If the lines x - 1 2 = 2 - y - 3 = z - 3 and x - 4 5 = y - 1 2 = z intersect, then the magnitude of the minimum value of 8 is _______ . [2023]

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Solution

Q.
If the lines $\frac{x - 1}{2} = \frac{2 - y}{- 3} = \frac{z - 3}{α}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β}$ intersect, then the magnitude of the minimum value of $8 α β$ is _______ . [2023]

Ans.
(18)

$Let \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{α} = λ (say)$ ...(i)

$and \frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{β} = μ (say)$ ...(ii)

$∴$ $Any point on line (i) and (ii) are of the forms (2 λ + 1, 3 λ + 2, α λ + 3)$ $and (5 μ + 4, 2 μ + 1, β μ) respectively .$

$The lines are intersecting.$

$∴$ $2 λ + 1 = 5 μ + 4, 3 λ + 2 = 2 μ + 1, α λ + 3 = β μ$

$Solving first two equations, we get λ = - 1 and μ = - 1$

$From third equation, we have$

$- α + 3 = - β \Rightarrow α - 3 = β$ ...(iii)

$Let y = 8 α β = 8 α (α - 3)$

$y' = 8 α + 8 (α - 3) = 16 α - 24$

$For maxima/minima, 16 α - 24 = 0 \Rightarrow α = \frac{3}{2}$

$Now, y'' = 16 > 0$

$∴$ $8 α β is minimum at α = \frac{3}{2}$

$So, minimum value of$ $| 8 α β |$ $=$ $| 8 \times \frac{3}{2} \times \frac{- 3}{2} |$ $= 18$

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