3D Geometry 3 Question 54
54. If the straight lines $\frac{x-1}{2}=\frac{y+1}{K}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is/are
(2012)
(a) $y+2 z=-1$
(b) $y+z=-1$
(c) $y-z=-1$
(d) $y-2 z=-1$
Numerical Value
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Answer:
Correct Answer: 54. (c, d)
Solution:
- PLAN If the straight lines are coplanar. They the should lie in same plane.
Description of Situation If straight lines are coplanar.
$ \Rightarrow \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 $
Since, $\quad \frac{x-1}{2}=\frac{y+1}{K}=\frac{z}{2}$
and $\quad \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar.
$ \begin{array}{lll} \Rightarrow & \left|\begin{array}{ccc} 2 & 0 & 0 \\ 2 & K & 2 \\ 5 & 2 & K \end{array}\right| & =0 \Rightarrow K^{2}=4 \Rightarrow K= \pm 2 \\ \therefore & & \mathbf{n}_ {1}=\mathbf{b}_ {1} \times \mathbf{d}_ {1}=6 \mathbf{j}-6 \mathbf{k}, \text { for } k=2 \\ \therefore & & \mathbf{n}_ {2}=\mathbf{b}_ {2} \times \mathbf{d}_ {2}=14 \mathbf{j}+14 \mathbf{k}, \text { for } k=-2 \end{array} $
So, equation of planes are $(\mathbf{r}-\mathbf{a}) \cdot \mathbf{n} _1=0$
$ \begin{array}{ll} \Rightarrow & y-z=-1 \text { and }(\mathbf{r}-\mathbf{a}) \cdot \mathbf{n} _2=0 \\ \Rightarrow & y+z=-1 \end{array} $
