If lines L1-x2-y-2-1-z-11 and L2-x-1-y-22-z-1 are coplanar- then the line L2 passes through points

Given lines : L_1:x2=y 2 1=z 11 and L_2:x 1α=y+22+α=z 1 D.R's of L_1=(a_1,b_1,c_1)=(2, 1,1) D.R's of L_2=(a_2,b_2,c_2)=(α,2+α, 1) For coplanar lines : x_2 x_1 ...

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Byju's Answer

Standard X

Mathematics

Solving Simultaneous Linear Equation Using Cramer's Rule

If lines L1:x...

Question

If lines $L_{1} : \frac{x}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $L_{2} : \frac{x - 1}{α} = \frac{y + 2}{2 + α} = \frac{z}{- 1}$ are coplanar, then the line $L_{2}$ passes through point(s)

(- 12, - 11, 13)

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(- 12, 1, - 8)

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(14, 4, - 13)

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(14, - 5, 8)

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Solution

The correct option is D $(14, - 5, 8)$
Given lines :
$L_{1} : \frac{x}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $L_{2} : \frac{x - 1}{α} = \frac{y + 2}{2 + α} = \frac{z}{- 1}$
$D . R^{'} s$ of $L_{1} = (a_{1}, b_{1}, c_{1}) = (2, - 1, 1)$
$D . R^{'} s$ of $L_{2} = (a_{2}, b_{2}, c_{2}) = (α, 2 + α, - 1)$
For coplanar lines :
$∣ ∣ ∣ ∣ \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 4 & - 1 2 & - 1 & 1 α & 2 + α & - 1 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow (1 - 2 - α) + 4 (- 2 - α) - (4 + 2 α + α) = 0 \Rightarrow α = - \frac{13}{8}$
so, any point on $L_{2} = (1 - \frac{13}{8} λ, - 2 + \frac{3}{8} λ, - λ)$
For $λ = 8, - 8$ points are $(- 12, 1, - 8)$ and $(14, - 5, 8)$

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