Two lines L1-x-5-y-3-z-2 and L2-x-y-1-z-2- are coplanar- Then- - can take values

If two straight lines are coplanar, i.e. x x_1/a_1=y y_1/b_1=z z_1/c_1 and x x_2/a_2=y y_2/b_2=z z_2/c_2 are coplanar Then, (x_2 x_1, y_2 y_1, z_2 z_1),(a_1,b_ ...

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Byju's Answer

Standard XI

Mathematics

Condition for Two Lines to be on the Same Plane

Two lines L1:...

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)

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Solution

The correct option is D
4

If two straight lines are coplanar,
i.e. $\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}}$ and $\frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}}$ are coplanar

Then, $(x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1}), (a_{1}, b_{2}, c_{1}) a n d (a_{2}, b_{2}, c_{2})$ are coplanar,
i.e. $∣ ∣ ∣ ∣ \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} a_{1} & b_{2} & c_{3} a_{2} & b_{2} & c_{2} \end{matrix} ∣ ∣ ∣ ∣$
Here,
$x = 5, \frac{y}{3 - α} = \frac{z}{- 2} \Rightarrow \frac{x - 5}{0} = \frac{y - 0}{- (α - 3)} = \frac{z - 0}{- 2} . . . (i) and x = α, \frac{y}{- 1} = \frac{z}{2 - α} \Rightarrow \frac{x - α}{0} = \frac{y - 0}{- 1} = \frac{z - 0}{2 - α} . . . (i i) \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 5 - α & 0 & 0 0 & 3 - α & - 2 0 & - 1 & 2 - α \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow (5 - α) [(3 - α) (2 - α) - 2] = 0 \Rightarrow (5 - α) [α^{2} - 5 α + 4] = 0 \Rightarrow (5 - α) (α - 1) (α - 4) = 0 ∴ α = 1, 4, 5$

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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)

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)