COMEDK · Maths · 34. Three Dimensional Geometry
Two lines \(\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}\) and \(\dfrac{x-3}{1}=\dfrac{y-k}{2}=\dfrac{z}{1}\) intersect at a point. Then the value of ' \(k\) ' is
- A \(-\dfrac{13}{2}\)
- B \(\dfrac{13}{2}\)
- C \(\dfrac{7}{2}\)
- D \(\dfrac{9}{2}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{9}{2}\)
Step-by-step Solution
Detailed explanation
Let the first line be \(L_1: \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4} = \lambda\). Any point on \(L_1\) is \(P(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)\). Let the second line be \(L_2: \dfrac{x-3}{1} = \dfrac{y-k}{2} = \dfrac{z}{1} = \mu\). Any point on \(L_2\) is…
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