The lines $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-4}{-\mathrm{k}}$ and…

The lines $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-4}{-\mathrm{k}}$ and $\frac{\mathrm{x}-1}{\mathrm{k}}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{1}$ are coplanar if
  1. $\mathrm{k}=3$ or $-2$
  2. $\mathrm{k}=0$ or $-1$
  3. $\mathrm{k}=1$ or $-1$
  4. $\mathrm{k}=0$ or $-3$

Solution

$\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{l}_1 & \mathrm{~m}_1 & \mathrm{n}_1 \\ \mathrm{l}_2 & \mathrm{~m}_2 & \mathrm{n}_2\end{array}\right|=0$ $\left|\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -\mathrm{k} \\ \mathrm{k} & 2 & 1\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}0 & 0 & -1 \\ 2 & 1+\mathrm{k} & -\mathrm{k} \\ \mathrm{k}+2 & 1 & 1\end{array}\right|=0$ $\mathrm{k}^2+3 \mathrm{k}^2=0 \Rightarrow \mathrm{k}(\mathrm{k}+3)=0$ or $\mathrm{k}=0$ or $-3$

Asked in: JEE Main 2003

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