Question:
The shortest distance between the lines
\(\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}\)
and
\(\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}\)
is :
The shortest distance between the lines
\(\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}\)
and
\(\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}\)
is :
\(\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}\)
and
\(\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}\)
is :
Show Hint
Use \(d=\dfrac{|(\vec{b}-\vec{a})\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}\) with the direction vectors \((4,5,3)\) and \((3,4,2)\).
Updated On: Nov 2, 2026
- \(3\sqrt{6}\)
- \(6\sqrt{2}\)
- \(6\sqrt{3}\)
- \(2\sqrt{6}\)
Show Solution
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The Correct Option is A
Solution and Explanation
The lines pass through \(A(4,-2,-3)\) and \(B(1,3,4)\) with direction vectors \(\vec{d_1}=(4,5,3)\), \(\vec{d_2}=(3,4,2)\).
\(\vec{d_1}\times\vec{d_2}=\big(5\cdot2-3\cdot4,\;3\cdot3-4\cdot2,\;4\cdot4-5\cdot3\big)=(-2,1,1)\), so \(|\vec{d_1}\times\vec{d_2}|=\sqrt{4+1+1}=\sqrt{6}\).
\(\overrightarrow{AB}=(1-4,\,3+2,\,4+3)=(-3,5,7)\).
\(\overrightarrow{AB}\cdot(\vec{d_1}\times\vec{d_2})=6+5+7=18\).
Shortest distance \(=\dfrac{18}{\sqrt{6}}=3\sqrt{6}\).
Answer: (1) \(3\sqrt{6}\)
\(\vec{d_1}\times\vec{d_2}=\big(5\cdot2-3\cdot4,\;3\cdot3-4\cdot2,\;4\cdot4-5\cdot3\big)=(-2,1,1)\), so \(|\vec{d_1}\times\vec{d_2}|=\sqrt{4+1+1}=\sqrt{6}\).
\(\overrightarrow{AB}=(1-4,\,3+2,\,4+3)=(-3,5,7)\).
\(\overrightarrow{AB}\cdot(\vec{d_1}\times\vec{d_2})=6+5+7=18\).
Shortest distance \(=\dfrac{18}{\sqrt{6}}=3\sqrt{6}\).
Answer: (1) \(3\sqrt{6}\)
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