If the shortest between the lines (x+√6)/2 = (y-√6)/3 = (z-√6)/4 and (x-λ)/3 = (y-2√6)/4 = (z+2√6)/5

Q: Wait, why do we have two values for $\lambda$? Shouldn't there be a unique solution?

That is a fantastic observation! The absolute value in the distance formula $|\dots| = d$ creates a bifurcation. Because the distance is a magnitude, the expression inside can be positive or negative, leading to two valid geometric configurations where the lines are separated by exactly 6 units. Both are physically valid!

Q: Is there a way to solve this without the cross product formula?

Technically, yes! You could define a vector connecting a general point on $L_1$ to a general point on $L_2$ and minimize its length using calculus or by setting its dot product with both direction vectors to zero. However, the formula is the 'express lane' for JEE Advanced. It saves time and reduces the chance of algebraic errors.

Q: How do I make sure I don't mess up the signs when extracting the points from the symmetric equations?

Always remember the standard form is $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$. If you see $x+\sqrt{6}$, think of it as $x - (-\sqrt{6})$. The negative sign is part of the formula, so the coordinate is $-\sqrt{6}$. Writing it out explicitly like this prevents those silly sign errors that haunt even the best students.

Question

If the shortest between the lines $\frac{x+\sqrt{6}}{2} = \frac{y-\sqrt{6}}{3} = \frac{z-\sqrt{6}}{4}$ and $\frac{x-\lambda}{3} = \frac{y-2\sqrt{6}}{4} = \frac{z+2\sqrt{6}}{5}$ is 6, then the square of sum of all possible values of $\lambda$ is

Sigma Percentile · Accepted Answer

### The Geometry of Skew Lines

Imagine you are standing in a vast, empty room. You have two thin, straight wires suspended in the air. They are not parallel, and they do not touch.

In the language of 3D geometry, we call these 'skew lines.' They are like two ships passing in the night, separated by a specific, unbridgeable gap.

Our goal today is to find the magnitude of that gap, given as $d=6$, and use it to unlock the mystery of the parameter $\lambda$. This isn't just about plugging numbers into a formula; it is about understanding the spatial relationship between two infinite lines.

### Decoding the Lines

Before we touch any complex algebra, we must translate the symmetric equations into the language of vectors. The symmetric form

$$\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$$

is a beautiful shorthand, but for calculations, we need the position vector $\vec{a}$ and the direction vector $\vec{b}$.

For line $L_1$, given by $\frac{x+\sqrt{6}}{2} = \frac{y-\sqrt{6}}{3} = \frac{z-\sqrt{6}}{4}$, we identify the point $\vec{a}_1 = -\sqrt{6}\hat{i} + \sqrt{6}\hat{j} + \sqrt{6}\hat{k}$ and the direction $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

For line $L_2$, given by $\frac{x-\lambda}{3} = \frac{y-2\sqrt{6}}{4} = \frac{z+2\sqrt{6}}{5}$, we identify the point $\vec{a}_2 = \lambda\hat{i} + 2\sqrt{6}\hat{j} - 2\sqrt{6}\hat{k}$ and the direction $\vec{b}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}$.

### The Common Normal

Why do we calculate the cross product $\vec{b}_1 	imes \vec{b}_2$? Because this vector is the 'common normal'—it is the only direction that is perpendicular to both lines simultaneously. It defines the orientation of the shortest distance segment.

Computing this determinant is where we find the elegance of the problem:

$$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$$

Its magnitude is $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}$.

### The Projection

The shortest distance is simply the projection of the vector connecting the two lines, $\vec{a}_2 - \vec{a}_1$, onto this common normal. The formula is:

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

Calculating $\vec{a}_2 - \vec{a}_1 = (\lambda + \sqrt{6})\hat{i} + \sqrt{6}\hat{j} - 3\sqrt{6}\hat{k}$.

The dot product yields:

$$(\lambda + \sqrt{6})(-1) + (\sqrt{6})(2) + (-3\sqrt{6})(-1) = -\lambda - \sqrt{6} + 2\sqrt{6} + 3\sqrt{6} = -\lambda + 4\sqrt{6}$$

### The Final Reveal

Setting this equal to $6$, we get:

$$6 = \frac{|-\lambda + 4\sqrt{6}|}{\sqrt{6}}$$

This simplifies to $|\lambda - 4\sqrt{6}| = 6\sqrt{6}$.

This leads to two solutions: 
1. $\lambda - 4\sqrt{6} = 6\sqrt{6} \Rightarrow \lambda = 10\sqrt{6}$
2. $\lambda - 4\sqrt{6} = -6\sqrt{6} \Rightarrow \lambda = -2\sqrt{6}$

The sum is $8\sqrt{6}$, and the square is $(8\sqrt{6})^2 = 64 	imes 6 = 384$.

**You have conquered the geometry! The final result is 384.**

Animated Solution for Mathematics - Three Dimensional Geometry: If the shortest between the lines $\frac{x + 6}{2} = \frac{y - 6}{3} = \frac{z - 6}{4}$ and $\frac{x - λ}{3} = \frac{y - 2 6}{4} = \frac{z + 2 6}{5}$ is 6, then the square of sum of all possible values of $λ$ is

Enter Numerical Value:

Visualized Solution

The 3D Geometry Setup

Components of Line $L_{1}$

Components of Line $L_{2}$

The Shortest Distance Formula

Calculate Cross Product $b_{1} \times b_{2}$

Magnitude of Cross Product

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

Calculate the Dot Product

Set up the Distance Equation

Solve for $λ$

Final Calculation

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of Skew Lines

Decoding the Lines

The Common Normal

The Projection

The Final Reveal

Similar Questions

If the shortest distance between the lines $\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{- 3}$ and $\frac{x - λ}{2} = \frac{y + 1}{4} = \frac{z - 2}{- 5}$ is $\frac{6}{5}$ , then the sum of all possible values of $λ$ is :

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{3}$ , then the sum of all possible values of $λ$ is :

The sum of all values of $α$ , for which the shortest distance between the lines $\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α}$ and $\frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $2$ , is

If the shortest distance between the lines $\frac{x - λ}{- 2} = \frac{y - 2}{1} = \frac{z - 1}{1}$ and $\frac{x - 3}{1} = \frac{y - 1}{- 2} = \frac{z - 2}{1}$ is 1, then the sum of all possible values of $λ$ is :

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{6}$ , then the sum of all possible values of $α$ is

If the shortest distance between the lines $\frac{x - λ}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $\frac{x + 2}{- 3} = \frac{y + 5}{2} = \frac{z - 4}{4}$ is $\frac{44}{30}$ , then the largest possible value of $∣ λ ∣$ is equal to _________

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 :

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 :

Let $λ$ be an integer. If the shortest distance between the lines $x - λ = 2 y - 1 = - 2 z$ and $x = y + 2 λ = z - λ$ is $\frac{7}{2 2}$ , then the value of $∣ λ ∣$ is

Enter Numerical Value:

Visualized Solution

The 3D Geometry Setup

Components of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Components of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

The Shortest Distance Formula

Calculate Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​×b2​