Question

Two lines $L_1 : x =5 , \frac{y}{3 - \alpha} = \frac{z}{-2}$ and $L_2 = x = \alpha , \frac{y}{-1} = \frac{z}{2 - \alpha}$ are coplanar. Then, $\alpha$ can take value (s)

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A), (D)

PLAN If two straight lines are coplanar,
i.e. $ \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$
and $\frac{x-x_{2}}{ a _{2}}=\frac{y-y_{2}}{ b _{2}}=\frac{z-z_{2}}{ c _{2}}$ are coplanar

Then, $\left(x_{2}-x_{1}, y_{2}-y_{1}, z_{2}-z_{1}\right),\left(a_{1}, b_{1}, c_{1}\right)$ and $\left(a_{2}, b_{2}, c_{2}\right)$ are coplanar,
i.e. $\begin{vmatrix}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{vmatrix}=0$
Here, $x = 5, \frac{y}{3 - \alpha} = \frac{z}{-2}$
$\Rightarrow \frac{x-5}{0}=\frac{y-0}{-(\alpha-3)}=\frac{z-0}{-2} ....$(i)
and $ x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha} $
$\Rightarrow \frac{x-\alpha}{0}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha} ....$(ii)
$\Rightarrow \begin{vmatrix} 5-\alpha & 0 & 0 \\ 0 & 3-\alpha & -2 \\ 0 & -1 & 2-\alpha\end{vmatrix}=0$
$\Rightarrow (5-\alpha)[(3-\alpha)(2-\alpha)-2]=0$
$\Rightarrow (5-\alpha)\left[\alpha^{2}-5 \alpha+4\right]=0$
$\Rightarrow (5-\alpha)(\alpha-1)(\alpha-4)=0$
$\therefore \alpha=1,4,5$