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Question

Find the shortest distance between the lines:
$$\dfrac{x}{5}-\dfrac{y-2}{2}=\dfrac{z-3}{3}$$ and $$\dfrac{x+3}{5}-\dfrac{y-1}{2}=\dfrac{z+4}{3}$$ 


A
3.5
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B
4.5
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C
5.5
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D
6.5
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Solution

The correct option is A $$3.5$$
$$\cfrac { x }{ 5 } -\left( \cfrac { y-2 }{ 2 }  \right)  = \cfrac { z-3 }{ 3 }$$
$$\Rightarrow \left( 2x-5y+10 \right) \times 3 = 10z-30$$
$$ \Rightarrow 6x-15y-10z+60 = 0  \longrightarrow (1)$$

$$ \cfrac { x+3 }{ 5 } -\cfrac { y-1 }{ 2 } =\cfrac { z+4 }{ 3 }$$
$$\Rightarrow (2x+6-5y+5)\times 3=10z+40$$
$$\Rightarrow 6x-15y-10z-7 = 0 \longrightarrow (2)$$

Parallel distance $$= \cfrac { \left| -67 \right|  }{ \sqrt { 36+225+100 }  } =\cfrac { \left| 67 \right|  }{ 19 }  \sim  3.5$$ units

Mathematics

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