If the lines (x-2/1)=(y-3/1)=(z-4/-k) and (x-1/k)=(y-4/2)=(z-5/1) are coplanar, then k can have :

Question

If the lines (x-2/1)=(y-3/1)=(z-4/-k) and (x-1/k)=(y-4/2)=(z-5/1) are coplanar, then k can have : (A) any value (B) exactly one value (C) exactly two values (D) exactly three values

Tardigrade Admin · Accepted Answer

Since given lines are coplanar ∴ | 2-1 &3-4 4-5 [0.3em] 1 1 -k [0.3em] k &2 1 | = 0 i.e., | 1 -1 -1 [0.3em] 1 1 -k [0.3em] k &2 1 | = 0 ⇒ (1 + 2k) + (1 + k2 ) - ( 2 - k) = 0 ⇒ k2 + 3k = 0 ⇒ k = 0, -3

Q. If the lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar, then $k$ can have :

any value

exactly one value

exactly two values

exactly three values

Since given lines are coplanar
$∴ ∣ ∣ 2 - 1 1 k 3 - 4 12 4 - 5 - k 1 ∣ ∣ = 0 i . e ., ∣ ∣ 11 k - 1 12 - 1 - k 1 ∣ ∣ = 0$
$\Rightarrow$ $(1 + 2 k) + (1 + k^{2}) - (2 - k)$ = 0
$\Rightarrow$ $k^{2} + 3 k = 0 \Rightarrow k = 0, - 3$

Q. If the lines 1x−2​=1y−3​=−kz−4​ and kx−1​=2y−4​=1z−5​ are coplanar, then k can have :