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

• A. 4
• B. 3
• C. \(\frac{\sqrt{2}}{3}\)
• D. \(\frac{5\sqrt{2}}{2}\)

Hint:

- Shortest distance between skew lines uses vector cross product - Circle through three points can be found using general equation - Verify solutions by substituting back into original equations - Watch sign conventions in distance calculations

Answer 1

The Correct Option is D

Step 1: Find shortest distance between lines
Given lines: \[ L_1: \vec{r}_1 = (1,2,3) + t(2,3,4) \] \[ L_2: \vec{r}_2 = (\lambda,4,5) + s(3,4,5) \] Shortest distance formula: \[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} \] where \(\vec{a}_1 = (1,2,3)\), \(\vec{a}_2 = (\lambda,4,5)\), \(\vec{b}_1 = (2,3,4)\), and \(\vec{b}_2 = (3,4,5)\). Calculate \(\vec{b}_1 \times \vec{b}_2\): \[ (-1, 2, -1) \] Thus: \[ d = \frac{|(\lambda-1, 2, 2) \cdot (-1, 2, -1)|}{\sqrt{6}} = \frac{|1 - \lambda + 4 - 2|}{\sqrt{6}} = \frac{|3 - \lambda|}{\sqrt{6}} = \frac{1}{\sqrt{6}} \] Solving: \[ |3 - \lambda| = 1 \Rightarrow \lambda_1 = 2, \lambda_2 = 4 \]

Step 2: Find the circle through (0,0), (2,4), (4,2)
Using the general circle equation \(x^2 + y^2 + 2gx + 2fy + c = 0\): \[ \begin{cases} c = 0 \\ 4 + 16 + 4g + 8f = 0 \\ 16 + 4 + 8g + 4f = 0 \end{cases} \] Solving gives \(g = -\frac{5}{2}\), \(f = -\frac{5}{2}\), and \(c = 0\). Radius: \[ r = \sqrt{g^2 + f^2 - c} = \sqrt{\frac{25}{4} + \frac{25}{4}} = \frac{5\sqrt{2}}{2} \]