, unbridgeable gap. Our mission today is to find the length of that gap—the shortest perpendicular bridge connecting these two lines. This is a classic JEE Advanced challenge, and it is a beautiful exercise in vector algebra.

### Phase 1: Standardizing the Chaos

Before we can perform any calculations, we must bring order to our equations. We are given two lines:

$$L_1: 3(x - 1) = 6(y - 2) = 2(z - 1)$$
$$L_2: 4(x - 2) = 2(y - \lambda) = (z - 3)$$

In their current state, they are messy. The coefficients of $x, y,$ and $z$ are not $1$. To use the standard distance formula, we must convert these into the symmetric form.

For $L_1$, we divide by the least common multiple of $3, 6,$ and $2$, which is $6$. This transforms $L_1$ into:

$$L_1: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$$

We repeat this for $L_2$, dividing by the least common multiple of $4, 2,$ and $1$, which is $4$:

$$L_2: \frac{x-2}{1} = \frac{y-\lambda}{2} = \frac{z-3}{4}$$

Now, we have our vital statistics. From $L_1$, we have a point $\vec{a}_1 = (1, 2, 1)$ and a direction vector $\vec{b}_1 = (2, 1, 3)$. From $L_2$, we have a point $\vec{a}_2 = (2, \lambda, 3)$ and a direction vector $\vec{b}_2 = (1, 2, 4)$.

### Phase 2: The Bridge of Perpendicularity

To find the shortest distance, we need a vector that is perpendicular to both lines. This is where the cross product shines. We calculate $\vec{b}_1 	imes \vec{b}_2$ using the determinant method:

$$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & 3 \ 1 & 2 & 4 \end{vmatrix}$$

Expanding this, we get:

$$\vec{b}_1 	imes \vec{b}_2 = \hat{i}(4 - 6) - \hat{j}(8 - 3) + \hat{k}(4 - 1) = -2\hat{i} - 5\hat{j} + 3\hat{k}$$

This vector, $-2\hat{i} - 5\hat{j} + 3\hat{k}$, is the direction of our shortest path. Its magnitude is:

$$|\vec{b}_1 	imes \vec{b}_2| = \sqrt{(-2)^2 + (-5)^2 + 3^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$$

### Phase 3: The Final Calculation

The shortest distance $d$ is the projection of the vector connecting the two points, $\vec{a}_2 - \vec{a}_1$, onto this perpendicular vector. The vector connecting the points is:

$$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (\lambda-2)\hat{j} + (3-1)\hat{k} = \hat{i} + (\lambda-2)\hat{j} + 2\hat{k}$$

Now, we apply the distance formula:

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

Substituting our values:

$$d = \frac{|(\hat{i} + (\lambda - 2)\hat{j} + 2\hat{k}) \cdot (-2\hat{i} - 5\hat{j} + 3\hat{k})|}{\sqrt{38}}$$
$$d = \frac{|1(-2) + (\lambda - 2)(-5) + 2(3)|}{\sqrt{38}} = \frac{|-2 - 5\lambda + 10 + 6|}{\sqrt{38}} = \frac{|14 - 5\lambda|}{\sqrt{38}}$$

We are given that $d = \frac{1}{\sqrt{38}}$. Thus:

$$\frac{|14 - 5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}} \Rightarrow |14 - 5\lambda| = 1$$

This gives us two possibilities: $14 - 5\lambda = 1$ or $14 - 5\lambda = -1$. Solving these, we get $\lambda = \frac{13}{5}$ or $\lambda = 3$.

The question asks for the **integral value** of $\lambda$. Since $\frac{13}{5}$ is not an integer, our answer is **$3$**.

Animated Solution for Mathematics - Three Dimensional Geometry: If the shortest distance between the straight lines $3 (x - 1) = 6 (y - 2) = 2 (z - 1)$ and $4 (x - 2) = 2 (y - λ) = (z - 3), λ \in R$ is $\frac{1}{38}$ , then the integral value of $λ$ is equal to :

Select Answer:

Visualized Solution

Visualizing Skew Lines

Standardizing Line $L_{1}$

Standardizing Line $L_{2}$

Extracting Points and Directions

Connecting the Points

Finding the Common Perpendicular

Calculating $b_{1} \times b_{2}$

Magnitude of the Perpendicular Vector

The Shortest Distance Formula

Evaluating the Dot Product

Solving for $λ$

Selecting the Integral Value

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of Skew Lines

Phase 1

Phase 2

Phase 3

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