Question

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 2, then which of the following relation(s) is/are true?

• 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. $2\alpha -\beta +2\gamma -8=0$

Answer 1

Answer: (B), (D)

The correct options are
B

$2\alpha -\beta +2\gamma +4=0$

D

$2\alpha -\beta +2\gamma -8=0$

Here, $P_3:(x+z-1)+\lambda y=0i.e.P_3:x+\lambda y+z-1=0......\left(i\right)$
whose distance from (0, 1, 0) is 1.
$\therefore \frac{|0+\lambda +0-1|}{\sqrt{1+{\lambda }^2+1}}=1\Rightarrow |\lambda -1|=\sqrt{{\lambda }^2+2}\Rightarrow {\lambda }^2-2\lambda +1={\lambda }^2+2\Rightarrow \lambda =-\frac{1}{2}\therefore Equationof{P}_{3}is2x-y+2z-2=0.\because Distancefrom(\alpha ,\beta ,\gamma )is2.\therefore \frac{|2\alpha -\beta +2\gamma -2|}{\sqrt{4+1+4}}=2\Rightarrow 2\alpha -\beta +2\gamma -2=\pm 6\Rightarrow 2\alpha -\beta +2\gamma =8and2\alpha -\beta +2\gamma =-4$