Question:
If the line, \(\dfrac{x-3}{1}=\dfrac{y+2}{-1}=\dfrac{z+\lambda}{-2}\) lies in the plane, \(2x-4y+3z=2\), then the shortest distance between this line and the line, \(\dfrac{x-1}{12}=\dfrac y9=\dfrac z4\) is :
If the line, \(\dfrac{x-3}{1}=\dfrac{y+2}{-1}=\dfrac{z+\lambda}{-2}\) lies in the plane, \(2x-4y+3z=2\), then the shortest distance between this line and the line, \(\dfrac{x-1}{12}=\dfrac y9=\dfrac z4\) is :
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Line 1 passes through \((3,-2,-\lambda)\) with direction \(d_1=(1,-1,-2)\). It lies in the plane, so the point satisfies it: \(6+8-3\lambda=2\Rightarrow\lambda=4\) (and \(d_1\cdot n=2+4-6=0\) is consistent).
Point 1 \(=(3,-2,-4)\). Line 2: point \((1,0,0)\), \(d_2=(12,9,4)\).
\(d_1\times d_2=(14,-28,21)\). Vector joining points \(=(2,-2,-4)\).
Scalar triple product \(=28+56-84=0\).
Hence the lines are coplanar (they intersect) and the shortest distance is 0.
Answer: (3) 0
Point 1 \(=(3,-2,-4)\). Line 2: point \((1,0,0)\), \(d_2=(12,9,4)\).
\(d_1\times d_2=(14,-28,21)\). Vector joining points \(=(2,-2,-4)\).
Scalar triple product \(=28+56-84=0\).
Hence the lines are coplanar (they intersect) and the shortest distance is 0.
Answer: (3) 0
Updated On: Nov 11, 2026
- \(2\)
- \(1\)
- \(0\)
- \(3\)
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The Correct Option is C
Solution and Explanation
Find \(\lambda\) from the condition that the line lies in the plane, then use the shortest-distance formula (scalar triple product).
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