Two lines L1: x =5 , (y/3 - α) = (z/-2) and L2 = x = α , (y/-1) = (z/2 - α) are coplanar. Then, α can take value (s)

Question

Two lines L1: x =5 , (y/3 - α) = (z/-2) and L2 = x = α , (y/-1) = (z/2 - α) are coplanar. Then, α can take value (s) (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

PLAN 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_{1}, c_{1})

and

(a_{2}, b_{2}, c_{2})

are coplanar,
i.e.

∣ ∣ x_{2} - x_{1} a_{1} a_{2} y_{2} - y_{1} b_{1} b_{2} z_{2} - z_{1} c_{1} c_{2} ∣ ∣ = 0

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} ....

(ii)

\Rightarrow ∣ ∣ 5 - α 00 0 3 - α - 1 0 - 2 2 - α ∣ ∣ = 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

Q. Two lines L1​:x=5,3−αy​=−2z​ and L2​=x=α,−1y​=2−αz​ are coplanar. Then, α can take value (s)