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Question: The line \(\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}\) and \(\frac{x - 1}{k} = \frac{y -...

The line x−21=y−31=z−4−k\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k} and x−1k=y−42=z−51\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}are coplanar if

A

k = 0 or –1

B

k = 0 or 1

C

k = 0 or – 3

D

k = 3 or – 3

Answer

k = 0 or – 3

Explanation

Solution

For the lines to be coplanar, the distance between then is zero.

∣x2−x1y2−ylz2−z1a1b1c1a2b2c2∣\left| \begin{matrix} x_{2} - x_{1} & y_{2} - y_{l} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix} \right| = 0

∣1−24−35−411−kk21∣\left| \begin{matrix} 1 - 2 & 4 - 3 & 5 - 4 \\ 1 & 1 & - k \\ k & 2 & 1 \end{matrix} \right| = 0

Ž k = 0 or k = – 3