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# If the lines $x=1+s,\:y=-3-\lambda s,\:z=1+\lambda s$ and $x=t/2,\:y=1+t,\:z=2-t$ with parameters $s\:and\:t$ respectively, are coplanar, then $\lambda=?$

$\begin{array}{1 1}\large(a)\:\:0\:\:\:\qquad\:\:(b)\:\:-1\:\:\:\qquad\:\:(c)\:\:-\large\frac{1}{2}\:\:\:\qquad\:\:(d)\:\:-2. \end{array}$</p

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• The lines $\large\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}=\lambda_1$ and $\large\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}=\lambda_2$ are coplanar if $\left |\begin {array}{ccc} x_1-x_2 & y_1-y_2 & z_1-z_2 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end {array} \right |=0$
The given lines are written in standard form as
$\large\frac{x-1}{1}=\frac{y+3}{-\lambda}=\frac{z-1}{\lambda}=$$s\:\:and\:\:\large\frac{x}{1/2}=\frac{y-1}{1}=\frac{z-2}{-1}=$$t$
Condition for two lines to be coplanar is $\left |\begin {array}{ccc} 1 & -4 & -1 \\ 1 & -\lambda & \lambda \\ 1/2 & 1 & -1\end {array} \right |=0$
$\Rightarrow\:(\lambda-\lambda)+4(-1-\lambda/2)-(1+\lambda/2)=0$
$\Rightarrow\:\lambda=-2$
edited Jan 4, 2014