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JEE Advanced 2013Paper 2
Exam
JEE Advanced
Year
2013
Paper
Paper 2
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Advanced 2013 · Paper 2 · Mathematics · Three Dimensional Geometry

Two lines L _ 1 : x = 5 , y 3 - α = z - 2 and L _ 2 : x = α , y - 1...

Two lines L1:x=5,y3−α=z−2{\text{L}}_{1} \text{:} \text{x} = 5 \text{,} \dfrac{\text{y}}{3 - \alpha} = \dfrac{\text{z}}{- 2} and L2:x=α,y−1=z2−α{\text{L}}_{2} \text{:} \text{x} = \alpha \text{,} \dfrac{\text{y}}{- 1} = \dfrac{\text{z}}{2 - \alpha} are coplanar. Then α\alpha can take value(s)

Options

  1. A1
  2. B2
  3. C3
  4. D4

Correct answer

A. 1

Step-by-step solution

\dfrac{\text{x} - 5}{0} = \dfrac{- \text{y}}{α - 3} = \dfrac{\text{z}}{- 2}
\dfrac{\text{x} - α}{0} = \dfrac{\text{y}}{- 1} = \dfrac{\text{z}}{2 - α}
\left|{5−alpha;0003−alpha;−20−12−alpha;\begin{cases}5 - &alpha; & 0 & 0 \\ 0 & 3 - &alpha; & - 2 \\ 0 & - 1 & 2 - &alpha;\end{cases}\right| = 0<br />\left(5 - &alpha;\right) \left(\left(3 - &alpha;\right) \left(2 - &alpha;\right) - 2\right) = 0
\left(α - 5\right) \left({α}^{2} - 5 α + 6 - 2\right) = 0<br />\left(&alpha; - 5\right) \left({&alpha;}^{2} - 5 &alpha; + 4\right) = 0
&alpha; = 1 \text{,} 4 \text{,} 5
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