The lines: $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{\mathcal{z} - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{\mathcal{z} - 5}{1}$ are coplanar if:

Question

Dr. Abubacker · Accepted Answer

(b)
The lines $\ \frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ (i)
and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ (ii)
are coplanar if $\left| \begin{matrix}
x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \
l_{1} & m_{1} & n_{1} \
l_{2} & m_{1} & n_{2}
\end{matrix} ight| = 0$
or $\left| \begin{matrix}
1 & - 1 & - 1 \
1 & 1 & - k \
k & 2 & 1
\end{matrix} ight| = 0$
$\Rightarrow \ k^{2} + 3k = 0$
$\Rightarrow k = 0$ or $- 3$

Q. The lines: $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{\mathcal{z} - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{\mathcal{z} - 5}{1}$ are coplanar if:

Explanation

Reviewed by Dr. Abubacker

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