Two lines L1 : x = 5, \frac{y}{3 - \alpha} = \frac{z}{- 2} and L2 : x = \alpha , \frac{y}{- 1} = \frac{z}{2 - \alpha} are coplanar. Then a can take value(s)

Engineering

Mathematics

Equation of Straight Line in 3D and Distance Of a Point From Line

Question

Two lines L₁ : x = 5, $\frac{y}{3 - α} = \frac{z}{- 2}$ and L₂ : $x = α, \frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar. Then a can take value(s)

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$L_{1} : \frac{x - 5}{0} = \frac{y - 0}{3 - α} = \frac{z - 0}{- 2}$

$L_{2} : \frac{x - α}{0} = \frac{y - 0}{- 1} = \frac{z - 0}{2 - α}$

As L₁ and L₂ are coplanar, so $| \begin{matrix} (5 - α) & 0 & 0 \\ 0 & (3 - α) & - 2 \\ 0 & - 1 & (2 - α) \end{matrix} | = 0$

⇒ – (α – 5) [(α – 3)(α – 2) – 2] = 0

⇒ – (α – 5) [(α² – 5α + 4) = 0

⇒ (α – 5) (α – 1) (α – 4) = 0

⇒ α = 1, 4, 5