3D Geometry Ques 12
12. 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)
(2013 Adv.)
(a) 1
(b) 2
(c) 3
(d) 4
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Answer:
Correct Answer: 12.(a, d)
Solution:
Formula:
- Key Idea If two straight lines are coplanar,
$ \begin{aligned} & \text { i.e. } \quad \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}} \\ & \text { and } \quad \frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}} \text { are coplanar } \end{aligned} $
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. $\quad\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|=0$
Here, $\quad x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$
$\Rightarrow \quad \frac{x-5}{0}=\frac{y-0}{-(\alpha-3)}=\frac{z-0}{-2}$
and
$ x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha} $
$\Rightarrow \quad \frac{x-\alpha}{0}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha}$
$ \begin{aligned} & \Rightarrow \quad\left|\begin{array}{ccc}12-\alpha & 0 & 0 \\ 0 & 3-\alpha & -2 \\ 0 & -1 & 2-\alpha \end{array}\right|=0 \\ & \Rightarrow \quad(5-\alpha)[(3-\alpha)(2-\alpha)-2]=0 \\ & \Rightarrow \quad(5-\alpha)\left[\alpha^{2}-5 \alpha+4\right]=0 \\ & \Rightarrow \quad(5-\alpha)(\alpha-1)(\alpha-4)=0 \\ & \therefore \quad \alpha=1,4,5 \end{aligned} $
BETA
