Question:
The shortest distance between the lines
\(\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}\)
and \(x+y+z+1=0,\ 2x-y+z+3=0\) is :
The shortest distance between the lines
\(\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}\)
and \(x+y+z+1=0,\ 2x-y+z+3=0\) is :
\(\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}\)
and \(x+y+z+1=0,\ 2x-y+z+3=0\) is :
Show Hint
Find the direction of the second line as the cross product of the two plane normals, then use the skew-lines distance formula.
Updated On: Nov 5, 2026
- 1
- \(\frac{1}{2}\)
- \(\frac{1}{\sqrt{2}}\)
- \(\frac{1}{\sqrt{3}}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Line 1: point \(P_1(1,-1,0)\), direction \(\vec d_1=(0,-1,1)\).
Line 2: normals \((1,1,1)\) and \((2,-1,1)\), so direction \(\vec d_2=(1,1,1)\times(2,-1,1)=(1\cdot1-1\cdot(-1),\ 1\cdot2-1\cdot1,\ 1\cdot(-1)-1\cdot2)=(2,1,-3)\). A point on it (put \(z=0\)): \(x+y=-1,\ 2x-y=-3\Rightarrow P_2\left(-\frac{4}{3},\frac{1}{3},0\right)\).
\(\vec d_1\times\vec d_2=(0,-1,1)\times(2,1,-3)=(3-1,\ 2-0,\ 0+2)=(2,2,2)\), magnitude \(2\sqrt{3}\).
\(\overrightarrow{P_1P_2}=\left(-\frac{7}{3},\frac{4}{3},0\right)\); dot with \((2,2,2)\): \(\frac{-14+8}{3}=-2\).
Distance \(=\frac{|-2|}{2\sqrt{3}}=\frac{1}{\sqrt{3}}\).
Answer: (4) \(\frac{1}{\sqrt{3}}\)
Line 2: normals \((1,1,1)\) and \((2,-1,1)\), so direction \(\vec d_2=(1,1,1)\times(2,-1,1)=(1\cdot1-1\cdot(-1),\ 1\cdot2-1\cdot1,\ 1\cdot(-1)-1\cdot2)=(2,1,-3)\). A point on it (put \(z=0\)): \(x+y=-1,\ 2x-y=-3\Rightarrow P_2\left(-\frac{4}{3},\frac{1}{3},0\right)\).
\(\vec d_1\times\vec d_2=(0,-1,1)\times(2,1,-3)=(3-1,\ 2-0,\ 0+2)=(2,2,2)\), magnitude \(2\sqrt{3}\).
\(\overrightarrow{P_1P_2}=\left(-\frac{7}{3},\frac{4}{3},0\right)\); dot with \((2,2,2)\): \(\frac{-14+8}{3}=-2\).
Distance \(=\frac{|-2|}{2\sqrt{3}}=\frac{1}{\sqrt{3}}\).
Answer: (4) \(\frac{1}{\sqrt{3}}\)
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