Let P be the point of intersection of the lines and . Then, the shortest distance of P from the line 4x = 2y = z is [2024]

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Solution

Q.
Let P be the point of intersection of the lines $\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}$ and $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}$ . Then, the shortest distance of P from the line 4x = 2y = z is [2024]

1 $\frac{\sqrt{14}}{7}$

2 $\frac{5 \sqrt{14}}{7}$

3 $\frac{6 \sqrt{14}}{7}$

4 $\frac{3 \sqrt{14}}{7}$

Ans.
(4)

$L_{1} : \frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1} = λ (say)$

$L_{2} : \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2} = μ (say)$

For point of intersection P,

$(λ + 2, 5 λ + 4, λ + 2) = (2 μ + 3, 3 μ + 2, 2 μ + 3)$

$\Rightarrow λ + 2 = 2 μ + 3, 5 λ + 4 = 3 μ + 2, λ + 2 = 2 μ + 3$

$\Rightarrow λ = - 1, μ = - 1$

$∴$ Point P is (1, –1, 1)

Distance of point P from $L_{3}$ : 4x = 2y = z

$L_{3} : \frac{x}{\frac{1}{4}} = \frac{y}{\frac{1}{2}} = \frac{z}{1}$

Any point on $L_{3}$ be $R (\frac{α}{4}, \frac{α}{2}, α)$

D.r.'s of $P R : (\frac{α}{4} - 1, \frac{α}{2} + 1, α - 1) ∵ P R ⊥ (\frac{1}{4}, \frac{1}{2}, 1)$

$\Rightarrow (\frac{α}{4} - 1) \frac{1}{4} + \frac{1}{2} (\frac{α}{2} + 1) + (α - 1) = 0$

$\Rightarrow \frac{α}{16} - \frac{1}{4} + \frac{α}{4} + \frac{1}{2} + α - 1 = 0 \Rightarrow \frac{21}{16} α = \frac{3}{4} \Rightarrow α = \frac{4}{7}$

$∴ R (\frac{1}{7}, \frac{2}{7}, \frac{4}{7})$

Now, $R P = \sqrt{{(\frac{1}{7} - 1)}^{2} + {(\frac{2}{7} + 1)}^{2} + {(\frac{4}{7} - 1)}^{2}}$

$= \sqrt{\frac{36}{49} + \frac{81}{49} + \frac{9}{49}} = \frac{\sqrt{126}}{7} = \frac{3 \sqrt{14}}{7}$ .

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