Let the values of λ for which the shortest distance between the lines (x-1)/2 = (y-2)/3 = (z-3)/4 an

Q: Wait, why can't we just use the distance formula between two points? Why is this so complicated?

That is a fantastic question! The standard distance formula is for the distance between two specific points. Here, we are dealing with lines, which are infinite sets of points. We need the shortest distance between the *sets*, which requires the common perpendicular approach. Think of it as finding the gap between two bridges that don't cross; you need to find the shortest vertical drop between them.

Q: Why does the center of the circle lie on $y=x$? Is this always true?

It is not always true, but it is true here because of the symmetry of the points! The points $(0,0), (2,4),$ and $(4,2)$ are symmetric with respect to the line $y=x$. If you reflect the point $(2,4)$ across $y=x$, you get $(4,2)$. Since the circle passes through these symmetric points, its center must also be symmetric with respect to the same line, placing it on $y=x$. It is a beautiful geometric shortcut!

Q: Is there a simpler way to find the radius without solving the quadratic equation?

While you could use the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ and solve a system of three equations, utilizing the symmetry of the points is far more elegant and less prone to algebraic errors. In JEE Advanced, spotting these symmetries is often the key to solving problems quickly and accurately.

Question

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the points $(0, 0), (\lambda_1, \lambda_2)$ and (\lambda_2, \lambda_1) is

Options:

(A) $\frac{5\sqrt{2}}{3}$

(B) $4$

(C) $\frac{\sqrt{2}}{3}$

(D) $3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. Before you, there are two skew lines—they never meet, they never run parallel, and they exist in their own separate worlds. Our mission is to find the shortest distance between them.

We identify the DNA of these lines: their fixed points and their direction vectors. For the first line, we have the point $A_1(1, 2, 3)$ and the direction vector $\vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

For the second line, we have $A_2(\lambda, 4, 5)$ and the direction vector $\vec{b_2} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.

### The Master Equation

The bridge we seek must be perpendicular to both lines. We invoke the power of the cross product: $\vec{b_1} 	imes \vec{b_2}$.

By calculating the determinant:

$$
\begin{vmatrix} 
\hat{i} & \hat{j} & \hat{k} \ 
2 & 3 & 4 \ 
3 & 4 & 5 
\end{vmatrix} = -\hat{i} + 2\hat{j} - \hat{k}
$$

The magnitude of this normal vector is $\sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}$.

### The Shortest Distance

We connect our two lines with a vector $\vec{A_1A_2} = (\lambda-1)\hat{i} + 2\hat{j} + 2\hat{k}$. The shortest distance ($SD$) is the projection of this vector onto our common perpendicular:

$$
SD = \frac{|(\vec{A_2}-\vec{A_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}
$$

When we perform the dot product, the expression simplifies to:

$$
SD = \frac{|3-\lambda|}{\sqrt{6}}
$$

Given that this distance is $\frac{1}{\sqrt{6}}$, the $\sqrt{6}$ terms cancel, leaving $|3-\lambda| = 1$. This yields two solutions: **$\lambda_1 = 2$** and **$\lambda_2 = 4$**.

### Symmetry and the Circle

We now transition to a 2D plane with three points: the origin $O(0,0)$, $P(2,4)$, and $Q(4,2)$. We need the radius of the circle passing through these points.

Points $P$ and $Q$ are reflections of each other across the line $y=x$. Since the origin also lies on this line, the center of the circle $C(h, k)$ must also lie on $y=x$, implying $h=k$.

The radius squared, $R^2$, is the distance from $C(h, h)$ to the origin:

$$
R^2 = h^2 + h^2 = 2h^2
$$

It is also the distance from $C(h, h)$ to $P(2,4)$:

$$
R^2 = (h-2)^2 + (h-4)^2
$$

### Final Calculation

Equating these expressions, we have $2h^2 = (h-2)^2 + (h-4)^2$. Expanding this gives:

$$
2h^2 = h^2 - 4h + 4 + h^2 - 8h + 16
$$

The quadratic terms $2h^2$ cancel, leaving $0 = -12h + 20$, which simplifies to $h = \frac{5}{3}$. The center of the circle is $(\frac{5}{3}, \frac{5}{3})$.

Finally, the radius $R$ is:

$$
R = \sqrt{2h^2} = h\sqrt{2} = \mathbf{\frac{5\sqrt{2}}{3}}
$$

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Visualized Solution

$G eo m e t r y o f S k e w L in es$

$P o in t s an d D i r ec t i o nV ec t or s$

$C o mm o n P er p e n d i c u l a r V ec t or$

$S h or t es t D i s t an ce F or m u l a$

$S o l v in g f or λ$

$T r an s i t i o n t o t h e C i r c l e P r o b l e m$

$S y mm e t r y o f t h e C i r c l e$

$D i s t an ce F or m u l a f or R a d i u s$

$E q u a t in g t h eD i s t an ces$

$F in d in g t h e C e n t er$

$C a l c u l a t in g F ina l R a d i u s$

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Master Equation

The Shortest Distance

Symmetry and the Circle

Final Calculation

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 :

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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 $\frac{x - λ}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ 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 _________

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$ and $\frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$ is:

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Visualized Solution

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