Question:
The shortest distance between the \(z\)-axis and the line \(x+y+2z-3=0=2x+3y+4z-4\), is :
The shortest distance between the \(z\)-axis and the line \(x+y+2z-3=0=2x+3y+4z-4\), is :
Show Hint
Find a point and direction of the line, then use the skew-lines distance formula; or notice the line lies in the plane y = -2.
Updated On: Nov 11, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Direction of the line \(=(1,1,2)\times(2,3,4)=(4-6,\,4-4,\,3-2)=(-2,0,1)\).
Step 2: Point on the line (put \(z=0\)): \(x+y=3,\ 2x+3y=4\Rightarrow y=-2,\ x=5\), i.e. \((5,-2,0)\).
Step 3: The z-axis has point \((0,0,0)\) and direction \((0,0,1)\). \((-2,0,1)\times(0,0,1)=(0,2,0)\), magnitude 2.
Step 4: Distance \(=\frac{|(5,-2,0)\cdot(0,2,0)|}{2}=\frac{4}{2}=2\).
Answer: (2) 2
Step 2: Point on the line (put \(z=0\)): \(x+y=3,\ 2x+3y=4\Rightarrow y=-2,\ x=5\), i.e. \((5,-2,0)\).
Step 3: The z-axis has point \((0,0,0)\) and direction \((0,0,1)\). \((-2,0,1)\times(0,0,1)=(0,2,0)\), magnitude 2.
Step 4: Distance \(=\frac{|(5,-2,0)\cdot(0,2,0)|}{2}=\frac{4}{2}=2\).
Answer: (2) 2
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