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
B
4.5
C
5.5
D
6.5

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

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