If for some - the lines L1-x-12-y-2-1-z-11 and L2-x-2-y-15-z-11 coplanar- then the line L2 passes through the point-

A( 1, 2, 1), B( 2, 1, 1) [ AB b_1 b_2 ]=0 ⇒ 1 amp; 3 nbsp; amp; 2 nbsp; 2 amp; 1 amp;1 nbsp; α amp;5 α nbsp; amp;1 nbsp; =0 ⇒ ...

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Byju's Answer

Standard X

Mathematics

Solving Simultaneous Linear Equation Using Cramer's Rule

If for some α...

Question

If for some $α \in R,$ the lines $L_{1} : \frac{x + 1}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $L_{2} : \frac{x + 2}{α} = \frac{y + 1}{5 - α} = \frac{z + 1}{1}$ coplanar, then the line $L_{2}$ passes through the point:

(2, - 10, - 2)

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(10, - 2, - 2)

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(10, 2, 2)

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(- 2, 10, 2)

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Solution

The correct option is A $(2, - 10, - 2)$
$A (- 1, 2, 1), B (- 2, - 1, - 1)$
$[- - \to A B_{1}_{2}] = 0$
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} - 1 & - 3 & - 2 2 & - 1 & 1 α & 5 - α & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow - 1 (- 1 + α - 5) + 3 (2 - α) - 2 (10 - 2 α + α) = 0$
$\Rightarrow 6 - α + 6 - 3 α + 2 α - 20 = 0$
$\Rightarrow - 8 - 2 α = 0$
$\Rightarrow α = - 4$
Checking Point satisfies $L_{2}$
$L_{2} : \frac{x + 2}{- 4} = \frac{y + 1}{9} = \frac{z + 1}{1}$
Any point on $L_{2}$ is $(- 4 k - 2, 9 k - 1, k - 1)$

From given options, for point $(2, - 10, - 2) :$
$\frac{4}{- 4} = \frac{- 9}{9} = \frac{- 1}{1}$
Therefore, option $(a)$ is correct.

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Similar questions

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)

Q. Let

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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)

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If for some α∈R, the lines L1:x+12=y−2−1=z−11 and L2:x+2α=y+15−α=z+11 coplanar, then the line L2 passes through the point:

If for some $α \in R,$ the lines $L_{1} : \frac{x + 1}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $L_{2} : \frac{x + 2}{α} = \frac{y + 1}{5 - α} = \frac{z + 1}{1}$ coplanar, then the line $L_{2}$ passes through the point: