The lines x - 2 1 = y - 3 1 = z - 4 - k and x - 1 k = y - 4 2 = z - 5 1 are coplanar if:

Question

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 if:

Smart Achievers · Accepted Answer

(4)

Ans. $k = 0$ or $- 3$

Explanation:

We know, if two lines are coplanar then

$| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} |$ $= 0$

Therefore, $| \begin{matrix} 1 - 2 & 4 - 3 & 5 - 4 \\ 1 & 1 & - k \\ k & 2 & 1 \end{matrix} |$ $= 0$

$\Rightarrow$ $| \begin{matrix} - 1 & 1 & 1 \\ 1 & 1 & - k \\ k & 2 & 1 \end{matrix} |$ $= 0$

Applying $C_{1} \to C_{1} + C_{2}$ and $C_{2} \to C_{2} - C_{3}$

$\Rightarrow$ $| \begin{matrix} 0 & 0 & 1 \\ 2 & 1 + k & - k \\ k + 2 & 1 & 1 \end{matrix} |$ $= 0$

Expanding along $C_{1}$ ,

$1 {2 - (1 + k) (2 + k)} = 0$

$2 - (2 + 3 k + k^{2}) = 0$

$k^{2} + 3 k = 0$

$k (k + 3) = 0$

$∴$ $k = 0, - 3$

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