3D Geometry 3 Question 14
####14. The equation of a plane containing the line of intersection of the planes $2 x-y-4=0$ and $y+2 z-4=0$ and passing through the point $(1,1,0)$ is
(2019 Main, 8 April I)
(a) $x-3 y-2 z=-2$
(b) $2 x-z=2$
(c) $x-y-z=0$
(d) $x+3 y+z=4$
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Answer:
(c)
Solution:
- Equations of given planes are
$ \text { and } \quad \begin{aligned} & 2 x-y-4=0 \\ & y+2 z-4=0 \end{aligned} $
Now, equation of family of planes passes through the line of intersection of given planes (i) and (ii) is
$ (2 x-y-4)+\lambda(y+2 z-4)=0 $
According to the question,
Plane (iii) passes through the point $(1,1,0)$, so
$ \begin{array}{lc} & (2-1-4)+\lambda(1+0-4)=0 \\ \Rightarrow & -3-3 \lambda=0 \\ \Rightarrow & \lambda=-1 \end{array} $
$ \Rightarrow $
Now, equation of required plane can be obtained by putting $\lambda=-1$ in the equation of plane (iii).
$ \begin{array}{lrl} \Rightarrow & (2 x-y-4)-1(y+2 z-4) & =0 \\ \Rightarrow & 2 x-y-4-y-2 z+4 & =0 \\ \Rightarrow & 2 x-2 y-2 z & =0 \\ \Rightarrow & x-y-z & =0 \end{array} $
