First slide

Planes in 3D

Question

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

Moderate

A

1
B

2
C

3
D

4

Solution

$L_{1} : \frac{x - 5}{0} = \frac{y}{3 - α} = \frac{z}{- 2}$
$L_{2} : \frac{x - a}{0} = \frac{y}{- 1} = \frac{z}{2 - α}$
As $L_{1}, L_{2}$ are coplanar, therefore $|\begin{matrix} 5 - α & 0 & 0 \\ 0 & 3 - α & - 2 \\ 0 & - 1 & 2 - α \end{matrix}| = 0$
$\Rightarrow (5 - α) [6 - 5 α + α^{2} - 2] = 0$
$\Rightarrow (5 - α) (α - 1) (α - 4) = 0$
$\Rightarrow α = 1, 4, 5$

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