COMEDK · Maths · 34. Three Dimensional Geometry
If two lines \(L_1: \dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}\) and \(L_2: \dfrac{x-3}{1}=\dfrac{y-k}{2}=z\) intersect at a point, then \(2 k\) is equal to
- A 9
- B \(\dfrac{9}{2}\)
- C 1
- D \(\dfrac{1}{2}\)
Answer & Solution
Correct Answer
(A) 9
Step-by-step Solution
Detailed explanation
The lines are given by \(L_1: \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4} = \lambda\) and \(L_2: \dfrac{x-3}{1} = \dfrac{y-k}{2} = \dfrac{z}{1} = \mu\). Any point on \(L_1\) is \((2\lambda + 1, 3\lambda - 1, 4\lambda + 1)\) and any point on \(L_2\) is…
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