3D Geometry 3 Question 36
36. The equation of the plane containing the lines $2 x-5 y+z=3, x+y+4 z=5$ and parallel to the plane $x+3 y+6 z=1$ is
(2015 Main)
(a) $2 x+6 y+12 z=13$
(b) $x+3 y+6 z=-7$
(c) $x+3 y+6 z=7$
(d) $2 x+6 y+12 x=-13$
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Answer:
Correct Answer: 36. (c)
Solution:
- Let equation of plane containing the lines
$ \begin{aligned} 2 x-5 y+z=3 \text { and } x+y+4 z & =5 \text { be } \\ (2 x-5 y+z-3)+\lambda(x+y+4 z-5) & =0 \\ \Rightarrow(2+\lambda) x+(\lambda-5) y+(4 \lambda+1) z-3-5 \lambda & =0 \end{aligned} $
This plane is parallel to the plane $x+3 y+6 z=1$.
$\therefore \quad \frac{2+\lambda}{1}=\frac{\lambda-5}{3}=\frac{4 \lambda+1}{6}$
On taking first two equations, we get
$ 6+3 \lambda=\lambda-5 \Rightarrow 2 \lambda=-11 \Rightarrow \lambda=-\frac{11}{2} $
On taking last two equations, we get
$ 6 \lambda-30=3+12 \lambda \Rightarrow-6 \lambda=33 \Rightarrow \lambda=-\frac{11}{2} $
So, the equation of required plane is
$ \begin{aligned} & 2-\frac{11}{2} x+\frac{-11}{2}-5 y+-\frac{44}{2}+1 z-3+5 \times \frac{11}{2}=0 \\ & \Rightarrow \quad-\frac{7}{2} x-\frac{21}{2} y-\frac{42}{2} z+\frac{49}{2}=0 \\ & \Rightarrow \quad x+3 y+6 z-7=0 \end{aligned} $
