Let the values of for which the shortest distance between the lines and is be and . Then the radius of the circle passing through the points (0, 0), and is [2025]

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Solution

Q.
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 $\frac{x - λ}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{\sqrt{6}}$ be $λ_{1}$ and $λ_{2}$ . Then the radius of the circle passing through the points (0, 0), $(λ_{1}, λ_{2})$ and $(λ_{2}, λ_{1})$ is [2025]

1 3

2 4

3 $\frac{\sqrt{2}}{3}$

4 $\frac{5 \sqrt{2}}{3}$

Ans.
(4)

$\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} and \vec{b} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$

represents direction vector of the given lines.

$∴ \vec{a} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{matrix} |$

$= \hat{i} (15 - 16) - \hat{j} (10 - 12) + \hat{k} (8 - 9) = - \hat{i} + 2 \hat{j} - \hat{k}$

The given lines passes through ${\vec{p}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and ${\vec{p}}_{2} = λ \hat{i} + 4 \hat{j} + 5 \hat{k}$

$∴ {\vec{p}}_{2} - {\vec{p}}_{1} = (λ - 1) \hat{i} + 2 \hat{j} + 2 \hat{k}$

Shortest distance between given lines $=$ $| \frac{({\vec{p}}_{2} - {\vec{p}}_{1}) \cdot (\vec{a} \times \vec{b})}{| \vec{a} \times \vec{b} |} |$

$\Rightarrow \frac{1}{\sqrt{6}} =$ $| \frac{- λ + 1 + 4 - 2}{\sqrt{1 + 4 + 1}} |$

$\Rightarrow$ $| - λ + 3 |$ $= 1 \Rightarrow λ = 3 \pm 1 \Rightarrow λ = 4, 2$

Radius of circle passing through points (0, 0), (4, 2), (2, 4) is given by $\frac{a b c}{4 △}$ .

$= \frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2} | \begin{matrix} 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 \end{matrix} |} = \frac{20 \times 2 \sqrt{2}}{2 \times 12} = \frac{5 \sqrt{2}}{3}$

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