Question

Consider the following plane equations,
$P_1:-2x-y+z=5P_2:3x-2y+4z=6P_3:4x+2y-2z=6.$
Then which of the following statement(s) is/are correct ?

• A. $P_1$ and $P_2$ are parallel.
• B. $P_1$ and $P_2$ are perpendicular.
• C. $P_1$ and $P_3$ are parallel.
• D. $P_2$ and $P_3$ are perpendicular

Answer 1

Answer: (B), (C), (D)

$P_2$ and $P_3$ are perpendicular

Given planes :
$P_1:-2x-y+z=5P_2:3x-2y+4z=6P_3:4x+2y-2z=6.$
$D.r^{\prime }s$ of normal to plane $P_1=(a_1,b_1,c_1)=(-2,-1,1)$
$D.r^{\prime }s$ of normal to plane $P_2=(a_2,b_2,c_2)=(3,-2,4)$
$D.r^{\prime }s$ of normal to plane $P_3=(a_3,b_3,c_3)=(4,2,-2)$
As, $\frac{a_1}{a_3}=\frac{b_1}{b_3}=\frac{c_1}{c_3}$
$\Rightarrow$ Planes $P_1$ and $P_3$ are parallel.
and $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
So, planes $P_1$ and $P_2$ are perpendicular.
Similarly,
$a_2{a}_3+b_2{b}_3+c_2{c}_3=0$
So, planes $P_2$ and $P_3$ are perpendicular.