The shortest distance between the lines (x-3)/(3)=(y-8)/(-1)=(z-3)/(1) and (x+3)/(-3)=(y+7)/(2)=(z-6)/(4) is :

3D Geometry: The shortest distance between the lines… | JEE Main 2020

JEE Main 2020 (Online) 8th January Morning ShiftQuestion 1

The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is :

3D Geometry

$2\sqrt{30}$$3\sqrt{30}$✓ Correct answer$3$$\frac72\sqrt{30}$

Solution

Points $(3,8,3)$, $(-3,-7,6)$; directions $(3,-1,1)$, $(-3,2,4)$. Cross $=(-4-2,\,-3-12,\,6-3)=(-6,-15,3)$, magnitude $=\sqrt{270}=3\sqrt{30}$. Connecting vector $(-6,-15,3)$: dot $=36+225+9=270$. $SD=\frac{270}{3\sqrt{30}}=3\sqrt{30}$. Option (B).

3D Geometry - JEE Mathematics question (JEE Main 2020 (Online) 8th January Morning Shift)

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