3D Geometry Ques 60
60. In $R^{3}$, consider the planes $P_{1}: y=0$ and $P_{2}: x+z=1$. Let $P_{3}$ be a plane, different from $P_{1}$ and $P_{2}$, which passes through the intersection of $P_{1}$ and $P_{2}$. If the distance of the point $(0,1,0)$ from $P_{3}$ is 1 and the distance of a point $(\alpha, \beta, \gamma)$ from $P_{3}$ is 2 , then which of the following relation(s) is/are true?
(2015 Adv.)
(a) $2 \alpha+\beta+2 \gamma+2=0$
(b) $2 \alpha-\beta+2 \gamma+4=0$
(c) $2 \alpha+\beta-2 \gamma-10=0$
(d) $2 \alpha-\beta+2 \gamma-8=0$
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Answer:
Correct Answer: 60.(b, d)
Solution:
Formula:
- Here, $P_{3}:(x+z-1)+\lambda y=0$
i.e. $\quad P_{3}: x+\lambda y+z-1=0$
whose distance from $(0,1,0)$ is 1 .
$ \begin{aligned} & \therefore \quad \frac{|0+\lambda+0-1|}{\sqrt{1+\lambda^{2}+1}}=1 \\ & \Rightarrow \quad|\lambda-1|=\sqrt{\lambda^{2}+2} \\ & \Rightarrow \quad \lambda^{2}-2 \lambda+1=\lambda^{2}+2 \Rightarrow \lambda=-\frac{1}{2} \end{aligned} $
$\therefore \quad$ Equation of $P_{3}$ is $2 x-y+2 z-2=0$.
$\because$ Distance from $(\alpha, \beta, \gamma)$ is 2 .
$ \begin{array}{lc} \therefore & \frac{|2 \alpha-\beta+2 \gamma-2|}{\sqrt{4+1+4}}=2 \\ \Rightarrow & 2 \alpha-\beta+2 \gamma-2= \pm 6 \\ \Rightarrow & 2 \alpha-\beta+2 \gamma=8 \text { and } 2 \alpha-\beta+2 \gamma=-4 \end{array} $
BETA
