Let the shortest distance between the lines L : x - 5 - 2 = y - 0 = z + 1 , 0 and L 1 : x + 1 = y - 1 = 4 - z be 2 6 . If lies on L, then which of the following is NOT possible? [2023]

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Solution

Q.
Let the shortest distance between the lines $L : \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1}, λ \geq 0$ and $L_{1} : x + 1 = y - 1 = 4 - z$ be $2 \sqrt{6}$ . If $(α, β, γ)$ lies on L, then which of the following is NOT possible [2023]

1 $α + 2 γ = 24$

2 $2 α + γ = 7$

3 $α - 2 γ = 19$

4 $2 α - γ = 9$

Ans.
(1)

$Since shortest distance between lines L and L_{1} is 2 \sqrt{6} .$

$Here, L : \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1}, λ \geq 0$

$and L_{1} : \frac{x + 1}{1} = \frac{y - 1}{1} = \frac{z - 4}{- 1}$

$Shortest distance between L and L_{1} is given by$

$d =$ $| \frac{| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} |}{\sqrt{{(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2} + {(a_{1} b_{2} - a_{2} b_{1})}^{2}}} |$

$∴$ $2 \sqrt{6} =$ $| \frac{| \begin{matrix} - 1 - 5 & 1 - λ & 4 + λ \\ - 2 & 0 & 1 \\ 1 & 1 & - 1 \end{matrix} |}{\sqrt{{(0 - 1)}^{2} + {(1 - 2)}^{2} + {(2 - 0)}^{2}}} |$

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

$=$ $| \frac{6 + λ - 1 - 2 λ - 8}{\sqrt{6}} |$ $=$ $| \frac{- λ - 3}{\sqrt{6}} |$

$\Rightarrow \frac{- 3 - λ}{\sqrt{6}} = \pm 2 \sqrt{6} \Rightarrow \frac{- 3 - λ}{\sqrt{6}} = 2 \sqrt{6} and \frac{- 3 - λ}{\sqrt{6}} = - 2 \sqrt{6}$

$- 3 - λ = 12 \Rightarrow λ = - 15$ and $- 3 - λ = - 12 \Rightarrow λ = 9$

$Since λ \geq 0, therefore λ = 9$

$Now, from line L, \frac{x - 5}{- 2} = \frac{y - λ}{0} = \frac{z + λ}{1} = r (say)$

$\Rightarrow x = - 2 r + 5, y = 0 + λ, z = r - λ$

$\Rightarrow x = - 2 r + 5, y = 9, z = r - 9$

$Since (α, β, γ) lies on L,$ then $α = - 2 r + 5, β = 9, γ = r - 9$

$Now, α + 2 γ = - 2 r + 5 + 2 r - 18 = - 13$

   $2 α + γ = - 4 r + 10 + r - 9 = - 3 r + 1, r \in R$

   $α - 2 γ = - 2 r + 5 - 2 r + 18 = - 4 r + 23, r \in R$

   $2 α - γ = - 4 r + 10 - r + 9 = - 5 r + 19, r \in R$

$Hence, α + 2 γ = 24 is not possible.$

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