Let the values of lambda for which the shortest distance between the lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-lambda)/3 = (y-4)/4 = (z-5)/5 is 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

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q15 – JEE-MAIN 2025

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Home Maths 3D GeometryQuestion #30

Let the values of

λ

for which the shortest distance between the lines

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}

and

\frac{x - λ}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}

\frac{1}{\sqrt{6}}

λ_{1}

and

λ_{2}

. Then the radius of the circle passing through the points

(0, 0)

(λ_{1}, λ_{2})

and

(λ_{2}, λ_{1})

Clue 1 of 3: Geometric Interpretation

The problem involves two main geometric concepts: the shortest distance between two skew lines in 3D space and the radius of a circle passing through three given points.

🥷

Ninja Strategy•Magnitude Estimation

Estimate the approximate size of the circumradius by comparing it to the side lengths of the triangle formed by the three points. The radius must be at least half the longest side.

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Step 1: Determine

λ_{1}

and

λ_{2}

using the shortest distance formula✦ Active

The given lines are $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $L_{2} : \frac{x - λ}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ . From $L_{1}$ , $\vec{a_{1}} = (1, 2, 3)$ and $\vec{b_{1}} = (2, 3, 4)$ . From $L_{2}$ , $\vec{a_{2}} = (λ, 4, 5)$ and $\vec{b_{2}} = (3, 4, 5)$ . Calculate $\vec{a_{2}} - \vec{a_{1}} = (λ - 1, 2, 2)$ . Calculate $\vec{b_{1}} \times \vec{b_{2}} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix} | = (- 1, 2, - 1)$ . The magnitude is $| \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{(- 1)^{2} + 2^{2} + (- 1)^{2}} = \sqrt{6}$ . The scalar triple product is $(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{b_{1}} \times \vec{b_{2}}) = (λ - 1) (- 1) + 2 (2) + 2 (- 1) = - λ + 1 + 4 - 2 = 3 - λ$ . Given shortest distance $d = \frac{1}{\sqrt{6}}$ . Using the formula $d = \frac{| (3 - λ) |}{\sqrt{6}}$ , we get $| 3 - λ | = 1$ . This implies $3 - λ = 1$ or $3 - λ = - 1$ . Solving these, we get $λ = 2$ or $λ = 4$ . Thus, $λ_{1} = 2$ and $λ_{2} = 4$ (or vice versa).

Step 2: Formulate the circle equation and find its parameters○ Expand

The three points are $P_{1} (0, 0)$ , $P_{2} (λ_{1}, λ_{2}) = (2, 4)$ , and $P_{3} (λ_{2}, λ_{1}) = (4, 2)$ . Let the equation of the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Substituting $P_{1} (0, 0)$ : $0^{2} + 0^{2} + 2 g (0) + 2 f (0) + c = 0 ⟹ c = 0$ . Substituting $P_{2} (2, 4)$ : $2^{2} + 4^{2} + 2 g (2) + 2 f (4) + 0 = 0 ⟹ 4 + 16 + 4 g + 8 f = 0 ⟹ 4 g + 8 f = - 20 ⟹ g + 2 f = - 5$ . Substituting $P_{3} (4, 2)$ : $4^{2} + 2^{2} + 2 g (4) + 2 f (2) + 0 = 0 ⟹ 16 + 4 + 8 g + 4 f = 0 ⟹ 8 g + 4 f = - 20 ⟹ 2 g + f = - 5$ . Solving the system of equations: $g + 2 f = - 5$ and $2 g + f = - 5$ . Subtracting the first from the second gives $(2 g + f) - (g + 2 f) = 0 ⟹ g - f = 0 ⟹ g = f$ . Substitute $g = f$ into $g + 2 f = - 5 ⟹ f + 2 f = - 5 ⟹ 3 f = - 5 ⟹ f = - \frac{5}{3}$ . Therefore, $g = - \frac{5}{3}$ and $f = - \frac{5}{3}$ .

Step 3: Calculate the radius of the circle○ Expand

The radius of the circle is $R = \sqrt{g^{2} + f^{2} - c}$ . Substituting the values $g = - \frac{5}{3}$ , $f = - \frac{5}{3}$ , and $c = 0$ : $R = \sqrt{{(- \frac{5}{3})}^{2} + {(- \frac{5}{3})}^{2} - 0}$ . $R = \sqrt{\frac{25}{9} + \frac{25}{9}} = \sqrt{\frac{50}{9}}$ . $R = \frac{\sqrt{50}}{\sqrt{9}} = \frac{5 \sqrt{2}}{3}$ .

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Maths Question 15 – JEE-MAIN 2025

Clue 1 of 3: Geometric Interpretation

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