If the shortest distance between the lines (x-λ)/-2 = (y-2)/1 = (z-1)/1 and (x-√3)/1 = (y-1)/-2 = (z

Q: Why do we need the cross product? Can't we just find the distance between the two points given in the equations?

That is a classic trap! The points $(\lambda, 2, 1)$ and $(\sqrt{3}, 1, 2)$ are just arbitrary points on the lines. The distance between them is not the shortest distance. The shortest distance is unique and occurs only along the common perpendicular. The cross product $\vec{b}_1 \times \vec{b}_2$ gives us the direction of this unique common perpendicular, which is essential for the projection.

Q: I solved the equation $|\sqrt{3} - \lambda| = \sqrt{3}$ and got $\lambda = 0$. Why is there another value?

Remember that the modulus function $|x| = a$ implies $x = a$ OR $x = -a$. When you remove the absolute value bars, you must account for both the positive and negative possibilities. Ignoring the negative case is a common way to lose marks in JEE Advanced! Always check for both branches.

Q: Is there a geometric way to do this without the formula?

Absolutely! The formula is just a mathematical shorthand for the geometric projection of the vector connecting the two lines onto the common normal. If you ever forget the formula, you can find the common normal vector, find the vector connecting the two points, and project the latter onto the former. The result will be identical!

Question

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

Options:

(A) 0

(B) $2\sqrt{3}$

(C) $3\sqrt{3}$

(D) $-2\sqrt{3}$

Sigma Percentile · Accepted Answer

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

Welcome, future engineers. Today, we are not just solving a problem; we are stepping into the third dimension. Imagine you are standing in a vast, empty room where two lines are suspended in the air.

These lines are not parallel, and they do not intersect; these are what we call **skew lines**. They are like two airplanes flying at different altitudes and different headings—they never meet, yet they have a unique distance that separates them at their closest point.

Our mission is to find the value of $\lambda$ that dictates the position of one of these lines, given that this shortest distance is exactly $1$.

### Phase 1: The Toolkit

In the realm of 3D geometry, we rely on the projection of the vector connecting any two points on the lines onto the common perpendicular. The formula is our compass:

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

Here, $\vec{a}_1$ and $\vec{a}_2$ are position vectors of points on the lines, and $\vec{b}_1$ and $\vec{b}_2$ are the direction vectors. Let us break this down into manageable pieces.

### Phase 2: Extracting the DNA of the Lines

For line $L_1$, given by $\frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}$, we identify the point $\vec{a}_1 = (\lambda, 2, 1)$ and the direction vector $\vec{b}_1 = (-2, 1, 1)$.

For line $L_2$, given by $\frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}$, we identify the point $\vec{a}_2 = (\sqrt{3}, 1, 2)$ and the direction vector $\vec{b}_2 = (1, -2, 1)$.

We now find the vector connecting these two points, $\vec{a}_2 - \vec{a}_1$:

$$\vec{a}_2 - \vec{a}_1 = (\sqrt{3} - \lambda, 1 - 2, 2 - 1) = (\sqrt{3} - \lambda, -1, 1)$$

### Phase 3: The Common Perpendicular

To find the direction perpendicular to both lines, we perform the cross product $\vec{b}_1 	imes \vec{b}_2$. This is the heart of the calculation:

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

With this, the magnitude of our common perpendicular is:

$$|\vec{b}_1 	imes \vec{b}_2| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$$

### Phase 4: The Final Convergence

Next, we calculate the dot product of our connecting vector and the common perpendicular:

$$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2) = (\sqrt{3} - \lambda)(3) + (-1)(3) + (1)(3) = 3\sqrt{3} - 3\lambda$$

Substituting everything into our shortest distance formula, where $S.D. = 1$:

$$1 = \frac{|3\sqrt{3} - 3\lambda|}{3\sqrt{3}} \Rightarrow 1 = \frac{3|\sqrt{3} - \lambda|}{3\sqrt{3}} \Rightarrow |\sqrt{3} - \lambda| = \sqrt{3}$$

When we remove the modulus, we consider both cases:

1. $\sqrt{3} - \lambda = \sqrt{3} \Rightarrow \lambda = 0$
2. $\sqrt{3} - \lambda = -\sqrt{3} \Rightarrow \lambda = 2\sqrt{3}$

The sum of these values is $0 + 2\sqrt{3} = **2\sqrt{3}**$.

Animated Solution for Mathematics - Three Dimensional Geometry: 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 :

Select Answer:

Visualized Solution

Visualizing Skew Lines

Shortest Distance Formula

Extracting Data for Line $L_{1}$

Extracting Data for Line $L_{2}$

Finding the Connecting Vector

Calculating the Cross Product

Magnitude of the Cross Product

Calculating the Dot Product

Setting up the Equation

Solving for $λ$

Final Sum of Values

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Dance of Skew Lines

Phase 1

Phase 2

Phase 3

Phase 4

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