3D Geometry 3 Question 63
63. Find the equations of the plane containing the line $2 x-y+z-3=0,3 x+y+z=5$ and at a distance of $\frac{1}{\sqrt{6}}$ from the point $(2,1,-1)$.
$(2005,2 M)$
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Answer:
Correct Answer: 63. $\frac{9}{2}$ cu units
Solution:
- Equation of plane containing the lines
$ \begin{aligned} & 2 x-y+z-3=0 \text { and } 3 x+y+z=5 \text { is } \\ & (2 x-y+z-3)+\lambda(3 x+y+z-5)=0 \\ & \Rightarrow \quad(2+3 \lambda) x+(\lambda-1) y+(\lambda+1) z-3-5 \lambda=0 \end{aligned} $
Since, distance of plane from $(2,1,-1)$ to above plane is $1 / \sqrt{6}$.
$\therefore \frac{6 \lambda+4+\lambda-1-\lambda-1-3-5 \lambda}{\sqrt{(3 \lambda+2)^{2}+(\lambda-1)^{2}+(\lambda+1)^{2}}}=\frac{1}{\sqrt{6}}$
$ \begin{aligned} \Rightarrow & & 6(\lambda-1)^{2} & =11 \lambda^{2}+12 \lambda+6 \\ & \Rightarrow & \lambda & =0,-\frac{24}{5} \end{aligned} $
$\therefore$ Equations of planes are
$ 2 x-y+z-3=0 \text { and } 62 x+29 y+19 z-105=0 $
