3D Geometry Ques 11
- 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.)
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Answer:
Correct Answer: 11.(a, d)
Solution:
Formula:
- Key Idea If two straight lines are coplanar, they must intersect at exactly one point.
$$ \begin{alignedat} & \text { i.e. } \quad \frac{x-x _1}{a _1}=\frac{y-y _1}{b _1}=\frac{z-z _1}{c _1} \\ & \text { and } \quad \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2} \text { are coplanar } \end{aligned} $$
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, $\quad x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$
$\Rightarrow \quad \frac{x-5}{1}=\frac{y-0}{-(\alpha-3)}=\frac{z-0}{-2}$
and
$$ x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha} $$
$\Rightarrow \quad \frac{x-\alpha}{0}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha}$
$$ \begin{alignedat} & \Rightarrow \quad\left|\begin{array}{ccc}11-\alpha & 0 & 0 \\ 0 & 3-\alpha & -2 \ 0 & -1 & 2-\alpha \end{array}\right|=0 \\ & \Rightarrow \quad(5-\alpha)[(3-\alpha)(2-\alpha)-2]=0 \\ & \Rightarrow \quad(5-\alpha)\left[\alpha^{2}-5 \alpha+4\right]=0 \\ & \Rightarrow \quad(5-\alpha)(\alpha-1)(\alpha-4)=0 \\ & \therefore \quad \alpha=1,4,5 \end{aligned} $$
BETA
